Chapter 5 Complex Numbers And Quadratic Equations

Given:

The expression to evaluate is $(1 \;+\; i)^6 + (1 \;–\; i)^3$.

To Find:

The value of the given expression.

Solution:

Let's evaluate each part of the expression separately.

Consider the first part: $(1 + i)^6$.

We know that $(1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i$.

So, $(1 + i)^6 = ((1 + i)^2)^3 = (2i)^3$.

$(2i)^3 = 2^3 \times i^3 = 8 \times (-i) = -8i$.

Thus, $(1 + i)^6 = -8i$.

Now consider the second part: $(1 – i)^3$.

We can use the binomial expansion $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a=1$ and $b=i$.

$(1 – i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i^2) - i^3$.

$(1 – i)^3 = 1 - 3i + 3(-1) - (-i)$.

$(1 – i)^3 = 1 - 3i - 3 + i$.

Combine the real and imaginary parts:

$(1 – i)^3 = (1 - 3) + (-3i + i)$.

$(1 – i)^3 = -2 - 2i$.

Now, add the results of the two parts:

$(1 \;+\; i)^6 + (1 \;–\; i)^3 = (-8i) + (-2 - 2i)$.

$(1 \;+\; i)^6 + (1 \;–\; i)^3 = -8i - 2 - 2i$.

Combine the imaginary parts:

$(1 \;+\; i)^6 + (1 \;–\; i)^3 = -2 + (-8i - 2i)$.

$(1 \;+\; i)^6 + (1 \;–\; i)^3 = -2 - 10i$.

The value of the expression $(1 \;+\; i)^6 + (1 \;–\; i)^3$ is $-2 - 10i$.

Given:

$(x\;+\;iy)^{\frac{1}{3}} = a + ib$, where $x, y, a, b \in \mathbb{R}$.

To Show:

$\frac{x}{a} - \frac{y}{b} = -2 (a^2 + b^2)$.

Solution:

We are given the equation:

$(x + iy)^{\frac{1}{3}} = a + ib$

Cube both sides of the equation:

$((x + iy)^{\frac{1}{3}})^3 = (a + ib)^3$

$x + iy = (a + ib)^3$

Expand the right side using the binomial formula $(A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$:

$(a + ib)^3 = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3$

Recall that $i^2 = -1$ and $i^3 = i^2 \cdot i = -1 \cdot i = -i$. Substitute these values:

$(a + ib)^3 = a^3 + 3ia^2b + 3a(-b^2) + (-i)b^3$

$(a + ib)^3 = a^3 + 3ia^2b - 3ab^2 - ib^3$

Group the real and imaginary terms on the right side:

$(a + ib)^3 = (a^3 - 3ab^2) + i(3a^2b - b^3)$

Now equate the real and imaginary parts of the equation $x + iy = (a^3 - 3ab^2) + i(3a^2b - b^3)$:

Equating the real parts:

$x = a^3 - 3ab^2$

Equating the imaginary parts:

$y = 3a^2b - b^3$

Now, consider the expression $\frac{x}{a} - \frac{y}{b}$. Substitute the expressions for $x$ and $y$ (assuming $a \neq 0$ and $b \neq 0$):

$\frac{x}{a} = \frac{a^3 - 3ab^2}{a}$

Factor out $a$ from the numerator:

$\frac{x}{a} = \frac{a(a^2 - 3b^2)}{a}$

Cancel the common factor $a$ (since $a \neq 0$):

$\frac{x}{a} = a^2 - 3b^2$

Similarly for $\frac{y}{b}$:

$\frac{y}{b} = \frac{3a^2b - b^3}{b}$

Factor out $b$ from the numerator:

$\frac{y}{b} = \frac{b(3a^2 - b^2)}{b}$

Cancel the common factor $b$ (since $b \neq 0$):

$\frac{y}{b} = 3a^2 - b^2$

Now subtract $\frac{y}{b}$ from $\frac{x}{a}$:

$\frac{x}{a} - \frac{y}{b} = (a^2 - 3b^2) - (3a^2 - b^2)$

Remove the parentheses:

$\frac{x}{a} - \frac{y}{b} = a^2 - 3b^2 - 3a^2 + b^2$

Combine like terms ($a^2$ terms and $b^2$ terms):

$\frac{x}{a} - \frac{y}{b} = (a^2 - 3a^2) + (-3b^2 + b^2)$

$\frac{x}{a} - \frac{y}{b} = -2a^2 - 2b^2$

Factor out $-2$ from the right side:

$\frac{x}{a} - \frac{y}{b} = -2(a^2 + b^2)$

This is the required result.

Given:

The equation $z^2 = \overline{z}$, where $z = x + iy$, with $x, y \in \mathbb{R}$.

To Solve:

Find all complex numbers $z$ that satisfy the given equation.

Solution:

We are given the equation $z^2 = \overline{z}$.

Let $z = x + iy$, where $x$ and $y$ are real numbers.

First, calculate $z^2$:

$z^2 = (x + iy)^2$

$z^2 = x^2 + 2(x)(iy) + (iy)^2$

$z^2 = x^2 + 2ixy + i^2y^2$

Since $i^2 = -1$, we have:

$z^2 = x^2 + 2ixy - y^2$

Group the real and imaginary parts of $z^2$:

$z^2 = (x^2 - y^2) + i(2xy)$

Next, find the conjugate of $z$, $\overline{z}$:

$\overline{z} = \overline{x + iy} = x - iy$

Now, substitute these expressions for $z^2$ and $\overline{z}$ into the given equation $z^2 = \overline{z}$:

$(x^2 - y^2) + i(2xy) = x - iy$

For two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal.

Equating the real parts:

Equating the imaginary parts:

Now, we need to solve this system of two equations for $x$ and $y$.

Consider equation (2):

$2xy = -y$

Move $-y$ to the left side:

$2xy + y = 0$

Factor out $y$:

$y(2x + 1) = 0$

This equation implies that either $y = 0$ or $2x + 1 = 0$.

Case 1: $y = 0$

Substitute $y = 0$ into equation (1):

$x^2 - (0)^2 = x$

$x^2 = x$

$x^2 - x = 0$

Factor out $x$:

$x(x - 1) = 0$

This implies that either $x = 0$ or $x - 1 = 0$, which gives $x = 1$.

So, when $y = 0$, we have two possible values for $x$: $x=0$ and $x=1$.

If $x=0$ and $y=0$, then $z = 0 + i(0) = 0$.

If $x=1$ and $y=0$, then $z = 1 + i(0) = 1$.

Case 2: $2x + 1 = 0$

This implies $2x = -1$, so $x = -\frac{1}{2}$.

Substitute $x = -\frac{1}{2}$ into equation (1):

$(-\frac{1}{2})^2 - y^2 = -\frac{1}{2}$

$\frac{1}{4} - y^2 = -\frac{1}{2}$

Solve for $y^2$:

$-y^2 = -\frac{1}{2} - \frac{1}{4}$

$-y^2 = -\frac{2}{4} - \frac{1}{4}$

$-y^2 = -\frac{3}{4}$

$y^2 = \frac{3}{4}$

Taking the square root of both sides:

$y = \pm \sqrt{\frac{3}{4}}$

$y = \pm \frac{\sqrt{3}}{\sqrt{4}}$

$y = \pm \frac{\sqrt{3}}{2}$

So, when $x = -\frac{1}{2}$, we have two possible values for $y$: $y=\frac{\sqrt{3}}{2}$ and $y=-\frac{\sqrt{3}}{2}$.

If $x = -\frac{1}{2}$ and $y = \frac{\sqrt{3}}{2}$, then $z = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. This is a complex cube root of unity, often denoted as $\omega$ or $e^{i2\pi/3}$.

If $x = -\frac{1}{2}$ and $y = -\frac{\sqrt{3}}{2}$, then $z = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. This is another complex cube root of unity, often denoted as $\omega^2$ or $e^{i4\pi/3}$.

The solutions for $z$ are the values obtained from these two cases.

From Case 1, $z = 0$ and $z = 1$.

From Case 2, $z = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $z = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$.

These four values are the solutions to the equation $z^2 = \overline{z}$.

Given:

The imaginary part of the complex expression $\frac{2z \;+\; 1}{iz \;+\; 1}$ is $-2$, where $z$ is a complex number.

To Show:

The locus of the point representing $z$ in the argand plane is a straight line.

Solution:

Let the complex number $z$ be represented as $z = x + iy$, where $x$ and $y$ are real numbers representing the coordinates of the point in the argand plane.

Substitute $z = x + iy$ into the given expression:

$\frac{2z + 1}{iz + 1} = \frac{2(x + iy) + 1}{i(x + iy) + 1}$

Simplify the numerator and the denominator:

Numerator: $2x + 2iy + 1 = (2x + 1) + i(2y)$

Denominator: $ix + i^2y + 1 = ix - y + 1 = (1 - y) + i(x)$

So the expression becomes:

$\frac{(2x + 1) + i(2y)}{(1 - y) + i(x)}$

To find the real and imaginary parts of this complex fraction, we multiply the numerator and the denominator by the conjugate of the denominator.

The conjugate of the denominator $(1 - y) + i(x)$ is $(1 - y) - i(x)$.

$\frac{(2x + 1) + i(2y)}{(1 - y) + i(x)} \times \frac{(1 - y) - i(x)}{(1 - y) - i(x)}$

Calculate the denominator:

$((1 - y) + i(x))((1 - y) - i(x)) = (1 - y)^2 - (i(x))^2 = (1 - y)^2 - i^2x^2$

Since $i^2 = -1$, the denominator is $(1 - y)^2 - (-1)x^2 = (1 - y)^2 + x^2 = 1 - 2y + y^2 + x^2 = x^2 + y^2 - 2y + 1$.

Note that the denominator $x^2 + y^2 - 2y + 1 = x^2 + (y-1)^2$ is zero only if $x=0$ and $y=1$, which corresponds to $z=i$. If $z=i$, the original expression $iz+1 = i(i)+1 = i^2+1 = -1+1=0$, so the expression is undefined. Thus, the point $z=i$ (or $(0, 1)$ in the argand plane) is not part of the locus.

Calculate the numerator:

$((2x + 1) + i(2y))((1 - y) - i(x))$

$= (2x + 1)(1 - y) + i(2y)(1 - y) - (2x + 1)i(x) - i(2y)i(x)$

$= (2x - 2xy + 1 - y) + i(2y - 2y^2) - i(2x^2 + x) - i^2(2xy)$

$= (2x - 2xy + 1 - y) + i(2y - 2y^2) - i(2x^2 + x) + 2xy$

Group the real and imaginary parts of the numerator:

Real part of numerator: $(2x - 2xy + 1 - y) + 2xy = 2x - y + 1$

Imaginary part of numerator: $(2y - 2y^2) - (2x^2 + x) = 2y - 2y^2 - 2x^2 - x$

The expression is therefore:

$\frac{(2x - y + 1) + i(2y - 2y^2 - 2x^2 - x)}{x^2 + y^2 - 2y + 1}$

The imaginary part of this expression is $\frac{2y - 2y^2 - 2x^2 - x}{x^2 + y^2 - 2y + 1}$.

We are given that the imaginary part is $-2$.

$\frac{2y - 2y^2 - 2x^2 - x}{x^2 + y^2 - 2y + 1} = -2$

Multiply both sides by the denominator (which is non-zero for points on the locus):

$2y - 2y^2 - 2x^2 - x = -2(x^2 + y^2 - 2y + 1)$

$2y - 2y^2 - 2x^2 - x = -2x^2 - 2y^2 + 4y - 2$

Move all terms to the left side of the equation:

$2y - 2y^2 - 2x^2 - x + 2x^2 + 2y^2 - 4y + 2 = 0$

Combine like terms:

$(-2x^2 + 2x^2) + (-2y^2 + 2y^2) + (2y - 4y) - x + 2 = 0$

$0 + 0 - 2y - x + 2 = 0$

$-x - 2y + 2 = 0$

Multiply by $-1$ to make the coefficient of $x$ positive:

$x + 2y - 2 = 0$

This equation is in the form $Ax + By + C = 0$, which is the standard equation of a straight line in the Cartesian coordinate system.

The locus of the point representing $z = x + iy$ is the set of all points $(x, y)$ satisfying $x + 2y - 2 = 0$, excluding the point $(0, 1)$. This equation describes a straight line in the argand plane.

Given:

The equation $|z^2 - 1| = |z|^2 + 1$.

To Show:

$z$ lies on the imaginary axis (i.e., the real part of $z$ is zero).

Solution:

Let the complex number $z$ be represented as $z = x + iy$, where $x$ and $y$ are real numbers.

We need to express both sides of the given equation in terms of $x$ and $y$.

First, consider $|z|^2$:

$|z|^2 = |x + iy|^2 = x^2 + y^2$

So the right side of the given equation is $|z|^2 + 1 = x^2 + y^2 + 1$.

Next, consider the left side, $|z^2 - 1|$.

First, calculate $z^2$:

$z^2 = (x + iy)^2 = x^2 + 2xyi + (iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + i(2xy)$

Now calculate $z^2 - 1$:

$z^2 - 1 = (x^2 - y^2) + i(2xy) - 1 = (x^2 - y^2 - 1) + i(2xy)$

The modulus of $z^2 - 1$ is:

$|z^2 - 1| = |(x^2 - y^2 - 1) + i(2xy)| = \sqrt{(x^2 - y^2 - 1)^2 + (2xy)^2}$

Now substitute these expressions back into the given equation $|z^2 - 1| = |z|^2 + 1$:

$\sqrt{(x^2 - y^2 - 1)^2 + (2xy)^2} = x^2 + y^2 + 1$

Square both sides of the equation to eliminate the square root:

$(x^2 - y^2 - 1)^2 + (2xy)^2 = (x^2 + y^2 + 1)^2$

Expand the terms.

Left side: $(x^2 - (y^2 + 1))^2 + 4x^2y^2$

$= (x^2)^2 - 2x^2(y^2 + 1) + (y^2 + 1)^2 + 4x^2y^2$

$= x^4 - 2x^2y^2 - 2x^2 + (y^4 + 2y^2 + 1) + 4x^2y^2$

$= x^4 - 2x^2y^2 - 2x^2 + y^4 + 2y^2 + 1 + 4x^2y^2$

Combine $x^2y^2$ terms:

$= x^4 + y^4 + 2x^2y^2 - 2x^2 + 2y^2 + 1$

This can be rearranged as $(x^2 + y^2)^2 - 2x^2 + 2y^2 + 1$.

Right side: $((x^2 + y^2) + 1)^2$

$= (x^2 + y^2)^2 + 2(x^2 + y^2)(1) + 1^2$

$= (x^2 + y^2)^2 + 2x^2 + 2y^2 + 1$

Equating the expanded left and right sides:

$(x^2 + y^2)^2 - 2x^2 + 2y^2 + 1 = (x^2 + y^2)^2 + 2x^2 + 2y^2 + 1$

Subtract $(x^2 + y^2)^2 + 2y^2 + 1$ from both sides of the equation:

$-2x^2 = 2x^2$

Add $2x^2$ to both sides:

$0 = 2x^2 + 2x^2$

$0 = 4x^2$

Divide by 4:

$x^2 = 0$

Taking the square root of both sides:

$x = 0$

Since $z = x + iy$ and we found that $x = 0$, the real part of $z$ is zero.

A complex number with a real part equal to zero lies on the imaginary axis in the argand plane.

Therefore, the locus of the point representing $z$ is the imaginary axis.

Given:

1. $\overline{z_{1}} + i\overline{z_{2}} = 0$

2. $\text{arg} (z_{1} z_{2}) = \pi$

where $z_1$ and $z_2$ are complex numbers.

To Find:

$\text{arg} (z_{1})$

Solution:

From the first given condition, we have:

$\overline{z_{1}} + i\overline{z_{2}} = 0$

$\overline{z_{1}} = -i\overline{z_{2}}$

Taking the complex conjugate of both sides:

$\overline{(\overline{z_{1}})} = \overline{(-i\overline{z_{2}})}$

$z_{1} = \overline{(-i)} \overline{(\overline{z_{2}})}$

Since $\overline{-i} = i$ and $\overline{(\overline{z_{2}})} = z_{2}$, we get:

$z_{1} = i z_{2}$

Let $\theta_1$ be an argument of $z_1$, i.e., $\theta_1 = \text{arg}(z_1)$.

Let $\theta_2$ be an argument of $z_2$, i.e., $\theta_2 = \text{arg}(z_2)$.

From $z_{1} = i z_{2}$, we can relate their arguments. The argument of the product of two complex numbers is the sum of their arguments (modulo $2\pi$).

$\text{arg}(z_{1}) = \text{arg}(i z_{2})$

$\text{arg}(z_{1}) = \text{arg}(i) + \text{arg}(z_{2}) + 2k\pi$, where $k$ is an integer.

An argument of $i$ is $\frac{\pi}{2}$. So,

This congruence can be written as $\theta_1 \equiv \theta_2 + \frac{\pi}{2} \pmod{2\pi}$.

The second given condition is $\text{arg}(z_{1} z_{2}) = \pi$. This means $\pi$ is an argument of $z_1 z_2$.

Using the property of arguments of a product:

$\text{arg}(z_{1} z_{2}) = \text{arg}(z_{1}) + \text{arg}(z_{2}) + 2m\pi$, where $m$ is an integer.

So, we have:

This congruence can be written as $\theta_1 + \theta_2 \equiv \pi \pmod{2\pi}$.

Now we solve the system of congruences:

(1) $\theta_1 \equiv \theta_2 + \frac{\pi}{2} \pmod{2\pi}$

(2) $\theta_1 + \theta_2 \equiv \pi \pmod{2\pi}$

From (1), we can write $\theta_2 \equiv \theta_1 - \frac{\pi}{2} \pmod{2\pi}$.

Substitute this into (2):

$\theta_1 + (\theta_1 - \frac{\pi}{2}) \equiv \pi \pmod{2\pi}$

$2\theta_1 - \frac{\pi}{2} \equiv \pi \pmod{2\pi}$

$2\theta_1 \equiv \pi + \frac{\pi}{2} \pmod{2\pi}$

$2\theta_1 \equiv \frac{3\pi}{2} \pmod{2\pi}$

This means $2\theta_1 = \frac{3\pi}{2} + 2n\pi$ for some integer $n$.

Divide by 2:

$\theta_1 = \frac{3\pi}{4} + n\pi$

The set of all possible arguments for $z_1$ is given by $\{\frac{3\pi}{4} + n\pi \mid n \in \mathbb{Z}\}$.

If "arg(z1)" refers to the principal argument, which is typically in the interval $(-\pi, \pi]$, we find the values of $n$ for which $\theta_1$ falls in this interval.

For $n = 0$: $\theta_1 = \frac{3\pi}{4}$. This is in $(-\pi, \pi]$.

For $n = 1$: $\theta_1 = \frac{3\pi}{4} + \pi = \frac{7\pi}{4}$. This is not in $(-\pi, \pi]$.

For $n = -1$: $\theta_1 = \frac{3\pi}{4} - \pi = -\frac{\pi}{4}$. This is in $(-\pi, \pi]$.

For other integer values of $n$, $\theta_1$ will fall outside the interval $(-\pi, \pi]$.

Thus, if "arg(z1)" refers to the principal argument in $(-\pi, \pi]$, there are two possible values: $\frac{3\pi}{4}$ and $-\frac{\pi}{4}$.

Both values satisfy the original conditions when properties of principal arguments are carefully applied.

For example, if $\text{Arg}(z_1) = 3\pi/4$, then from (1), $\text{Arg}(z_2) = 3\pi/4 - \pi/2 = \pi/4$. $\text{Arg}(z_1)+\text{Arg}(z_2) = 3\pi/4 + \pi/4 = \pi$. Since $\pi \in (-\pi, \pi]$, $\text{Arg}(z_1 z_2) = \pi$, which matches the given condition.

If $\text{Arg}(z_1) = -\pi/4$, then from (1), $\text{Arg}(z_2) = -\pi/4 - \pi/2 = -3\pi/4$. $\text{Arg}(z_1)+\text{Arg}(z_2) = -\pi/4 + (-3\pi/4) = -\pi$. Since $-\pi \notin (-\pi, \pi]$, $\text{Arg}(z_1 z_2) = -\pi + 2\pi = \pi$, which matches the given condition.

Since the question asks for "arg(z1)" (singular) in a short answer context, and both principal values are valid, one of these values is expected.

Answer: $\text{arg} (z_{1}) = \frac{3\pi}{4}$ (or $-\frac{\pi}{4}$).

Given:

Let $z_1$ and $z_2$ be two complex numbers such that $|z_1 + z_2| = |z_1| + |z_2|$.

To Describe:

The relationship between the complex numbers $z_1$ and $z_2$ that satisfies the given equation.

Solution:

We are given the equation:

$|z_1 + z_2| = |z_1| + |z_2|$

This equation is the condition for equality in the triangle inequality for complex numbers.

Square both sides of the equation:

$|z_1 + z_2|^2 = (|z_1| + |z_2|)^2$

Using the property that for any complex number $w$, $|w|^2 = w\overline{w}$ where $\overline{w}$ is the complex conjugate of $w$, and also $|w_1 w_2| = |w_1||w_2|$:

$(z_1 + z_2)(\overline{z_1 + z_2}) = |z_1|^2 + 2|z_1||z_2| + |z_2|^2$

Using the property $\overline{w_1 + w_2} = \overline{w_1} + \overline{w_2}$:

$(z_1 + z_2)(\overline{z_1} + \overline{z_2}) = |z_1|^2 + 2|z_1||z_2| + |z_2|^2$

Expand the left side by distributing:

$z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1} + z_2\overline{z_2} = |z_1|^2 + 2|z_1||z_2| + |z_2|^2$

Using the property $z\overline{z} = |z|^2$ on the first and fourth terms of the left side:

$|z_1|^2 + z_1\overline{z_2} + z_2\overline{z_1} + |z_2|^2 = |z_1|^2 + 2|z_1||z_2| + |z_2|^2$

Subtract $|z_1|^2 + |z_2|^2$ from both sides of the equation:

Let $w = z_1\overline{z_2}$. Then the term $z_2\overline{z_1}$ is the complex conjugate of $w$, i.e., $z_2\overline{z_1} = \overline{z_1\overline{z_2}} = \overline{w}$.

Substitute $w$ and $\overline{w}$ into equation (1):

$$w + \overline{w} = 2|z_1||z_2|$$

Recall that for any complex number $w$, the sum of $w$ and its conjugate $\overline{w}$ is equal to twice its real part: $w + \overline{w} = 2 \text{Re}(w)$.

Applying this property:

$$2 \text{Re}(z_1\overline{z_2}) = 2|z_1||z_2|$$

Divide both sides by 2:

We also know that for any complex number $u$, $\text{Re}(u) \le |u|$. In this case, $u = z_1\overline{z_2}$.

$$|z_1\overline{z_2}| = |z_1||\overline{z_2}|$$

Since the modulus of a conjugate is equal to the modulus of the original number, $|\overline{z_2}| = |z_2|$.

So, $|z_1\overline{z_2}| = |z_1||z_2|$.

Equation (2) is $\text{Re}(z_1\overline{z_2}) = |z_1\overline{z_2}|$.

This condition $\text{Re}(u) = |u|$ holds for a complex number $u$ if and only if $u$ is a non-negative real number. That is, $u = k$ where $k \in \mathbb{R}$ and $k \ge 0$.

Thus, $z_1\overline{z_2}$ must be a non-negative real number.

Let $z_1\overline{z_2} = k$, where $k \ge 0$ is a real number.

If $z_2 \neq 0$, we can multiply by $z_2$ and divide by $|z_2|^2$ (since $|z_2|^2 = z_2\overline{z_2}$ and $|z_2|^2 > 0$).

$$z_1\overline{z_2} = k$$

$$z_1\overline{z_2} z_2 = k z_2$$

$$z_1 (z_2\overline{z_2}) = k z_2$$

$$z_1 |z_2|^2 = k z_2$$

$$z_1 = \frac{k}{|z_2|^2} z_2$$

Let $c = \frac{k}{|z_2|^2}$. Since $k \ge 0$ and $|z_2|^2 > 0$, $c$ is a non-negative real number ($c \ge 0$).

So, if $z_2 \neq 0$, the condition implies $z_1 = c z_2$ where $c \ge 0$ is real.

If $z_2 = 0$, the original equation $|z_1 + 0| = |z_1| + |0|$ becomes $|z_1| = |z_1|$, which is always true. In this case, $z_2 = 0 \cdot z_1$, which is of the form $z_2 = c z_1$ with $c=0$, a non-negative real number.

Therefore, the condition $|z_1 + z_2| = |z_1| + |z_2|$ holds if and only if one of the complex numbers is a non-negative real multiple of the other. This means $z_1 = c z_2$ for some real $c \ge 0$, or $z_2 = c z_1$ for some real $c \ge 0$.

Geometrically, this means that the points representing $z_1$, $z_2$, and the origin are collinear, and $z_1$ and $z_2$ lie on the same ray emanating from the origin (or one/both is the origin).

Given:

$z_1, z_2, z_3$ are complex numbers such that:

1. $|z_1| = |z_2| = |z_3| = 1$

2. $\left| \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} \right| = 1$

To Find:

The value of $|z_1 + z_2 + z_3|$.

Solution:

From the first given condition, $|z_k| = 1$ for $k = 1, 2, 3$.

We know that for any complex number $z$, $|z|^2 = z\overline{z}$.

So, $|z_k|^2 = z_k \overline{z_k} = 1^2 = 1$.

From $z_k \overline{z_k} = 1$, we can write $\overline{z_k} = \frac{1}{z_k}$ for $k = 1, 2, 3$.

Now consider the second given condition:

$\left| \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} \right| = 1$

Substitute $\frac{1}{z_k} = \overline{z_k}$ into this equation:

$|\overline{z_{1}} + \overline{z_{2}} + \overline{z_{3}}| = 1$

Using the property that the modulus of a sum of complex numbers is equal to the modulus of the sum of their conjugates, i.e., $|w_1 + w_2 + w_3| = |\overline{w_1 + w_2 + w_3}| = |\overline{w_1} + \overline{w_2} + \overline{w_3}|$.

In our case, $w_k = z_k$, so $\overline{w_k} = \overline{z_k}$.

$|\overline{z_{1}} + \overline{z_{2}} + \overline{z_{3}}| = |\overline{z_{1} + z_{2} + z_{3}}|$

Also, the modulus of a conjugate is equal to the modulus of the original number, i.e., $|\overline{w}| = |w|$.

So, $|\overline{z_{1} + z_{2} + z_{3}}| = |z_{1} + z_{2} + z_{3}|$.

Combining these results, we have:

$\left| \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} \right| = |\overline{z_{1}} + \overline{z_{2}} + \overline{z_{3}}| = |z_{1} + z_{2} + z_{3}|$

Since we are given that $\left| \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} \right| = 1$, it follows that:

$|z_{1} + z_{2} + z_{3}| = 1$

The value of $|z_1 + z_2 + z_3|$ is 1.

Given:

A complex number $z$ lies in the interior or on the boundary of a circle with radius 3 and center at $(-4, 0)$ in the argand plane.

To Find:

The greatest and least values of $|z + 1|$.

Solution:

Let $z = x + iy$, where $x$ and $y$ are real numbers.

The center of the circle is the point $(-4, 0)$, which corresponds to the complex number $-4 + 0i = -4$.

The condition that $z$ lies in the interior or on the boundary of the circle with center $-4$ and radius 3 can be expressed using the modulus of complex numbers. The distance between $z$ and the center $-4$ is $|z - (-4)| = |z + 4|$. This distance must be less than or equal to the radius 3.

We want to find the greatest and least values of $|z + 1|$.

Let $w = z + 1$. We want to find the range of $|w|$.

We can rewrite $w$ in terms of $z + 4$ by adding and subtracting 3:

$w = z + 1 = (z + 4) - 3$

Now, we can use the triangle inequality for complex numbers, which states that for any complex numbers $u$ and $v$, $||u| - |v|| \le |u + v| \le |u| + |v|$.

Let $u = z + 4$ and $v = -3$. Then $u + v = (z + 4) + (-3) = z + 1$.

Applying the triangle inequality to $|(z + 4) + (-3)|$:

$||z + 4| - |-3|| \le |(z + 4) - 3| \le |z + 4| + |-3|$

We know that $|-3| = 3$. So,

$||z + 4| - 3| \le |z + 1| \le |z + 4| + 3$

From the given condition (1), $|z + 4| \le 3$.

Using the upper bound of the inequality:

$|z + 1| \le |z + 4| + 3$

Since $|z + 4| \le 3$, the maximum value of $|z + 4| + 3$ occurs when $|z + 4|$ is maximum, which is 3.

So, $|z + 1| \le 3 + 3 = 6$.

The greatest value of $|z + 1|$ is 6. This occurs when $z + 4$ is in the same direction as $3$ (i.e., $z+4 = 3k$ for $k \ge 0$) and $|z+4|=3$. Specifically, when $z+4$ is a positive multiple of 1, i.e., when $z$ is on the boundary of the circle such that $z$, $-4$, and $-1$ are collinear with $-1$ between $-4$ and $z$, and $z+4$ points towards $-3$ (or opposite to the vector from $-4$ to $-1$). This happens at the point on the circle boundary furthest from $-1$. The point $-1$ corresponds to $( -1, 0)$. The center is $(-4, 0)$. The point on the circle furthest from $(-1, 0)$ along the real axis is $(-4-3, 0) = (-7, 0)$, which is $z = -7$. Let's check: $|-7 + 4| = |-3| = 3 \le 3$. $|-7 + 1| = |-6| = 6$. This confirms the maximum value.

Using the lower bound of the inequality:

$||z + 4| - 3| \le |z + 1|$

Since $0 \le |z + 4| \le 3$, the value $|z + 4| - 3$ is between $0 - 3 = -3$ and $3 - 3 = 0$. So, $|z + 4| - 3 \le 0$.

The absolute value $||z + 4| - 3|$ is equal to $-(|z + 4| - 3) = 3 - |z + 4|$.

So, $3 - |z + 4| \le |z + 1|$.

The minimum value of $3 - |z + 4|$ occurs when $|z + 4|$ is maximum, which is 3.

Minimum value of $3 - |z + 4|$ is $3 - 3 = 0$. In this case, the inequality gives $0 \le |z + 1|$, which is always true for a modulus.

To find the minimum value of $|z + 1|$, consider the geometric interpretation. $|z + 1|$ is the distance between the point $z$ and the point representing $-1$ in the argand plane.

The point $-1$ is located at $(-1, 0)$. The circle is centered at $(-4, 0)$ with radius 3.

The distance between the center of the circle $(-4, 0)$ and the point $(-1, 0)$ is $|-4 - (-1)| = |-4 + 1| = |-3| = 3$.

Since the distance between the center and the point $-1$ is equal to the radius of the circle, the point $-1$ lies on the boundary of the circle $|z + 4| = 3$.

The points $z$ are inside or on the boundary of the circle $|z + 4| \le 3$. The minimum distance from a point $z$ in this region to the point $-1$ occurs at the point $-1$ itself, since $-1$ is on the boundary of the region.

When $z = -1$, $|z + 1| = |-1 + 1| = |0| = 0$.

So the least value of $|z + 1|$ is 0.

This occurs when $z = -1$, which satisfies $|-1 + 4| = |3| = 3 \le 3$.

The greatest value of $|z + 1|$ is 6.

The least value of $|z + 1|$ is 0.

Given:

The condition $3 < |z| < 4$ for a complex number $z$.

To Locate:

The points in the argand plane that satisfy the given condition.

Solution:

Let $z = x + iy$, where $x$ and $y$ are real numbers representing the coordinates of the point in the argand plane.

The modulus of $z$ is given by $|z| = |x + iy| = \sqrt{x^2 + y^2}$.

The given condition is $3 < |z| < 4$.

Substituting the expression for $|z|$, we get:

$3 < \sqrt{x^2 + y^2} < 4$

This inequality can be broken into two parts:

Part 1: $3 < \sqrt{x^2 + y^2}$

Square both sides:

$3^2 < (\sqrt{x^2 + y^2})^2$

$9 < x^2 + y^2$

This inequality $x^2 + y^2 > 9$ represents the region outside the circle centered at the origin $(0, 0)$ with radius $\sqrt{9} = 3$. The points on the circle $x^2 + y^2 = 9$ are not included because the inequality is strict ($>$).

Part 2: $\sqrt{x^2 + y^2} < 4$

Square both sides:

$(\sqrt{x^2 + y^2})^2 < 4^2$

$x^2 + y^2 < 16$

This inequality $x^2 + y^2 < 16$ represents the region inside the circle centered at the origin $(0, 0)$ with radius $\sqrt{16} = 4$. The points on the circle $x^2 + y^2 = 16$ are not included because the inequality is strict ($<$).

The condition $3 < |z| < 4$ means that both inequalities must be satisfied simultaneously.

The points must be outside the circle of radius 3 centered at the origin AND inside the circle of radius 4 centered at the origin.

This region is an annulus (a ring shape) centered at the origin.

The locus of points satisfying $3 < |z| < 4$ is the set of all points in the argand plane whose distance from the origin is greater than 3 and less than 4.

This is the open annular region between the circle $|z|=3$ and the circle $|z|=4$. Neither boundary circle is included in the locus.

In summary, the points for which $3 < |z| < 4$ are located in the region between the circle with center $(0,0)$ and radius 3, and the circle with center $(0,0)$ and radius 4. The circles themselves are not part of the region.

Given:

The polynomial $P(x) = 2x^4 + 5x^3 + 7x^2 – x + 41$.

The value of $x = – 2 \;–\; \sqrt{3}i$.

To Find:

The value of the polynomial $P(x)$ when $x = – 2 \;–\; \sqrt{3}i$.

Solution:

We are given $x = – 2 – \sqrt{3}i$.

Rearrange the equation to isolate the term with $i$:

$x + 2 = – \sqrt{3}i$

Square both sides of the equation:

$(x + 2)^2 = (– \sqrt{3}i)^2$

$x^2 + 4x + 4 = (– \sqrt{3})^2 \cdot i^2$

$x^2 + 4x + 4 = 3 \cdot (–1)$

$x^2 + 4x + 4 = –3$

Move all terms to one side to form a quadratic equation:

$x^2 + 4x + 4 + 3 = 0$

Since $x = – 2 – \sqrt{3}i$ is a root of the quadratic equation $x^2 + 4x + 7 = 0$, the value of the polynomial $P(x) = 2x^4 + 5x^3 + 7x^2 – x + 41$ for this value of $x$ is equal to the remainder obtained when $P(x)$ is divided by $x^2 + 4x + 7$.

Let's perform polynomial long division:

Divide $2x^4 + 5x^3 + 7x^2 – x + 41$ by $x^2 + 4x + 7$.

From the long division, we can write the polynomial as:

$2x^4 + 5x^3 + 7x^2 – x + 41 = (x^2 + 4x + 7)(2x^2 - 3x + 5) + 6$

Now substitute the value of $x = – 2 – \sqrt{3}i$. From equation (1), we know that for this value of $x$, $x^2 + 4x + 7 = 0$.

So, $P(– 2 – \sqrt{3}i) = (0)(2(– 2 – \sqrt{3}i)^2 - 3(– 2 – \sqrt{3}i) + 5) + 6$

$P(– 2 – \sqrt{3}i) = 0 + 6$

$P(– 2 – \sqrt{3}i) = 6$

The value of the polynomial when $x = – 2 \;–\; \sqrt{3}i$ is 6.

Given:

The quadratic equation $x^2 – Px + 8 = 0$.

The difference of the roots of the equation is 2.

To Find:

The value of $P$.

Solution:

Let the roots of the quadratic equation $x^2 – Px + 8 = 0$ be $\alpha$ and $\beta$.

For a quadratic equation of the form $ax^2 + bx + c = 0$, the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product of the roots is $\alpha \beta = \frac{c}{a}$.

In the given equation $x^2 – Px + 8 = 0$, we have $a=1$, $b=-P$, and $c=8$.

Sum of the roots:

Product of the roots:

We are given that the difference of the roots is 2. The difference can be $|\alpha - \beta| = 2$. Without loss of generality, let's assume $\alpha - \beta = 2$ or $\beta - \alpha = 2$. Squaring the difference will cover both cases.

$(\alpha - \beta)^2 = 2^2$

$(\alpha - \beta)^2 = 4$

We know the algebraic identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$.

Substitute this into the equation:

$(\alpha + \beta)^2 - 4\alpha \beta = 4$

Now, substitute the expressions for $(\alpha + \beta)$ and $\alpha \beta$ from equations (1) and (2):

$P^2 - 4(8) = 4$

$P^2 - 32 = 4$

Add 32 to both sides:

$P^2 = 4 + 32$

$P^2 = 36$

Take the square root of both sides:

$P = \pm \sqrt{36}$

$P = \pm 6$

Thus, the possible values for $P$ are 6 and -6.

If $P=6$, the equation is $x^2 - 6x + 8 = 0$. Factoring, $(x-2)(x-4) = 0$, roots are 2 and 4. Difference is $|4-2|=2$.

If $P=-6$, the equation is $x^2 + 6x + 8 = 0$. Factoring, $(x+2)(x+4) = 0$, roots are -2 and -4. Difference is $|-2 - (-4)| = |-2+4| = |2|=2$.

Both values of $P$ are valid.

Given:

The quadratic equation is $x^2 – (a – 2) x – (a + 1) = 0$.

To Find:

The value of the real number $a$ such that the sum of the squares of the roots of the equation is least.

Solution:

Let the roots of the quadratic equation $x^2 – (a – 2) x – (a + 1) = 0$ be $\alpha$ and $\beta$.

For a quadratic equation of the form $Ax^2 + Bx + C = 0$, the sum of the roots is $\alpha + \beta = -\frac{B}{A}$ and the product of the roots is $\alpha \beta = \frac{C}{A}$.

In the given equation, we have $A=1$, $B=-(a-2)$, and $C=-(a+1)$.

The sum of the roots is:

$\alpha + \beta = -\frac{-(a - 2)}{1} = a - 2$

The product of the roots is:

$\alpha \beta = \frac{-(a + 1)}{1} = -(a + 1)$

We need to find the value of $a$ for which the sum of the squares of the roots, $\alpha^2 + \beta^2$, is least.

The sum of the squares of the roots can be expressed in terms of the sum and product of the roots using the identity:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$

Substitute the expressions for $(\alpha + \beta)$ and $\alpha \beta$ in terms of $a$:

$\alpha^2 + \beta^2 = (a - 2)^2 - 2(-(a + 1))$

Expand the expression:

$\alpha^2 + \beta^2 = (a^2 - 4a + 4) + 2(a + 1)$

$\alpha^2 + \beta^2 = a^2 - 4a + 4 + 2a + 2$

Combine like terms:

$\alpha^2 + \beta^2 = a^2 - 2a + 6$

Let $S(a) = \alpha^2 + \beta^2 = a^2 - 2a + 6$. This is a quadratic function of $a$. Since the coefficient of $a^2$ is $1$ (which is positive), the graph of this function is a parabola opening upwards, meaning it has a minimum value.

To find the value of $a$ for which $S(a)$ is least, we can complete the square for the quadratic expression $a^2 - 2a + 6$:

$S(a) = (a^2 - 2a) + 6$

To complete the square for $a^2 - 2a$, we add and subtract $(\frac{-2}{2})^2 = (-1)^2 = 1$:

$S(a) = (a^2 - 2a + 1) - 1 + 6$

Recognize the perfect square trinomial $(a^2 - 2a + 1) = (a - 1)^2$:

$S(a) = (a - 1)^2 + 5$

Since $(a - 1)^2$ is the square of a real number, its minimum possible value is 0, which occurs when $a - 1 = 0$, i.e., $a = 1$.

When $(a - 1)^2 = 0$, the value of $S(a)$ is $0 + 5 = 5$. This is the least value of the sum of the squares of the roots.

The minimum occurs when $a = 1$.

The value of $a$ such that the sum of the squares of the roots is least is 1.

Solution:

The given equation is:

$\left| 1-\overline{z_{1}}z_{2} \right|^{2}$ - |z₂ - z₁|² = k(1 - |z₁|²) (1 - |z₂|^2})$

We use the property that for any complex number $z$, $|z|^2 = z\overline{z}$. Also, we use the properties of conjugation: $\overline{z_1 \pm z_2} = \overline{z_1} \pm \overline{z_2}$, $\overline{z_1 z_2} = \overline{z_1} \overline{z_2}$, and $\overline{\overline{z}} = z$.

Let's expand the Left Hand Side (LHS):

$LHS = \left| 1-\overline{z_{1}}z_{2} \right|^{2} - |z_{2} - z_{1}|^{2}$

Consider the first term:

$\left| 1-\overline{z_{1}}z_{2} \right|^{2} = (1-\overline{z_{1}}z_{2}) \overline{(1-\overline{z_{1}}z_{2})}$

$= (1-\overline{z_{1}}z_{2})(1 - \overline{(\overline{z_{1}}z_{2})})$

$= (1-\overline{z_{1}}z_{2})(1 - \overline{\overline{z_{1}}}\overline{z_{2}})$

$= (1-\overline{z_{1}}z_{2})(1 - z_{1}\overline{z_{2}})$

Expanding the product:

$= 1 \cdot (1 - z_{1}\overline{z_{2}}) - \overline{z_{1}}z_{2} \cdot (1 - z_{1}\overline{z_{2}})$

$= 1 - z_{1}\overline{z_{2}} - \overline{z_{1}}z_{2} + (\overline{z_{1}}z_{2}) (z_{1}\overline{z_{2}})$

$= 1 - z_{1}\overline{z_{2}} - \overline{z_{1}}z_{2} + (\overline{z_{1}}z_{1}) (z_{2}\overline{z_{2}})$

Using $|z|^2 = z\overline{z}$:

$= 1 - z_{1}\overline{z_{2}} - \overline{z_{1}}z_{2} + |z_{1}|^2 |z_{2}|^2$

Consider the second term:

$|z_{2} - z_{1}|^{2} = (z_{2} - z_{1}) \overline{(z_{2} - z_{1})}$

$= (z_{2} - z_{1})(\overline{z_{2}} - \overline{z_{1}})$

Expanding the product:

$= z_{2}(\overline{z_{2}} - \overline{z_{1}}) - z_{1}(\overline{z_{2}} - \overline{z_{1}})$

$= z_{2}\overline{z_{2}} - z_{2}\overline{z_{1}} - z_{1}\overline{z_{2}} + z_{1}\overline{z_{1}}$

Using $|z|^2 = z\overline{z}$:

$= |z_{2}|^2 - z_{2}\overline{z_{1}} - z_{1}\overline{z_{2}} + |z_{1}|^2$

$= |z_{1}|^2 + |z_{2}|^2 - (z_{2}\overline{z_{1}} + z_{1}\overline{z_{2}})$

Now, subtract the second term from the first term to find the LHS:

$LHS = (1 - z_{1}\overline{z_{2}} - \overline{z_{1}}z_{2} + |z_{1}|^2 |z_{2}|^2) - (|z_{1}|^2 + |z_{2}|^2 - z_{2}\overline{z_{1}} - z_{1}\overline{z_{2}})$

$LHS = 1 - z_{1}\overline{z_{2}} - \overline{z_{1}}z_{2} + |z_{1}|^2 |z_{2}|^2 - |z_{1}|^2 - |z_{2}|^2 + z_{2}\overline{z_{1}} + z_{1}\overline{z_{2}}$

Observe that the terms $- z_{1}\overline{z_{2}}$ and $+ z_{1}\overline{z_{2}}$ cancel out.

Also, the terms $- \overline{z_{1}}z_{2}$ and $+ z_{2}\overline{z_{1}}$ cancel out (since $z_{2}\overline{z_{1}} = \overline{z_{1}}\overline{\overline{z_{2}}} = \overline{\overline{z_{1}}z_{2}}$, which is the conjugate of $\overline{z_1}z_2$).

So, the LHS simplifies to:

$LHS = 1 + |z_{1}|^2 |z_{2}|^2 - |z_{1}|^2 - |z_{2}|^2$

Now let's expand the Right Hand Side (RHS):

$RHS = k(1 - |z_{1}|^2) (1 - |z_{2}|^2)$

Expand the product of the two factors:

$(1 - |z_{1}|^2) (1 - |z_{2}|^2) = 1 \cdot (1 - |z_{2}|^2) - |z_{1}|^2 \cdot (1 - |z_{2}|^2)$

$= 1 - |z_{2}|^2 - |z_{1}|^2 + |z_{1}|^2 |z_{2}|^2$

So, the RHS is:

$RHS = k(1 - |z_{1}|^2 - |z_{2}|^2 + |z_{1}|^2 |z_{2}|^2)$

Equating the LHS and RHS based on the given equation:

$1 - |z_{1}|^2 - |z_{2}|^2 + |z_{1}|^2 |z_{2}|^2 = k(1 - |z_{1}|^2 - |z_{2}|^2 + |z_{1}|^2 |z_{2}|^2)$

This equation is given to hold for all complex numbers $z_{1}$ and $z_{2}$.

Let $A = 1 - |z_{1}|^2 - |z_{2}|^2 + |z_{1}|^2 |z_{2}|^2$. The equation can be written as $A = kA$.

Note that $A$ can be factored as $A = (1 - |z_{1}|^2)(1 - |z_{2}|^2)$.

The equation is $(1 - |z_{1}|^2)(1 - |z_{2}|^2) = k(1 - |z_{1}|^2)(1 - |z_{2}|^2)$.

This equality must hold for any choice of $z_{1}, z_{2}$. If we can find just one pair $(z_{1}, z_{2})$ for which the term $(1 - |z_{1}|^2)(1 - |z_{2}|^2)$ is non-zero, we can determine $k$.

For example, choose $z_{1} = 0$ and $z_{2} = 0$. Then $|z_{1}| = 0$ and $|z_{2}| = 0$.

In this case, $(1 - |z_{1}|^2)(1 - |z_{2}|^2) = (1 - 0)(1 - 0) = 1 \cdot 1 = 1$, which is non-zero.

Substituting these values into the equation $A = kA$ (or the expanded form):

$1 - 0^2 - 0^2 + 0^2 \cdot 0^2 = k(1 - 0^2 - 0^2 + 0^2 \cdot 0^2)$

$1 = k(1)$

$k = 1$

Since the value of $k=1$ satisfies the equation for all $z_{1}, z_{2}$ (as shown by the expansion of LHS and RHS yielding the same expression), the value of $k$ is uniquely determined as 1.

Final Answer:

The value of $k$ is 1.

Solution:

We are given that both complex numbers $z_1$ and $z_2$ satisfy the equation $z + \overline{z} = 2|z - 1|$.

Let $z = x + iy$, where $x, y$ are real numbers.

Then $\overline{z} = x - iy$.

Substitute $z$ and $\overline{z}$ into the given equation:

$(x + iy) + (x - iy) = 2|(x + iy) - 1|$

$2x = 2|(x - 1) + iy|$

$x = \sqrt{(x - 1)^2 + y^2}$

For this equation to hold, the Left Hand Side must be non-negative, so $x \ge 0$.

Squaring both sides (valid for $x \ge 0$):

$x^2 = (x - 1)^2 + y^2$

$x^2 = x^2 - 2x + 1 + y^2$

$0 = -2x + 1 + y^2$

$y^2 = 2x - 1$

For $y^2$ to be a real number, we must have $2x - 1 \ge 0$, which means $x \ge \frac{1}{2}$. This condition $x \ge \frac{1}{2}$ is consistent with $x \ge 0$.

Thus, a complex number $z = x + iy$ satisfies the given equation if and only if $y^2 = 2x - 1$ and $x \ge \frac{1}{2}$.

Since $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$ both satisfy this condition, we have:

$y_1^2 = 2x_1 - 1$ and $x_1 \ge \frac{1}{2}$

$y_2^2 = 2x_2 - 1$ and $x_2 \ge \frac{1}{2}$

We are also given $\arg (z_{1} - z_{2}) = \frac{\pi}{4}$.

Let $z_{1} - z_{2} = (x_1 + iy_1) - (x_2 + iy_2) = (x_1 - x_2) + i(y_1 - y_2)$.

The argument of $(x_1 - x_2) + i(y_1 - y_2)$ being $\frac{\pi}{4}$ means that the complex number $z_1 - z_2$ lies in the first quadrant on the line $y=x$.

Therefore, the real part and the imaginary part must be equal and positive:

$x_1 - x_2 > 0$

$y_1 - y_2 > 0$

And their ratio is 1:

$\frac{y_1 - y_2}{x_1 - x_2} = \tan\left(\frac{\pi}{4}\right) = 1$

This implies:

$y_1 - y_2 = x_1 - x_2$

Now, subtract the equation for $z_2$ from the equation for $z_1$:

$y_1^2 - y_2^2 = (2x_1 - 1) - (2x_2 - 1)$

$(y_1 - y_2)(y_1 + y_2) = 2x_1 - 2x_2$

$(y_1 - y_2)(y_1 + y_2) = 2(x_1 - x_2)$

Substitute $x_1 - x_2 = y_1 - y_2$ into the above equation:

$(y_1 - y_2)(y_1 + y_2) = 2(y_1 - y_2)$

Since $y_1 - y_2 > 0$, we know that $y_1 - y_2$ is not zero. We can divide both sides by $(y_1 - y_2)$:

$y_1 + y_2 = 2$

We need to find $\text{Im}(z_{1} + z_{2})$.

$z_{1} + z_{2} = (x_1 + iy_1) + (x_2 + iy_2) = (x_1 + x_2) + i(y_1 + y_2)$

The imaginary part of $z_{1} + z_{2}$ is $\text{Im}(z_{1} + z_{2}) = y_1 + y_2$.

Using the result $y_1 + y_2 = 2$, we find:

$\text{Im}(z_{1} + z_{2}) = 2$

Final Answer:

The value of Im $(z_1 + z_2 )$ is 2.

Solution for (i):

Let the given complex number be $Z = 3i^3 - 2ai^2 + (1 - a)i + 5$.

We know that $i^2 = -1$ and $i^3 = i^2 \cdot i = -1 \cdot i = -i$.

Substitute these values into the expression for $Z$:

$Z = 3(-i) - 2a(-1) + (1 - a)i + 5$

$Z = -3i + 2a + (1 - a)i + 5$

Group the real and imaginary parts:

$Z = (2a + 5) + (-3 + 1 - a)i$

$Z = (2a + 5) + (-2 - a)i$

For $Z$ to be a real number, its imaginary part must be zero.

Im$(Z) = -2 - a$

Set the imaginary part to zero:

$-2 - a = 0$

$-a = 2$

$a = -2$

The real value of 'a' for which the expression is real is -2.

Solution for (ii):

We are given $|z| = 2$ and $\arg(z) = \frac{\pi}{4}$.

A complex number $z$ can be written in polar form as $z = |z| (\cos(\arg(z)) + i \sin(\arg(z)))$.

Substitute the given values:

$z = 2 \left(\cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right)\right)$

We know that $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

$z = 2 \left(\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right)$

$z = 2 \cdot \frac{1}{\sqrt{2}} + 2 \cdot i \frac{1}{\sqrt{2}}$

$z = \sqrt{2} + i \sqrt{2}$

If $|z| = 2$ and arg (z) = $\frac{\pi}{4}$, then z = $\sqrt{2} + i\sqrt{2}$.

Solution for (iii):

The argument of a complex number $z = x + iy$ (where $z \ne 0$) is the angle $\theta$ that the line segment from the origin to the point $(x, y)$ makes with the positive real axis.

Given $\arg(z) = \frac{\pi}{3}$. This means that the angle is $60^\circ$ with the positive real axis.

For a complex number $z = x+iy$ to have an argument of $\frac{\pi}{3}$, it must satisfy $y = x \tan\left(\frac{\pi}{3}\right)$ and $x > 0$ (since the angle is in the first quadrant). Note that $z \ne 0$, so $(0,0)$ is excluded.

$y = x \sqrt{3}$ with $x > 0$.

This equation describes a straight line passing through the origin with a positive slope $\sqrt{3}$. The condition $x > 0$ means we only consider the part of the line in the first quadrant.

Thus, the locus is a ray (half-line) starting from the origin (excluding the origin itself) and making an angle of $\frac{\pi}{3}$ with the positive x-axis.

The locus of z satisfying arg (z) = $\frac{\pi}{3}$ is the ray starting from the origin (excluding the origin) making an angle of $\frac{\pi}{3}$ with the positive real axis.

Solution for (iv):

We need to find the value of $\left( -\sqrt{-1} \right)^{4n-3}$, where $n \in N$.

We know that $\sqrt{-1} = i$.

So, the expression is $(-i)^{4n-3}$.

Using the property $(ab)^m = a^m b^m$ and $a^{m-n} = a^m / a^n$:

$(-i)^{4n-3} = (-1 \cdot i)^{4n-3} = (-1)^{4n-3} \cdot i^{4n-3}$

Since $n \in N$, $4n$ is an even number. Therefore, $4n-3$ is an odd number.

So, $(-1)^{4n-3} = -1$.

Now consider $i^{4n-3}$. Using the property $i^{m-k} = i^m / i^k$:

$i^{4n-3} = \frac{i^{4n}}{i^3}$

We know that $i^4 = (i^2)^2 = (-1)^2 = 1$. So $i^{4n} = (i^4)^n = 1^n = 1$.

And $i^3 = -i$.

So, $i^{4n-3} = \frac{1}{-i}$.

To simplify $\frac{1}{-i}$, multiply the numerator and denominator by $i$:

$\frac{1}{-i} = \frac{1 \cdot i}{-i \cdot i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i$.

Therefore, $(-i)^{4n-3} = (-1) \cdot i = -i$.

The value of $\left( -\sqrt{-1} \right)^{4n-3}$ , where n $\in$ N, is -i.

Solution for (v):

We need to find the conjugate of the complex number $\frac{1 - i}{1 + i}$.

First, simplify the complex number by multiplying the numerator and denominator by the conjugate of the denominator ($1-i$).

$\frac{1 - i}{1 + i} = \frac{(1 - i)(1 - i)}{(1 + i)(1 - i)}$

Numerator: $(1 - i)(1 - i) = 1^2 - 2(1)(i) + i^2 = 1 - 2i - 1 = -2i$.

Denominator: $(1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$.

So, $\frac{1 - i}{1 + i} = \frac{-2i}{2} = -i$.

Now, find the conjugate of $-i$. If $z = x + iy$, its conjugate is $\overline{z} = x - iy$.

Here, $z = -i = 0 + (-1)i$. So $x = 0$ and $y = -1$.

The conjugate is $\overline{z} = 0 - (-1)i = 0 + i = i$.

The conjugate of the complex number $\frac{1 \;-\; i}{1 \;+\; i}$ is i.

Solution for (vi):

Let the complex number be $z = x + iy$.

If $z$ lies in the third quadrant, then its real part $x$ is negative ($x < 0$) and its imaginary part $y$ is negative ($y < 0$).

The conjugate of $z$ is $\overline{z} = x - iy$.

The real part of $\overline{z}$ is $x$. Since $z$ is in the third quadrant, $x < 0$.

The imaginary part of $\overline{z}$ is $-y$. Since $z$ is in the third quadrant, $y < 0$, which means $-y > 0$.

So, the conjugate $\overline{z}$ has a negative real part ($x < 0$) and a positive imaginary part ($-y > 0$).

A complex number with a negative real part and a positive imaginary part lies in the second quadrant.

If a complex number lies in the third quadrant, then its conjugate lies in the second quadrant.

Solution for (vii):

We are given the equation $(2 + i) (2 + 2i) (2 + 3i) ... (2 + ni) = x + iy$.

Let $P = (2 + i) (2 + 2i) (2 + 3i) ... (2 + ni)$. So $P = x + iy$.

We need to find the value of $5 \cdot 8 \cdot 13 \cdot ... \cdot (4 + n^2)$.

Let's consider the magnitude of each term in the product $P$:

$|2 + i| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$.

$|2 + 2i| = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8}$.

$|2 + 3i| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.

... and so on.

$|2 + ni| = \sqrt{2^2 + n^2} = \sqrt{4 + n^2}$.

The magnitude of the product of complex numbers is the product of their magnitudes:

$|P| = |(2 + i) (2 + 2i) (2 + 3i) ... (2 + ni)|$

$|P| = |2 + i| |2 + 2i| |2 + 3i| ... |2 + ni|$

$|P| = \sqrt{5} \cdot \sqrt{8} \cdot \sqrt{13} \cdot ... \cdot \sqrt{4 + n^2}$

$|P| = \sqrt{5 \cdot 8 \cdot 13 \cdot ... \cdot (4 + n^2)}$

We are given $P = x + iy$. The magnitude of $P$ is $|P| = |x + iy| = \sqrt{x^2 + y^2}$.

Equating the two expressions for $|P|$:

$\sqrt{5 \cdot 8 \cdot 13 \cdot ... \cdot (4 + n^2)} = \sqrt{x^2 + y^2}$

Squaring both sides:

$5 \cdot 8 \cdot 13 \cdot ... \cdot (4 + n^2) = x^2 + y^2$

If $(2 + i) (2 + 2i) (2 + 3i) ... (2 + ni) = x + iy$, then 5.8.13 ... (4 + n²) = $x^2 + y^2$.

Solution for (i):

Let $z$ be a non-zero complex number. In polar form, $z = r(\cos\theta + i\sin\theta)$, where $r = |z| > 0$ and $\theta = \arg(z)$.

The complex number $i$ can be written in polar form as $i = 1(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2})$.

When we multiply $z$ by $i$, we get the product $z \cdot i$. The polar form of $z \cdot i$ is found by multiplying the magnitudes and adding the arguments:

$z \cdot i = (r \cdot 1) \left(\cos\left(\theta + \frac{\pi}{2}\right) + i\sin\left(\theta + \frac{\pi}{2}\right)\right)$

$z \cdot i = r \left(\cos\left(\theta + \frac{\pi}{2}\right) + i\sin\left(\theta + \frac{\pi}{2}\right)\right)$

The new argument is $\theta + \frac{\pi}{2}$. This represents a rotation of the original argument $\theta$ by an angle of $\frac{\pi}{2}$ (or $90^\circ$) in the anti-clockwise direction.

The statement is True.

Solution for (ii):

Let $z = \cos\theta + i\sin\theta$. This is a complex number in polar form.

The magnitude of $z$ is $|z| = \sqrt{\cos^2\theta + \sin^2\theta}$.

Using the fundamental trigonometric identity $\cos^2\theta + \sin^2\theta = 1$, we get:

$|z| = \sqrt{1} = 1$

For a complex number to be zero ($0 + 0i$), its magnitude must be zero. Since the magnitude of $\cos\theta + i\sin\theta$ is always 1 for any real value of $\theta$, it can never be equal to zero.

The statement is False.

Solution for (iii):

Let $z$ be a complex number, $z = x + iy$, where $x$ and $y$ are real numbers.

The conjugate of $z$ is $\overline{z} = x - iy$.

If a complex number coincides with its conjugate, then $z = \overline{z}$.

$x + iy = x - iy$

Subtracting $x$ from both sides gives:

$iy = -iy$

Adding $iy$ to both sides gives:

$2iy = 0$

Since 2 and $i$ are non-zero, this equation implies that $y$ must be zero ($y=0$).

If $y = 0$, the complex number becomes $z = x + i(0) = x$. This is a purely real number.

Purely real numbers lie on the real axis (the x-axis) in the complex plane. The imaginary axis is where purely imaginary numbers (those with $x=0$) lie.

Therefore, if a complex number coincides with its conjugate, it must lie on the real axis.

The statement is False.

Solution for (iv):

Let $z = (1 + i\sqrt{3})(1 + i)(\cos\theta + i\sin\theta)$.

The argument of a product of complex numbers is the sum of their individual arguments (modulo $2\pi$).

$\arg(z) = \arg(1 + i\sqrt{3}) + \arg(1 + i) + \arg(\cos\theta + i\sin\theta)$.

Let's find the argument of each factor:

For $z_1 = 1 + i\sqrt{3}$: This is in the first quadrant. Real part = 1, Imaginary part = $\sqrt{3}$.

$\arg(z_1) = \arctan\left(\frac{\sqrt{3}}{1}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3}$.

For $z_2 = 1 + i$: This is in the first quadrant. Real part = 1, Imaginary part = 1.

$\arg(z_2) = \arctan\left(\frac{1}{1}\right) = \arctan(1) = \frac{\pi}{4}$.

For $z_3 = \cos\theta + i\sin\theta$: This is already in polar form. The argument is $\theta$.

$\arg(z_3) = \theta$.

Now, sum the arguments:

$\arg(z) = \frac{\pi}{3} + \frac{\pi}{4} + \theta$

Combine the fractions:

$\frac{\pi}{3} + \frac{\pi}{4} = \frac{4\pi}{12} + \frac{3\pi}{12} = \frac{7\pi}{12}$

So, $\arg(z) = \frac{7\pi}{12} + \theta$.

The statement is True.

Solution for (v):

The inequality $|z + 1| < |z - 1|$ can be interpreted geometrically.

$|z + 1|$ represents the distance between the complex number $z$ and the point $-1$ in the complex plane.

$|z - 1|$ represents the distance between the complex number $z$ and the point $1$ in the complex plane.

The inequality $|z + 1| < |z - 1|$ means that the distance from $z$ to $-1$ is less than the distance from $z$ to $1$.

The set of points that are equidistant from two points forms the perpendicular bisector of the line segment joining those two points.

In this case, the points are $-1$ and $1$. The line segment joining $-1$ and $1$ lies on the real axis, from $x=-1$ to $x=1$. The midpoint is at $x=0$. The perpendicular bisector is the vertical line $x=0$, which is the imaginary axis.

The points satisfying $|z + 1| = |z - 1|$ lie on the imaginary axis ($x=0$).

The inequality $|z + 1| < |z - 1|$ means $z$ is closer to $-1$ than to $1$. This region is the half-plane to the left of the perpendicular bisector ($x=0$).

Let $z = x + iy$. The inequality becomes $\sqrt{(x+1)^2 + y^2} < \sqrt{(x-1)^2 + y^2}$.

Squaring both sides: $(x+1)^2 + y^2 < (x-1)^2 + y^2$

$x^2 + 2x + 1 + y^2 < x^2 - 2x + 1 + y^2$

$2x < -2x$

$4x < 0$

$x < 0$

The locus of points is the region where the real part of $z$ is negative, which is the left half-plane (excluding the imaginary axis). This is not the interior of a circle.

The statement is False.

Solution for (vi):

If three complex numbers $z_1, z_2, z_3$ are in Arithmetic Progression (A.P.), then the definition of A.P. states that the difference between consecutive terms is constant.

$z_2 - z_1 = z_3 - z_2$

Rearranging this equation, we get:

$2z_2 = z_1 + z_3$

$z_2 = \frac{z_1 + z_3}{2}$

This means that the complex number $z_2$ is the average of $z_1$ and $z_3$. Geometrically, if $z_1, z_2, z_3$ represent points $P_1, P_2, P_3$ in the complex plane, this equation means that $P_2$ is the midpoint of the line segment connecting $P_1$ and $P_3$.

If $P_2$ is the midpoint of the segment $P_1P_3$, then the three points $P_1, P_2, P_3$ must lie on the same straight line. Therefore, $z_1, z_2, z_3$ are collinear.

Points that are collinear lie on a straight line, not typically on a circle (unless all three points happen to lie on a circle, which is true for any set of three distinct non-collinear points, but the A.P. condition forces collinearity). The statement says "they lie on a circle", which is incorrect for three distinct complex numbers in A.P.

The statement is False.

Solution for (vii):

We need to find the value of $i^n + (i)^{n+1} + (i)^{n+2} + (i)^{n+3}$ for a positive integer $n$.

We can factor out $i^n$ from the expression:

$i^n + i^{n+1} + i^{n+2} + i^{n+3} = i^n(1 + i^1 + i^2 + i^3)$

Let's evaluate the sum inside the parenthesis using the basic powers of $i$:

$i^1 = i$

$i^2 = -1$

$i^3 = -i$

So, $1 + i^1 + i^2 + i^3 = 1 + i + (-1) + (-i) = 1 + i - 1 - i = 0$.

Substituting this back into the expression:

$i^n(1 + i + i^2 + i^3) = i^n(0) = 0$

This result holds true for any integer value of $n$, and specifically for any positive integer $n$.

The statement is True.

Solution:

Let's evaluate each statement in Column A and match it with the appropriate statement in Column B.

(a) The value of $1+i^2 + i^4 + i^6 + ... + i^{20}$ is

This is a sum of powers of $i$ with even exponents from 0 to 20. The powers are $i^{2k} = (i^2)^k = (-1)^k$.

The exponents are $0, 2, 4, \dots, 20$. These are $2 \times 0, 2 \times 1, 2 \times 2, \dots, 2 \times 10$. There are $10 - 0 + 1 = 11$ terms.

The sum is $1 + (-1) + 1 + (-1) + \dots + 1$. This is a geometric series with first term $a=1$, common ratio $r=i^2=-1$, and number of terms $n=11$.

The sum $S_{11} = a \frac{r^{11} - 1}{r - 1} = 1 \cdot \frac{(-1)^{11} - 1}{-1 - 1} = \frac{-1 - 1}{-2} = \frac{-2}{-2} = 1$.

The value is 1. A complex number $z = x+iy$ is purely real if $y=0$. Since $1 = 1 + 0i$, it is a purely real complex number.

Match: (a) - (ii)

(b) The value of $i^{-1097}$ is

$i^{-1097} = \frac{1}{i^{1097}}$. We find the remainder when 1097 is divided by 4: $1097 = 4 \times 274 + 1$.

So, $i^{1097} = i^1 = i$.

$i^{-1097} = \frac{1}{i}$. To simplify, we multiply the numerator and denominator by $i$:

$\frac{1}{i} = \frac{1 \cdot i}{i \cdot i} = \frac{i}{i^2} = \frac{i}{-1} = -i$.

The value is $-i$. A complex number $z = x+iy$ is purely imaginary if $x=0$ and $y \ne 0$. Since $-i = 0 + (-1)i$, it is a purely imaginary complex number.

Match: (b) - (i)

(c) Conjugate of $1+i$ lies in

Let $z = 1+i$. The conjugate of $z$ is $\overline{z} = \overline{1+i} = 1-i$.

For the complex number $1-i$, the real part is 1 (positive) and the imaginary part is -1 (negative).

A complex number with a positive real part and a negative imaginary part lies in the fourth quadrant.

Match: (c) - (iv)

(d) $\frac{1+2i}{1-i}$ lies in

We simplify the complex number by multiplying the numerator and denominator by the conjugate of the denominator ($1+i$).

$\frac{1+2i}{1-i} = \frac{(1+2i)(1+i)}{(1-i)(1+i)} = \frac{1 + i + 2i + 2i^2}{1^2 - i^2} = \frac{1 + 3i - 2}{1 - (-1)} = \frac{-1 + 3i}{2} = -\frac{1}{2} + \frac{3}{2}i$.

For the complex number $-\frac{1}{2} + \frac{3}{2}i$, the real part is $-\frac{1}{2}$ (negative) and the imaginary part is $\frac{3}{2}$ (positive).

A complex number with a negative real part and a positive imaginary part lies in the second quadrant.

Match: (d) - (iii)

(e) If $a, b, c \in \mathbb{R}$ and $b^2 – 4ac < 0$, then the roots of the equation $ax^2 + bx + c = 0$ are non real (complex) and

The roots of the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Given that $a, b, c$ are real numbers and the discriminant $\Delta = b^2 - 4ac < 0$.

When the discriminant is negative and the coefficients are real, the roots are always non-real complex numbers and they occur in conjugate pairs. For example, the roots are of the form $p \pm qi$ where $q \ne 0$. The conjugate of $p+qi$ is $p-qi$.

The roots must occur in conjugate pairs.

Match: (e) - (vi) (They "may occur" in conjugate pairs, which is true, as they always do under these conditions.)

(f) If $a, b, c \in \mathbb{R}$ and $b^2 – 4ac > 0$, and $b^2 – 4ac$ is a perfect square, then the roots of the equation $ax^2 + bx + c = 0$

Given that $a, b, c$ are real numbers and the discriminant $\Delta = b^2 - 4ac > 0$. This implies there are two distinct real roots.

Given that $\Delta$ is a perfect square. This implies $\sqrt{\Delta}$ is a rational number.

The roots are $x = \frac{-b \pm \sqrt{\Delta}}{2a}$. Since $a, b, c$ are real, and $\sqrt{\Delta}$ is a real number, the roots are real numbers. Since $b^2-4ac > 0$, the roots are distinct.

Real numbers are considered purely real complex numbers (their imaginary part is zero). The roots are real numbers (and rational in this specific case due to the perfect square condition, but the option just says "purely real").

Do these distinct real roots "occur in conjugate pairs" in the context complex numbers? A real number is its own conjugate. If the two distinct real roots are $r_1$ and $r_2$, they are not generally conjugates of each other ($r_1 \ne r_2$). So they do not form a non-trivial conjugate pair. They may not occur as a conjugate pair.

Match: (f) - (v) (They "may not occur" in conjugate pairs, referring to non-trivial complex conjugate pairs.)

Summary of Matches:

(a) - (ii)

(b) - (i)

(c) - (iv)

(d) - (iii)

(e) - (vi)

(f) - (v)

Given Expression:

$\frac{i^{4n+1}-i^{4n-1}}{2}$

To Find:

The value of the given expression.

Solution:

We need to simplify the expression $\frac{i^{4n+1}-i^{4n-1}}{2}$.

We use the properties of powers of $i$. We know that $i^4 = 1$.

Consider the term $i^{4n+1}$:

$i^{4n+1} = i^{4n} \cdot i^1$

$= (i^4)^n \cdot i$

Since $i^4 = 1$, we have:

$= 1^n \cdot i$

$= 1 \cdot i = i$

Consider the term $i^{4n-1}$:

$i^{4n-1} = i^{4n} \cdot i^{-1}$

$= (i^4)^n \cdot \frac{1}{i}$

$= 1^n \cdot \frac{1}{i} = \frac{1}{i}$

To simplify $\frac{1}{i}$, we multiply the numerator and denominator by $i$:

$\frac{1}{i} = \frac{1 \cdot i}{i \cdot i} = \frac{i}{i^2}$

Since $i^2 = -1$, we have:

$= \frac{i}{-1} = -i$

Now substitute the simplified terms back into the original expression:

$\frac{i^{4n+1}-i^{4n-1}}{2} = \frac{i - (-i)}{2}$

$= \frac{i + i}{2}$

$= \frac{2i}{2}$

$= i$

Final Answer:

The value of the expression is $i$.

Given Equation:

$(1 + i)^{2n} = (1 \;–\; i)^{2n}$

To Find:

The smallest positive integer $n$ for which the given equation holds.

Solution:

We are given the equation $(1 + i)^{2n} = (1 \;–\; i)^{2n}$.

Since $(1-i)^{2n}$ is not zero (as $1-i \ne 0$), we can divide both sides by $(1-i)^{2n}$:

$\frac{(1 + i)^{2n}}{(1 \;–\; i)^{2n}} = 1$

Using the property $\frac{a^m}{b^m} = \left(\frac{a}{b}\right)^m$, we can write:

$\left(\frac{1 + i}{1 \;–\; i}\right)^{2n} = 1$

Let's simplify the complex number inside the parenthesis, $\frac{1 + i}{1 \;–\; i}$. We multiply the numerator and denominator by the conjugate of the denominator, which is $1+i$:

$\frac{1 + i}{1 \;–\; i} = \frac{(1 + i)(1 + i)}{(1 \;–\; i)(1 + i)}$

Using $(a+b)^2 = a^2 + 2ab + b^2$ for the numerator and $(a-b)(a+b) = a^2 - b^2$ for the denominator:

Numerator: $(1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i$ (since $i^2 = -1$)

Denominator: $(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$

So, $\frac{1 + i}{1 \;–\; i} = \frac{2i}{2} = i$.

Substitute this back into the equation:

$(i)^{2n} = 1$

Using the property $(a^m)^k = a^{mk}$, we have:

$(i^2)^n = 1$

Since $i^2 = -1$, the equation becomes:

$(-1)^n = 1$

For $(-1)^n$ to be equal to 1, the exponent $n$ must be an even integer.

We are looking for the smallest positive integer $n$ that satisfies this condition.

The set of positive integers is $\{1, 2, 3, 4, 5, 6, \dots \}$.

The even integers in this set are $\{2, 4, 6, 8, \dots \}$.

The smallest positive integer in this set is 2.

Thus, the smallest positive integer $n$ for which the equation holds is 2.

Final Answer:

The smallest positive integer $n$ is 2.

Given Complex Number:

$z = 3 + \sqrt{7}i$

To Find:

The reciprocal of $z$.

Solution:

The reciprocal of a non-zero complex number $z$ is given by $\frac{1}{z}$.

So, the reciprocal of $3 + \sqrt{7}i$ is $\frac{1}{3 + \sqrt{7}i}$.

To express this in the standard form $x + iy$, we multiply the numerator and the denominator by the conjugate of the denominator.

The conjugate of the denominator $3 + \sqrt{7}i$ is $3 - \sqrt{7}i$.

Multiply the fraction by $\frac{3 - \sqrt{7}i}{3 - \sqrt{7}i}$:

Reciprocal $= \frac{1}{3 + \sqrt{7}i} \times \frac{3 - \sqrt{7}i}{3 - \sqrt{7}i}$

Numerator $= 1 \times (3 - \sqrt{7}i) = 3 - \sqrt{7}i$

Denominator $= (3 + \sqrt{7}i)(3 - \sqrt{7}i)$

Using the identity $(a+bi)(a-bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2$:

Denominator $= 3^2 + (\sqrt{7})^2$

$= 9 + 7$

$= 16$

Combine the numerator and denominator:

Reciprocal $= \frac{3 - \sqrt{7}i}{16}$

Write in the standard form $x+iy$:

Reciprocal $= \frac{3}{16} - \frac{\sqrt{7}}{16}i$

Final Answer:

The reciprocal of $3 + \sqrt{7}i$ is $\frac{3}{16} - \frac{\sqrt{7}}{16}i$.

Given:

$z_1 = \sqrt{3} + i\sqrt{3}$

$z_2 = \sqrt{3} + i$

To Find:

The quadrant in which $\left( \frac{z_{1}}{z_{2}} \right)$ lies.

Solution:

We need to calculate the complex number $\frac{z_1}{z_2}$ and determine the signs of its real and imaginary parts.

$\frac{z_1}{z_2} = \frac{\sqrt{3} + i\sqrt{3}}{\sqrt{3} + i}$

To simplify this fraction, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of the denominator $z_2 = \sqrt{3} + i$ is $\overline{z_2} = \sqrt{3} - i$.

Multiply the fraction by $\frac{\sqrt{3} - i}{\sqrt{3} - i}$:

$\frac{z_1}{z_2} = \frac{(\sqrt{3} + i\sqrt{3})(\sqrt{3} - i)}{(\sqrt{3} + i)(\sqrt{3} - i)}$

Let's calculate the numerator and the denominator separately.

Numerator: $(\sqrt{3} + i\sqrt{3})(\sqrt{3} - i)$

Using the distributive property (FOIL):

$(\sqrt{3})(\sqrt{3}) + (\sqrt{3})(-i) + (i\sqrt{3})(\sqrt{3}) + (i\sqrt{3})(-i)$

$= 3 - i\sqrt{3} + i(\sqrt{3})^2 - i^2\sqrt{3}$

Since $i^2 = -1$:

$= 3 - i\sqrt{3} + 3i - (-1)\sqrt{3}$

$= 3 - i\sqrt{3} + 3i + \sqrt{3}$

Group the real and imaginary terms:

$= (3 + \sqrt{3}) + i(3 - \sqrt{3})$

Denominator: $(\sqrt{3} + i)(\sqrt{3} - i)$

Using the identity $(a+bi)(a-bi) = a^2 + b^2$:

Denominatior $= (\sqrt{3})^2 + (1)^2$

$= 3 + 1 = 4$

Now, combine the numerator and denominator to get $\frac{z_1}{z_2}$:

$\frac{z_1}{z_2} = \frac{(3 + \sqrt{3}) + i(3 - \sqrt{3})}{4}$

Express in the standard form $x + iy$:

$\frac{z_1}{z_2} = \frac{3 + \sqrt{3}}{4} + i \frac{3 - \sqrt{3}}{4}$

Let $X = \frac{3 + \sqrt{3}}{4}$ be the real part and $Y = \frac{3 - \sqrt{3}}{4}$ be the imaginary part.

We need to determine the signs of $X$ and $Y$.

For $X = \frac{3 + \sqrt{3}}{4}$, since $3 > 0$ and $\sqrt{3} > 0$, their sum $3 + \sqrt{3}$ is positive. The denominator 4 is also positive. Therefore, $X > 0$.

For $Y = \frac{3 - \sqrt{3}}{4}$, since $3 \approx 3$ and $\sqrt{3} \approx 1.732$, we have $3 > \sqrt{3}$. Their difference $3 - \sqrt{3}$ is positive. The denominator 4 is also positive. Therefore, $Y > 0$.

The complex number $\left( \frac{z_{1}}{z_{2}} \right)$ has a positive real part ($X > 0$) and a positive imaginary part ($Y > 0$).

A complex number with both real and imaginary parts positive lies in the first quadrant of the complex plane.

Alternate Method (using argument):

We can also find the argument of $\frac{z_1}{z_2}$, which is $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$.

For $z_1 = \sqrt{3} + i\sqrt{3}$, the real part is $\sqrt{3}$ and the imaginary part is $\sqrt{3}$. It's in the first quadrant. $\tan(\arg(z_1)) = \frac{\sqrt{3}}{\sqrt{3}} = 1$. So, $\arg(z_1) = \frac{\pi}{4}$.

For $z_2 = \sqrt{3} + i$, the real part is $\sqrt{3}$ and the imaginary part is 1. It's in the first quadrant. $\tan(\arg(z_2)) = \frac{1}{\sqrt{3}}$. So, $\arg(z_2) = \frac{\pi}{6}$.

$\arg\left(\frac{z_1}{z_2}\right) = \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi}{12} - \frac{2\pi}{12} = \frac{\pi}{12}$.

Since the argument $\frac{\pi}{12}$ is between $0$ and $\frac{\pi}{2}$ ($0 < \frac{\pi}{12} < \frac{\pi}{2}$), the complex number lies in the first quadrant.

Final Answer:

The complex number $\left( \frac{z_{1}}{z_{2}} \right)$ lies in the first quadrant.

Given Expression:

$Z = \frac{\sqrt{5\;+\;12i}\;+\;\sqrt{5\;-\;12i}}{\sqrt{5\;+\;12i}\;-\;\sqrt{5\;-\;12i}}$

To Find:

The conjugate of $Z$, which is $\overline{Z}$.

Solution:

Let $a = \sqrt{5 + 12i}$ and $b = \sqrt{5 - 12i}$.

The expression can be written as $Z = \frac{a + b}{a - b}$.

We need to find the conjugate of $Z$, which is $\overline{Z} = \overline{\left(\frac{a + b}{a - b}\right)}$.

Using the property $\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}$, we get:

$\overline{Z} = \frac{\overline{a + b}}{\overline{a - b}}$

Using the property $\overline{z_1 \pm z_2} = \overline{z_1} \pm \overline{z_2}$, we get:

$\overline{Z} = \frac{\overline{a} + \overline{b}}{\overline{a} - \overline{b}}$

Now, let's find $\overline{a}$ and $\overline{b}$.

$a = \sqrt{5 + 12i}$. Then $\overline{a} = \overline{\sqrt{5 + 12i}}$.

Using the property $\overline{\sqrt{z}} = \sqrt{\overline{z}}$ (this property holds for the principal value of the square root), we have:

$\overline{a} = \sqrt{\overline{5 + 12i}} = \sqrt{5 - 12i}$.

Note that $\sqrt{5 - 12i} = b$. So, $\overline{a} = b$.

Similarly, $b = \sqrt{5 - 12i}$. Then $\overline{b} = \overline{\sqrt{5 - 12i}}$.

$\overline{b} = \sqrt{\overline{5 - 12i}} = \sqrt{5 + 12i}$.

Note that $\sqrt{5 + 12i} = a$. So, $\overline{b} = a$.

Substitute $\overline{a} = b$ and $\overline{b} = a$ into the expression for $\overline{Z}$:

$\overline{Z} = \frac{b + a}{b - a}$

We can rewrite the denominator as $-(a - b)$.

$\overline{Z} = \frac{a + b}{-(a - b)}$

$\overline{Z} = - \frac{a + b}{a - b}$

Recall that $Z = \frac{a + b}{a - b}$.

So, $\overline{Z} = -Z$.

This means that $Z + \overline{Z} = 0$. If $Z = x + iy$, then $\overline{Z} = x - iy$.

$(x + iy) + (x - iy) = 0$

$2x = 0$

$x = 0$

This implies that the original complex number $Z$ must be a purely imaginary number (or zero, but the denominator is non-zero). The conjugate of a purely imaginary number $iy$ is $-iy$.

Let's confirm by finding the square root of $5 + 12i$. Let $\sqrt{5 + 12i} = x + iy$.

$(x+iy)^2 = 5 + 12i$

$x^2 - y^2 + 2xyi = 5 + 12i$

Equating real and imaginary parts:

$x^2 - y^2 = 5$

$2xy = 12 \implies xy = 6$

We also know $|x+iy|^2 = |5+12i|$.

$x^2 + y^2 = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.

We have two equations: $x^2 - y^2 = 5$ and $x^2 + y^2 = 13$.

Adding the two equations: $(x^2 - y^2) + (x^2 + y^2) = 5 + 13 \implies 2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$.

If $x = 3$, then from $xy=6$, $3y=6 \implies y=2$. So $3+2i$ is a square root.

If $x = -3$, then from $xy=6$, $-3y=6 \implies y=-2$. So $-3-2i$ is another square root.

The principal value of $\sqrt{5+12i}$ has a positive real part, so let's take $\sqrt{5 + 12i} = 3 + 2i$.

Then $\sqrt{5 - 12i} = \sqrt{\overline{5 + 12i}} = \overline{3 + 2i} = 3 - 2i$.

Now substitute these values into the expression for $Z$:

$Z = \frac{(3 + 2i) + (3 - 2i)}{(3 + 2i) - (3 - 2i)}$

Numerator $= 3 + 2i + 3 - 2i = 6$

Denominator $= 3 + 2i - 3 + 2i = 4i$

$Z = \frac{6}{4i} = \frac{3}{2i}$

To simplify $\frac{3}{2i}$, multiply by $\frac{i}{i}$:

$Z = \frac{3i}{2i^2} = \frac{3i}{2(-1)} = \frac{3i}{-2} = -\frac{3}{2}i$.

So, $Z = -\frac{3}{2}i$. This is a purely imaginary number.

The conjugate of $Z = -\frac{3}{2}i$ is $\overline{Z} = \overline{-\frac{3}{2}i} = \frac{3}{2}i$.

Note that $\overline{Z} = -Z$, which matches our earlier derivation.

Final Answer:

The conjugate of the given expression is $\frac{3}{2}i$.

Given Complex Number:

$z = 1 - i$

To Find:

The principal value of the argument (amplitude) of $z = 1 - i$.

Solution:

Let $z = x + iy$. Here, $x = 1$ and $y = -1$.

The principal value of the argument of a complex number $z = x + iy$ is the unique value $\theta \in (-\pi, \pi]$ such that $z = |z|(\cos\theta + i\sin\theta)$.

First, determine the quadrant in which the complex number $1 - i$ lies.

The real part is $x = 1$ (positive) and the imaginary part is $y = -1$ (negative).

A complex number with a positive real part and a negative imaginary part lies in the fourth quadrant.

The angle $\theta$ can be related to $x$ and $y$ by $\tan\theta = \frac{y}{x}$, provided $x \ne 0$.

$\tan\theta = \frac{-1}{1} = -1$.

We need to find the angle $\theta$ in the interval $(-\pi, \pi]$ whose tangent is -1 and that lies in the fourth quadrant.

The angles whose tangent is -1 are of the form $\frac{3\pi}{4} + k\pi$ or $-\frac{\pi}{4} + k\pi$ for integer $k$.

In degrees, the angles are $135^\circ, 315^\circ, -45^\circ, \dots$

We need the angle in the fourth quadrant.

• $\frac{3\pi}{4}$ is in the second quadrant.
• $\frac{3\pi}{4} + \pi = \frac{7\pi}{4}$ is outside the interval $(-\pi, \pi]$.
• $-\frac{\pi}{4}$ is in the fourth quadrant. $-\frac{\pi}{4} \in (-\pi, \pi]$.
• $-\frac{\pi}{4} + \pi = \frac{3\pi}{4}$ is in the second quadrant.

The angle in the fourth quadrant with a tangent of -1 is $-\frac{\pi}{4}$. This angle is in the required interval $(-\pi, \pi]$.

Therefore, the principal value of the argument is $-\frac{\pi}{4}$.

Final Answer:

The principal value of amplitude of $1 \;–\; i$ is $-\frac{\pi}{4}$.

Given Complex Number:

$z = (i^{25})^3$

To Find:

The polar form of $z$.

Solution:

We need to simplify the given complex number first.

Using the property $(a^m)^n = a^{mn}$:

$z = i^{25 \times 3} = i^{75}$

Now, we need to evaluate the power of $i$. We find the remainder when 75 is divided by 4:

$75 = 4 \times 18 + 3$

So, $i^{75} = i^3$.

We know that $i^3 = -i$.

So, the complex number is $z = -i$.

Now, we need to express $z = -i$ in polar form, which is $z = r(\cos\theta + i\sin\theta)$, where $r = |z|$ and $\theta$ is the argument of $z$.

For $z = -i$, the real part is $x = 0$ and the imaginary part is $y = -1$.

Calculate the magnitude $r$:

$r = |z| = \sqrt{x^2 + y^2} = \sqrt{0^2 + (-1)^2} = \sqrt{0 + 1} = \sqrt{1} = 1$.

Calculate the argument $\theta$. Since $x = 0$, we cannot use $\tan\theta = \frac{y}{x}$.

The complex number $z = -i$ corresponds to the point $(0, -1)$ in the complex plane. This point lies on the negative imaginary axis.

The angle from the positive real axis to the negative imaginary axis is $-\frac{\pi}{2}$ (or $\frac{3\pi}{2}$). The principal value of the argument lies in the interval $(-\pi, \pi]$.

The angle $-\frac{\pi}{2}$ is in the interval $(-\pi, \pi]$.

So, $\theta = -\frac{\pi}{2}$.

The polar form is $z = r(\cos\theta + i\sin\theta)$:

$z = 1 \left(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right)$

We can verify this: $\cos\left(-\frac{\pi}{2}\right) = 0$ and $\sin\left(-\frac{\pi}{2}\right) = -1$.

$z = 1(0 + i(-1)) = -i$, which matches the original complex number.

Final Answer:

The polar form of $(i^{25})^3$ is $1 \left(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right)$ or $\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)$.

Given Condition:

Amplitude of $z \;–\; 2 \;–\; 3i$ is $\frac{\pi}{4}$.

This can be written as $\arg (z - (2 + 3i)) = \frac{\pi}{4}$.

To Find:

The locus of the complex number $z$ satisfying the given condition.

Solution:

Let $z = x + iy$, where $x$ and $y$ are real numbers.

The complex number inside the argument is $z - (2 + 3i)$.

$z - (2 + 3i) = (x + iy) - (2 + 3i)$

$= (x - 2) + i(y - 3)$

The argument of a complex number $w = X + iY$ is the angle the line segment from the origin to the point $(X, Y)$ makes with the positive real axis.

The condition $\arg ((x - 2) + i(y - 3)) = \frac{\pi}{4}$ means that the complex number $(x - 2) + i(y - 3)$ lies on a ray starting from the origin in the complex plane and making an angle of $\frac{\pi}{4}$ with the positive real axis.

Let $X = x - 2$ and $Y = y - 3$. The condition is $\arg(X + iY) = \frac{\pi}{4}$.

For a complex number $X+iY$ with $\arg(X+iY) = \frac{\pi}{4}$, it must satisfy:

$Y = X \tan\left(\frac{\pi}{4}\right)$

And since the argument is $\frac{\pi}{4}$ (which is in the first quadrant), the real part $X$ and the imaginary part $Y$ must both be positive (or both zero, but the argument of 0 is undefined, so $X$ and $Y$ cannot both be zero).

$\tan\left(\frac{\pi}{4}\right) = 1$. So, we have:

$Y = X$

And the conditions $X > 0$ and $Y > 0$.

Substitute back $X = x - 2$ and $Y = y - 3$:

$y - 3 = x - 2$

This simplifies to:

$y = x - 2 + 3$

$y = x + 1$

This is the equation of a straight line in the complex plane (treating it as the Cartesian plane).

Now, consider the conditions $X > 0$ and $Y > 0$:

$x - 2 > 0 \implies x > 2$

$y - 3 > 0 \implies y > 3$

We have the equation $y = x + 1$. If $x > 2$, then $y = x + 1 > 2 + 1 = 3$. So the condition $y > 3$ is automatically satisfied when $x > 2$ on the line $y = x + 1$.

The point where $x = 2$ on this line is $y = 2 + 1 = 3$. This corresponds to the point $(2, 3)$, which represents the complex number $z = 2 + 3i$. At this point, $z - (2 + 3i) = 0$, and $\arg(0)$ is undefined. Thus, the point $z = 2 + 3i$ must be excluded from the locus.

The locus of $z$ is the set of points $(x, y)$ satisfying the equation $y = x + 1$ with the restriction $x > 2$.

Geometrically, this is a ray (a half-line) starting from the point $(2, 3)$ but not including the point $(2, 3)$. The ray makes an angle of $\frac{\pi}{4}$ with the positive real axis direction.

Final Answer:

The locus of $z$ is a ray starting from the point representing $2 + 3i$ (excluding the point itself) and making an angle of $\frac{\pi}{4}$ with the positive direction of the real axis.

Given:

The equation is $x^2 + ax + b = 0$, where $a, b \in \mathbb{R}$.

One root of the equation is $1 - i$.

To Find:

The values of $a$ and $b$.

Solution Method 1: Using the property of conjugate roots

Since the coefficients $a$ and $b$ of the quadratic equation are real numbers, if a complex number is a root, its conjugate must also be a root.

The given root is $x_1 = 1 - i$.

The conjugate of $x_1$ is $\overline{x_1} = \overline{1 - i} = 1 + i$.

Since the coefficients are real, the other root must be $x_2 = 1 + i$.

For a quadratic equation $x^2 + ax + b = 0$, the sum of the roots is $-a$ and the product of the roots is $b$.

Sum of roots: $x_1 + x_2 = -a$

$(1 - i) + (1 + i) = -a$

$1 - i + 1 + i = -a$

$2 = -a$

$a = -2$

Product of roots: $x_1 \cdot x_2 = b$

$(1 - i)(1 + i) = b$

Using the identity $(u - v)(u + v) = u^2 - v^2$:

$1^2 - i^2 = b$

Since $i^2 = -1$:

$1 - (-1) = b$

$1 + 1 = b$

$b = 2$

Thus, the values are $a = -2$ and $b = 2$.

Solution Method 2: Substituting the root into the equation

Since $1 - i$ is a root of the equation $x^2 + ax + b = 0$, substituting $x = 1 - i$ into the equation must satisfy it.

$(1 - i)^2 + a(1 - i) + b = 0$

Evaluate $(1 - i)^2$:

$(1 - i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i - 1 = -2i$.

Substitute this back into the equation:

$-2i + a(1 - i) + b = 0$

$-2i + a - ai + b = 0$

Group the real and imaginary terms:

$(a + b) + (-2 - a)i = 0 + 0i$

For a complex number to be equal to zero, its real part and its imaginary part must both be zero.

Equating the real parts:

$a + b = 0$

Equating the imaginary parts:

$-2 - a = 0$

From equation (ii):

$-a = 2 \implies a = -2$

Substitute the value of $a = -2$ into equation (i):

$(-2) + b = 0$

$b = 2$

Both methods give the same result.

Final Answer:

The values are $a = -2$ and $b = 2$.

Solution:

The sum is $S = 1 + i^2 + i^4 + i^6 \;+ \;... \;+\; i^{2n}$.

This can be written as $S = \sum_{k=0}^n i^{2k} = \sum_{k=0}^n (i^2)^k$.

Since $i^2 = -1$, the sum becomes $S = \sum_{k=0}^n (-1)^k$.

$S = (-1)^0 + (-1)^1 + (-1)^2 + \dots + (-1)^n$

$S = 1 - 1 + 1 - \dots + (-1)^n$

This is a sum of $n+1$ terms.

If $n$ is an odd integer, the number of terms $n+1$ is even. The terms pair up to give $1 - 1 = 0$. So the sum is 0.

If $n$ is an even integer, the number of terms $n+1$ is odd. The pairs give 0, and the first term (or last term) remains. The sum is 1.

The value of the sum $1 + i^2 + i^4 + \dots + i^{2n}$ is 0 if $n$ is odd, and 1 if $n$ is even.

The possible values of the sum are 0 and 1.

Looking at the options:

• (A) positive: The sum can be 1, which is positive, but it is not always positive (it can be 0).
• (B) negative: The sum is never negative.
• (C) 0: The sum can be 0 (when $n$ is odd).
• (D) can not be evaluated: The sum can be evaluated (it's 0 or 1).

Since 0 is one of the possible values of the sum, option (C) is a correct statement about what the sum "is" for certain values of $n$. Given the multiple-choice format, option (C) is the most likely intended answer, indicating that 0 is a possible value.

The final answer is $\boxed{C}$.

Given Condition:

$|z + 1| = 1$

To Find:

The locus of the complex number $z$ satisfying the given condition.

Solution:

The condition $|z - z_0| = r$ represents a circle in the complex plane with center $z_0$ and radius $r$.

The given condition is $|z + 1| = 1$.

We can rewrite this as $|z - (-1)| = 1$.

Comparing this with the standard form $|z - z_0| = r$, we identify the center and radius of the circle:

The center of the circle is $z_0 = -1$. In the complex plane, this corresponds to the point $(-1, 0)$ in the Cartesian coordinate system.

The radius of the circle is $r = 1$.

Alternatively, let $z = x + iy$, where $x, y \in \mathbb{R}$.

Substitute $z = x + iy$ into the given condition:

$|(x + iy) + 1| = 1$

Group the real and imaginary parts:

$|(x + 1) + iy| = 1$

The magnitude of a complex number $X + iY$ is $|X + iY| = \sqrt{X^2 + Y^2}$.

Here, $X = x + 1$ and $Y = y$. So the condition becomes:

$\sqrt{(x + 1)^2 + y^2} = 1$

Squaring both sides of the equation:

$(x + 1)^2 + y^2 = 1^2$

$(x + 1)^2 + y^2 = 1$

This is the equation of a circle in the Cartesian coordinate system $(x, y)$ with center $(h, k) = (-1, 0)$ and radius $r = 1$.

Therefore, the locus of $z$ is a circle with center $(-1, 0)$ and radius 1.

Match with the given options:

• (A) x-axis: Incorrect.
• (B) circle with centre (1, 0) and radius 1: Incorrect, the center is $(-1, 0)$.
• (C) circle with centre (–1, 0) and radius 1: Correct.
• (D) y-axis: Incorrect.

The final answer is $\boxed{C}$.

Solution:

Let the three complex numbers representing the vertices of the triangle be $A = z$, $B = -iz$, and $C = z + iz$.

We can find the area of the triangle by translating one of the vertices to the origin. Let's translate vertex A to the origin.

The new vertices of the translated triangle will be:

$A' = A - A = z - z = 0$

$B' = B - A = -iz - z = z(-i - 1) = -z(1 + i)$

$C' = C - A = (z + iz) - z = iz$

The area of a triangle with vertices at the origin, $w_1$, and $w_2$ is given by the formula:

Area $= \frac{1}{2} |\text{Im}(\overline{w_1} w_2)|$

In our case, $w_1 = B' = -z(1 + i)$ and $w_2 = C' = iz$.

First, let's calculate $\overline{w_1}$:

$\overline{w_1} = \overline{-z(1 + i)}$

Using the property $\overline{z_1 z_2} = \overline{z_1} \overline{z_2}$ and $\overline{-1} = -1$:

$\overline{w_1} = \overline{-1} \cdot \overline{z} \cdot \overline{(1 + i)}$

$= (-1) \cdot \overline{z} \cdot (1 - i)$

$= -\overline{z}(1 - i)$

Now, calculate the product $\overline{w_1} w_2$:

$\overline{w_1} w_2 = [-\overline{z}(1 - i)](iz)$

$= -i \overline{z} (1 - i) z$

Using the commutative property of multiplication and the property $\overline{z}z = |z|^2$:

$= -i (\overline{z} z) (1 - i)$

$= -i |z|^2 (1 - i)$

Distribute $-i |z|^2$:

$= -i |z|^2 \cdot 1 - i |z|^2 \cdot (-i)$

$= -i |z|^2 + i^2 |z|^2$

Since $i^2 = -1$:

$= -i |z|^2 + (-1) |z|^2$

$= -i |z|^2 - |z|^2$

Rearrange the terms to identify the real and imaginary parts:

$= -|z|^2 - i |z|^2$

The imaginary part of $\overline{w_1} w_2$ is $\text{Im}(-|z|^2 - i |z|^2) = -|z|^2$.

Now, calculate the area of the triangle using the formula:

Area $= \frac{1}{2} |\text{Im}(\overline{w_1} w_2)|$

Area $= \frac{1}{2} |- |z|^2 |$

Since $|z|^2$ is always a non-negative real number, $|-|z|^2| = |z|^2$.

Area $= \frac{1}{2} |z|^2$

We also know that $|z| = |\overline{z}|$, so $|z|^2 = |\overline{z}|^2$.

Therefore, the area can also be written as $\frac{1}{2} |\overline{z}|^2$.

Match the calculated area with the given options:

• (A) $|z|^2$: Incorrect.
• (B) $\left| \overline{z} \right|^{2}$: Incorrect.
• (C) $\frac{\left| \overline{z} \right|^{2}}{2}$: Correct.
• (D) none of these: Incorrect.

The final answer is $\boxed{C}$.

Given Equation:

$|z + 1 - i| = |z - 1 + i|$

To Find:

The locus of the complex number $z$ satisfying the given equation.

Solution Method 1: Geometric Interpretation

The expression $|z - z_0|$ represents the distance between the complex number $z$ and the fixed point $z_0$ in the complex plane.

Rewrite the given equation in the form $|z - z_1| = |z - z_2|$.

$|z - (-1 + i)| = |z - (1 - i)|$

Let $z_1 = -1 + i$ and $z_2 = 1 - i$.

The equation states that the distance from $z$ to $z_1$ is equal to the distance from $z$ to $z_2$.

The set of points that are equidistant from two fixed points is the perpendicular bisector of the line segment joining those two points.

The locus of $z$ is the perpendicular bisector of the line segment joining the points representing $z_1 = -1 + i$ and $z_2 = 1 - i$ in the complex plane.

A perpendicular bisector is a straight line.

Solution Method 2: Algebraic Approach

Let $z = x + iy$, where $x, y \in \mathbb{R}$.

Substitute $z = x + iy$ into the given equation:

$|(x + iy) + 1 - i| = |(x + iy) - 1 + i|$

Group the real and imaginary parts on both sides:

$|(x + 1) + i(y - 1)| = |(x - 1) + i(y + 1)|$

The magnitude of a complex number $X + iY$ is $|X + iY| = \sqrt{X^2 + Y^2}$.

Applying this to both sides of the equation:

$\sqrt{(x + 1)^2 + (y - 1)^2} = \sqrt{(x - 1)^2 + (y + 1)^2}$

Square both sides to eliminate the square roots:

$(x + 1)^2 + (y - 1)^2 = (x - 1)^2 + (y + 1)^2$

Expand the squared terms:

$(x^2 + 2x + 1) + (y^2 - 2y + 1) = (x^2 - 2x + 1) + (y^2 + 2y + 1)$

$x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 - 2x + 1 + y^2 + 2y + 1$

$x^2 + 2x + y^2 - 2y + 2 = x^2 - 2x + y^2 + 2y + 2$

Subtract $x^2 + y^2 + 2$ from both sides:

$2x - 2y = -2x + 2y$

Move terms involving $x$ to one side and terms involving $y$ to the other:

$2x + 2x = 2y + 2y$

$4x = 4y$

Divide by 4:

$x = y$

The equation $x = y$ in the Cartesian coordinate system represents a straight line passing through the origin with a slope of 1.

Therefore, the locus of $z$ is a straight line.

Match with the given options:

• (A) straight line: Correct.
• (B) circle: Incorrect.
• (C) parabola: Incorrect.
• (D) hyperbola: Incorrect.

The final answer is $\boxed{A}$.

Given Equation:

$z^2 + |z|^2 = 0$

To Find:

The number of solutions of the given equation.

Solution:

Let $z = x + iy$, where $x, y \in \mathbb{R}$.

Then $z^2 = (x + iy)^2 = x^2 + 2xyi + (iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi$.

The magnitude of $z$ is $|z| = \sqrt{x^2 + y^2}$.

The magnitude squared is $|z|^2 = (\sqrt{x^2 + y^2})^2 = x^2 + y^2$.

Substitute these into the given equation $z^2 + |z|^2 = 0$:

$((x^2 - y^2) + 2xyi) + (x^2 + y^2) = 0$

$(x^2 - y^2 + x^2 + y^2) + 2xyi = 0$

$2x^2 + 2xyi = 0 + 0i$

For a complex number to be equal to zero, its real part and its imaginary part must both be zero.

Equating the real parts:

$2x^2 = 0$

$x^2 = 0$

This implies $x = 0$.

Equating the imaginary parts:

$2xy = 0$

Since $x = 0$ (from the real part equation), this equation becomes:

$2(0)y = 0$

$0 = 0$

This equation is true for any real value of $y$.

So, the solutions are of the form $z = x + iy$ where $x = 0$ and $y$ can be any real number.

$z = 0 + iy = iy$, where $y \in \mathbb{R}$.

The solutions are all purely imaginary numbers (including 0 when $y=0$).

Since $y$ can be any real number, there are infinitely many possible values for $y$.

Therefore, there are infinitely many solutions to the equation $z^2 + |z|^2 = 0$.

Match with the given options:

• (A) 1: Incorrect.
• (B) 2: Incorrect.
• (C) 3: Incorrect.
• (D) infinitely many: Correct.

The final answer is $\boxed{D}$.

Given Complex Number:

$z = \sin \frac{\pi}{5} + i \left(1 - \cos \frac{\pi}{5}\right)$

To Find:

The amplitude (principal argument) of the complex number $z$.

Solution:

Let $z = x + iy$, where $x = \sin \frac{\pi}{5}$ and $y = 1 - \cos \frac{\pi}{5}$.

Since $0 < \frac{\pi}{5} < \frac{\pi}{2}$, we have $\sin \frac{\pi}{5} > 0$ and $\cos \frac{\pi}{5} > 0$.

Also, $\cos \frac{\pi}{5} < 1$, so $1 - \cos \frac{\pi}{5} > 0$.

Thus, the real part $x > 0$ and the imaginary part $y > 0$. The complex number $z$ lies in the first quadrant.

The argument $\theta$ of a complex number $x+iy$ in the first quadrant is given by $\tan \theta = \frac{y}{x}$.

$\tan \theta = \frac{1 - \cos \frac{\pi}{5}}{\sin \frac{\pi}{5}}$

We use the half-angle trigonometric identities:

$1 - \cos A = 2 \sin^2 \frac{A}{2}$

$\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$

Let $A = \frac{\pi}{5}$. Then $\frac{A}{2} = \frac{\pi}{10}$.

Substitute these identities into the expression for $\tan \theta$:

$\tan \theta = \frac{2 \sin^2 \frac{\pi}{10}}{2 \sin \frac{\pi}{10} \cos \frac{\pi}{10}}$

Since $0 < \frac{\pi}{10} < \frac{\pi}{2}$, $\sin \frac{\pi}{10} \ne 0$. We can cancel $2 \sin \frac{\pi}{10}$ from the numerator and denominator:

$\tan \theta = \frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{10}}$

$\tan \theta = \tan \frac{\pi}{10}$

Since $z$ lies in the first quadrant, the principal value of the argument $\theta$ is $\frac{\pi}{10}$, as $0 < \frac{\pi}{10} < \frac{\pi}{2}$.

The amplitude of the complex number is $\frac{\pi}{10}$.

Compare this with the given options:

• (A) $\frac{2\pi}{5}$
• (B) $\frac{\pi}{5}$
• (C) $\frac{\pi}{15}$
• (D) $\frac{\pi}{10}$

Our calculated value matches option (D).

The final answer is $\boxed{D}$.

We are asked to find the value of $(1 - i)^n \left(1-\frac{1}{i}\right)^{n}$ for a positive integer $n$.

First, let's simplify the term $\frac{1}{i}$.

We know that $i^2 = -1$.

So, $\frac{1}{i} = \frac{1}{i} \times \frac{-i}{-i} = \frac{-i}{-i^2} = \frac{-i}{-(-1)} = \frac{-i}{1} = -i$.

Now, substitute this simplification into the given expression:

$(1 - i)^n \left(1-\frac{1}{i}\right)^{n} = (1 - i)^n (1 - (-i))^n$

$= (1 - i)^n (1 + i)^n$

Using the property $(ab)^n = a^n b^n$, we can write this as:

$= ((1 - i)(1 + i))^n$

Now, let's simplify the expression inside the parenthesis $(1 - i)(1 + i)$. This is in the form $(a-b)(a+b) = a^2 - b^2$.

$(1 - i)(1 + i) = (1)^2 - (i)^2 = 1 - (-1) = 1 + 1 = 2$.

Substitute this value back into the expression:

$((1 - i)(1 + i))^n = (2)^n$

Thus, the value of the given expression is $2^n$.

The final answer is $\mathbf{2^n}$.

We need to evaluate the sum $\sum\limits_{n=1}^{13} (i^n + i^{n+1})$.

The given sum can be written as:

$\sum\limits_{n=1}^{13} (i^n + i^{n+1}) = \sum\limits_{n=1}^{13} i^n + \sum\limits_{n=1}^{13} i^{n+1}$

Let's evaluate the first part of the sum, $\sum\limits_{n=1}^{13} i^n$:

$\sum\limits_{n=1}^{13} i^n = i^1 + i^2 + i^3 + i^4 + i^5 + i^6 + i^7 + i^8 + i^9 + i^{10} + i^{11} + i^{12} + i^{13}$

We know that the sum of four consecutive powers of $i$ is zero ($i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$).

We can group the terms in sets of four:

$(i^1 + i^2 + i^3 + i^4) + (i^5 + i^6 + i^7 + i^8) + (i^9 + i^{10} + i^{11} + i^{12}) + i^{13}$

Each group sums to 0:

$(0) + (0) + (0) + i^{13}$

Now, we evaluate $i^{13}$. We divide the exponent by 4 and look at the remainder.

$13 \div 4 = 3$ with a remainder of $1$.

So, $i^{13} = i^1 = i$.

Thus, $\sum\limits_{n=1}^{13} i^n = i$.

Now, let's evaluate the second part of the sum, $\sum\limits_{n=1}^{13} i^{n+1}$:

$\sum\limits_{n=1}^{13} i^{n+1} = i^{1+1} + i^{2+1} + ... + i^{13+1} = i^2 + i^3 + ... + i^{14}$

This sum has 13 terms.

$\sum\limits_{n=1}^{13} i^{n+1} = i^2 + i^3 + i^4 + i^5 + ... + i^{13} + i^{14}$

Again, we can group the terms in sets of four:

$(i^2 + i^3 + i^4 + i^5) + (i^6 + i^7 + i^8 + i^9) + (i^{10} + i^{11} + i^{12} + i^{13}) + i^{14}$

Each group sums to 0:

$(0) + (0) + (0) + i^{14}$

Now, we evaluate $i^{14}$. We divide the exponent by 4 and look at the remainder.

$14 \div 4 = 3$ with a remainder of $2$.

So, $i^{14} = i^2 = -1$.

Thus, $\sum\limits_{n=1}^{13} i^{n+1} = -1$.

Now, we add the two parts of the sum:

$\sum\limits_{n=1}^{13} (i^n + i^{n+1}) = \sum\limits_{n=1}^{13} i^n + \sum\limits_{n=1}^{13} i^{n+1} = i + (-1) = -1 + i$.

Alternatively, we can factor out $(1+i)$ from the original sum:

$\sum\limits_{n=1}^{13} (i^n + i^{n+1}) = \sum\limits_{n=1}^{13} i^n(1 + i) = (1+i) \sum\limits_{n=1}^{13} i^n$

We already found that $\sum\limits_{n=1}^{13} i^n = i$.

Substitute this back:

$(1+i) \sum\limits_{n=1}^{13} i^n = (1+i)(i) = i + i^2 = i + (-1) = -1 + i$.

The final answer is $\mathbf{-1 + i}$.

We are given the equation $\left( \frac{1+i}{1-i} \right)^{3} - \left( \frac{1-i}{1+i} \right)^{3} = x + iy$ and asked to find the values of $x$ and $y$.

First, let's simplify the complex fraction $\frac{1+i}{1-i}$. We multiply the numerator and denominator by the conjugate of the denominator, which is $1+i$.

$\frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)}$

Simplify the numerator and the denominator:

Numerator: $(1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i$

Denominator: $(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$

So, $\frac{1+i}{1-i} = \frac{2i}{2} = i$.

Now, let's simplify the second complex fraction $\frac{1-i}{1+i}$. This is the reciprocal of the first fraction.

$\frac{1-i}{1+i} = \frac{1}{\frac{1+i}{1-i}} = \frac{1}{i}$

We simplify $\frac{1}{i}$ by multiplying the numerator and denominator by $-i$ (or $i$). Using $-i$:

$\frac{1}{i} = \frac{1}{i} \times \frac{-i}{-i} = \frac{-i}{-i^2} = \frac{-i}{-(-1)} = \frac{-i}{1} = -i$.

Now substitute the simplified fractions back into the original equation:

$\left( \frac{1+i}{1-i} \right)^{3} - \left( \frac{1-i}{1+i} \right)^{3} = (i)^{3} - (-i)^{3}$

Evaluate the powers of $i$ and $-i$:

$i^3 = i^2 \times i = -1 \times i = -i$

$(-i)^3 = (-1 \times i)^3 = (-1)^3 \times i^3 = -1 \times (-i) = i$

Substitute these values back into the expression:

$(i)^{3} - (-i)^{3} = (-i) - (i) = -i - i = -2i$

We are given that the expression equals $x + iy$.

So, $-2i = x + iy$

To find $x$ and $y$, we compare the real and imaginary parts of the equation. The left side, $-2i$, can be written as $0 + (-2)i$.

$0 + (-2)i = x + iy$

Comparing the real parts: $x = 0$

Comparing the imaginary parts: $y = -2$

Therefore, $(x, y) = (0, -2)$.

The final answer is $\mathbf{(0, -2)}$.

We are given the equation $\frac{(1+i)^{2}}{2-i} = x + iy$ and asked to find the value of $x + y$.

First, we simplify the numerator of the complex fraction, $(1+i)^2$.

$(1+i)^2 = 1^2 + 2(1)(i) + i^2$

$(1+i)^2 = 1 + 2i + (-1)$

$(1+i)^2 = 1 + 2i - 1$

$(1+i)^2 = 2i$

Now substitute the simplified numerator back into the fraction:

$\frac{(1+i)^{2}}{2-i} = \frac{2i}{2-i}$

To express this fraction in the form $x+iy$, we multiply the numerator and the denominator by the conjugate of the denominator. The denominator is $2-i$, and its conjugate is $2+i$.

$\frac{2i}{2-i} = \frac{2i}{2-i} \times \frac{2+i}{2+i}$

Multiply the numerators:

$2i(2+i) = 2i(2) + 2i(i) = 4i + 2i^2 = 4i + 2(-1) = 4i - 2 = -2 + 4i$

Multiply the denominators using the formula $(a-b)(a+b) = a^2 - b^2$:

$(2-i)(2+i) = 2^2 - i^2 = 4 - (-1) = 4 + 1 = 5$

Now, combine the simplified numerator and denominator:

$\frac{-2 + 4i}{5} = -\frac{2}{5} + \frac{4}{5}i$

We are given that $\frac{(1+i)^{2}}{2-i} = x + iy$.

So, $-\frac{2}{5} + \frac{4}{5}i = x + iy$

By comparing the real and imaginary parts of the equation, we can find the values of $x$ and $y$.

$x = -\frac{2}{5}$

$y = \frac{4}{5}$

Finally, we need to find the value of $x + y$.

$x + y = -\frac{2}{5} + \frac{4}{5}$

$x + y = \frac{-2 + 4}{5}$

$x + y = \frac{2}{5}$

The final answer is $\mathbf{\frac{2}{5}}$.

We are given the equation $\left( \frac{1+i}{1-i} \right)^{100} = a + ib$ and asked to find the values of $a$ and $b$, which form the ordered pair $(a, b)$.

First, we simplify the complex fraction inside the parenthesis, $\frac{1+i}{1-i}$.

We multiply the numerator and the denominator by the conjugate of the denominator, which is $1+i$.

$\frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)}$

Let's simplify the numerator:

$(1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i + (-1) = 1 + 2i - 1 = 2i$

Let's simplify the denominator using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$:

$(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$

So, the simplified fraction is:

$\frac{1+i}{1-i} = \frac{2i}{2} = i$

Now, we need to raise this simplified expression to the power of 100:

$\left( \frac{1+i}{1-i} \right)^{100} = (i)^{100}$

To evaluate $i^{100}$, we use the property that the powers of $i$ repeat in a cycle of 4: $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$.

We divide the exponent, 100, by 4.

$100 \div 4 = 25$ with a remainder of $0$.

Since the remainder is $0$, $i^{100}$ is equivalent to $i^4$, which is $1$.

$i^{100} = (i^4)^{25} = (1)^{25} = 1$

So, we have:

$\left( \frac{1+i}{1-i} \right)^{100} = 1$

We are given that $\left( \frac{1+i}{1-i} \right)^{100} = a + ib$.

Therefore, $1 = a + ib$.

To find $a$ and $b$, we compare the real and imaginary parts of the equation. We can write $1$ as a complex number $1 + 0i$.

$1 + 0i = a + ib$

Comparing the real parts, we get $a = 1$.

Comparing the imaginary parts, we get $b = 0$.

Thus, the ordered pair $(a, b)$ is $(1, 0)$.

The final answer is $\mathbf{(1, 0)}$.

We are given $a = \cos \theta + i \sin \theta$ and asked to find the value of $\frac{1+a}{1-a}$.

Substitute the expression for $a$ into the given expression:

$\frac{1+a}{1-a} = \frac{1 + (\cos \theta + i \sin \theta)}{1 - (\cos \theta + i \sin \theta)}$

$= \frac{(1 + \cos \theta) + i \sin \theta}{(1 - \cos \theta) - i \sin \theta}$

To simplify this complex fraction, multiply the numerator and the denominator by the conjugate of the denominator. The denominator is $(1 - \cos \theta) - i \sin \theta$, and its conjugate is $(1 - \cos \theta) + i \sin \theta$.

$\frac{(1 + \cos \theta) + i \sin \theta}{(1 - \cos \theta) - i \sin \theta} \times \frac{(1 - \cos \theta) + i \sin \theta}{(1 - \cos \theta) + i \sin \theta}$

Calculate the numerator:

Numerator $= ((1 + \cos \theta) + i \sin \theta)((1 - \cos \theta) + i \sin \theta)$

$= (1 + \cos \theta)(1 - \cos \theta) + (1 + \cos \theta)(i \sin \theta) + (i \sin \theta)(1 - \cos \theta) + (i \sin \theta)(i \sin \theta)$

$= (1^2 - \cos^2 \theta) + i \sin \theta (1 + \cos \theta + 1 - \cos \theta) + i^2 \sin^2 \theta$

$= (1 - \cos^2 \theta) + i \sin \theta (2) - \sin^2 \theta$

Using the identity $1 - \cos^2 \theta = \sin^2 \theta$ and $i^2 = -1$:

$= \sin^2 \theta + 2i \sin \theta - \sin^2 \theta$

$= 2i \sin \theta$

Calculate the denominator using the formula $(X - Yi)(X + Yi) = X^2 + Y^2$, where $X = 1 - \cos \theta$ and $Y = \sin \theta$:

Denominator $= ((1 - \cos \theta) - i \sin \theta)((1 - \cos \theta) + i \sin \theta)$

$= (1 - \cos \theta)^2 + (\sin \theta)^2$

$= (1 - 2 \cos \theta + \cos^2 \theta) + \sin^2 \theta$

Using the identity $\cos^2 \theta + \sin^2 \theta = 1$:

$= 1 - 2 \cos \theta + 1$

$= 2 - 2 \cos \theta$

Now, form the simplified fraction:

$\frac{1+a}{1-a} = \frac{2i \sin \theta}{2 - 2 \cos \theta}$

$= \frac{2i \sin \theta}{2(1 - \cos \theta)}$

$= \frac{i \sin \theta}{1 - \cos \theta}$

We can simplify this further using half-angle trigonometric identities:

$\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$

$1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$

Substitute these identities into the expression:

$\frac{i (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})}{2 \sin^2 \frac{\theta}{2}}$

$= \frac{i \cancel{2} \cancel{\sin \frac{\theta}{2}} \cos \frac{\theta}{2}}{\cancel{2} \cancel{\sin \frac{\theta}{2}} \sin \frac{\theta}{2}}$

$= i \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}$

Using the identity $\frac{\cos x}{\sin x} = \cot x$:

$= i \cot \frac{\theta}{2}$

Note that this expression is defined provided the denominator $1 - \cos \theta - i \sin \theta$ is not zero, which means $a \neq 1$. This occurs when $\theta$ is not a multiple of $2\pi$ (i.e., $\theta \neq 2k\pi$ for any integer $k$), which ensures $\sin \frac{\theta}{2} \neq 0$.

The final answer is $\mathbf{i \cot \frac{\theta}{2}}$.

Given:

$(1 + i) z = (1 - i) \overline{z}$

To Show:

$z = - i \overline{z}$

Proof:

Start with the given equation:

To isolate $z$, we can divide both sides of the equation by $(1 + i)$, assuming $1+i \neq 0$, which is true.

$z = \frac{1 - i}{1 + i} \overline{z}$

Now, let's simplify the complex fraction $\frac{1 - i}{1 + i}$. We multiply the numerator and the denominator by the conjugate of the denominator, which is $1-i$.

$\frac{1 - i}{1 + i} = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(1 - i)^2}{(1 + i)(1 - i)}$

Simplify the numerator:

$(1 - i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i + (-1) = 1 - 2i - 1 = -2i$

Simplify the denominator using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$:

$(1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$

So, the simplified fraction is:

$\frac{1 - i}{1 + i} = \frac{-2i}{2} = -i$

Substitute this simplified value back into the equation for $z$:

$z = (-i) \overline{z}$

$z = -i \overline{z}$

This is the desired result.

Hence, shown that $\mathbf{z = - i \overline{z}}$.

We are given the equation $z\overline{z} + 2(z + \overline{z}) + b = 0$, where $z = x + iy$ and $b$ is a real number ($b \in \mathbb{R}$). We need to show that this equation represents a circle.

Let $z = x + iy$. Then the conjugate of $z$ is $\overline{z} = x - iy$.

First, let's evaluate the terms $z\overline{z}$ and $z + \overline{z}$ in terms of $x$ and $y$.

$z\overline{z} = (x + iy)(x - iy)$

Using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$:

$z\overline{z} = x^2 - (iy)^2 = x^2 - i^2y^2 = x^2 - (-1)y^2 = x^2 + y^2$

$z + \overline{z} = (x + iy) + (x - iy)$

$z + \overline{z} = x + iy + x - iy = 2x$

Now, substitute these expressions for $z\overline{z}$ and $z + \overline{z}$ into the given equation:

$(x^2 + y^2) + 2(2x) + b = 0$

$x^2 + y^2 + 4x + b = 0$

To determine if this equation represents a circle, we can rewrite it in the standard form of a circle's equation: $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

We complete the square for the $x$ terms.

$x^2 + 4x + y^2 + b = 0$

Group the $x$ terms: $(x^2 + 4x) + y^2 + b = 0$

To complete the square for $x^2 + 4x$, we add and subtract $(\frac{4}{2})^2 = 2^2 = 4$:

$(x^2 + 4x + 4) - 4 + y^2 + b = 0$

Rewrite the perfect square trinomial $(x^2 + 4x + 4)$ as $(x+2)^2$:

$(x+2)^2 + y^2 + b - 4 = 0$

Move the constant terms to the right side of the equation:

$(x+2)^2 + y^2 = 4 - b$

This equation is in the standard form $(x-h)^2 + (y-k)^2 = r^2$, where:

$h = -2$

$k = 0$

$r^2 = 4 - b$

For this equation to represent a circle, the radius squared, $r^2$, must be positive. That is, $4 - b > 0$.

If $4 - b > 0$ (i.e., $b < 4$), the equation represents a circle with center $(-2, 0)$ and radius $r = \sqrt{4-b}$.

If $4 - b = 0$ (i.e., $b = 4$), the equation becomes $(x+2)^2 + y^2 = 0$, which represents a single point $(-2, 0)$.

If $4 - b < 0$ (i.e., $b > 4$), the equation becomes $(x+2)^2 + y^2 = \text{negative number}$, which has no real solutions for $x$ and $y$, and thus represents no locus in the real plane.

Since there exists a condition on $b$ (namely $b < 4$) for which the equation represents a circle, the given equation $z\overline{z} + 2(z + \overline{z}) + b = 0$ represents a circle (or a point or no locus) depending on the value of $b$. However, the question asks to show it represents a circle, implying that it has the form of a circle's equation when interpreted geometrically in the complex plane as the real plane using $z=x+iy$. The derivation shows it takes the form $(x+2)^2 + y^2 = 4-b$, which is the standard form of a circle if $4-b > 0$.

Thus, the equation $\mathbf{z\overline{z} + 2(z + \overline{z}) + b = 0}$ represents a circle with center $\mathbf{(-2, 0)}$ and radius $\mathbf{\sqrt{4-b}}$, provided $\mathbf{b < 4}$.

Given:

The real part of $\frac{\overline{z}+2}{\overline{z}-1}$ is equal to 4.

$\text{Re}\left(\frac{\overline{z}+2}{\overline{z}-1}\right) = 4$

To Show:

The locus of the point representing $z$ in the complex plane is a circle.

Solution:

Let $z = x + iy$, where $x$ and $y$ are real numbers. The point representing $z$ in the complex plane is $(x, y)$.

The conjugate of $z$ is $\overline{z} = x - iy$.

Now, substitute $\overline{z}$ into the given complex expression:

$\frac{\overline{z}+2}{\overline{z}-1} = \frac{(x-iy)+2}{(x-iy)-1}$

$= \frac{(x+2)-iy}{(x-1)-iy}$

To find the real part of this complex fraction, we multiply the numerator and the denominator by the conjugate of the denominator. The denominator is $(x-1)-iy$, and its conjugate is $(x-1)+iy$.

$\frac{(x+2)-iy}{(x-1)-iy} = \frac{(x+2)-iy}{(x-1)-iy} \times \frac{(x-1)+iy}{(x-1)+iy}$

Calculate the numerator:

Numerator $= ((x+2)-iy)((x-1)+iy)$

$= (x+2)(x-1) + (x+2)(iy) - iy(x-1) - iy(iy)$

$= (x^2 - x + 2x - 2) + i(x+2) - i(x-1) - i^2y^2$

$= (x^2 + x - 2) + i(x+2 - x+1) + y^2$

$= (x^2 + y^2 + x - 2) + i(3)$

Calculate the denominator:

Denominator $= ((x-1)-iy)((x-1)+iy)$

$= (x-1)^2 + y^2$

$= x^2 - 2x + 1 + y^2$

So, the complex fraction is:

$\frac{\overline{z}+2}{\overline{z}-1} = \frac{(x^2 + y^2 + x - 2) + i(3)}{x^2 + y^2 - 2x + 1}$

$= \frac{x^2 + y^2 + x - 2}{x^2 + y^2 - 2x + 1} + i \frac{3}{x^2 + y^2 - 2x + 1}$

The real part of this expression is $\frac{x^2 + y^2 + x - 2}{x^2 + y^2 - 2x + 1}$.

We are given that the real part is 4:

$\frac{x^2 + y^2 + x - 2}{x^2 + y^2 - 2x + 1} = 4$

For the expression to be defined, the denominator cannot be zero, so $x^2 + y^2 - 2x + 1 \neq 0$. Note that $x^2 - 2x + 1 + y^2 = (x-1)^2 + y^2$. This is zero only if $x=1$ and $y=0$, which corresponds to the point $\overline{z} = 1$ or $z=1$. So, $z \neq 1$.

Multiply both sides by the denominator:

$x^2 + y^2 + x - 2 = 4(x^2 + y^2 - 2x + 1)$

$x^2 + y^2 + x - 2 = 4x^2 + 4y^2 - 8x + 4$

Move all terms to one side to obtain the general form of a conic section equation:

$0 = 4x^2 - x^2 + 4y^2 - y^2 - 8x - x + 4 + 2$

$0 = 3x^2 + 3y^2 - 9x + 6$

Divide the entire equation by 3:

$x^2 + y^2 - 3x + 2 = 0$

This equation is in the form $x^2 + y^2 + Dx + Ey + F = 0$, where $D = -3$, $E = 0$, and $F = 2$. This is the general equation of a circle.

To confirm it's a circle and find its properties, we can complete the square:

$(x^2 - 3x) + y^2 + 2 = 0$

$(x^2 - 3x + (-\frac{3}{2})^2) - (-\frac{3}{2})^2 + y^2 + 2 = 0$

$(x - \frac{3}{2})^2 - \frac{9}{4} + y^2 + 2 = 0$

$(x - \frac{3}{2})^2 + y^2 = \frac{9}{4} - 2$

$(x - \frac{3}{2})^2 + y^2 = \frac{9 - 8}{4}$

$(x - \frac{3}{2})^2 + y^2 = \frac{1}{4}$

This is the standard equation of a circle $(x-h)^2 + (y-k)^2 = r^2$ with center $(h, k) = (\frac{3}{2}, 0)$ and radius $r = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

Since the equation derived from the given condition is the equation of a circle in the $(x, y)$ plane, the locus of the point representing $z = x + iy$ is a circle.

Note that the point $(1, 0)$ is excluded from the locus because the denominator $\overline{z}-1$ cannot be zero. Evaluating the equation $x^2 + y^2 - 3x + 2 = 0$ at $(1, 0)$ gives $1^2 + 0^2 - 3(1) + 2 = 1 - 3 + 2 = 0$. So the point $(1,0)$ lies on the derived circle. Therefore, the locus is precisely this circle with the point $(1,0)$ removed.

Hence, shown that the locus of the point representing z is a circle (excluding the point (1,0)).

Given:

The complex number $z$ satisfies the condition $\arg \left( \frac{z-1}{z+1} \right) = \frac{\pi}{4}$.

To Show:

The locus of the point representing $z$ in the complex plane is a circle.

Solution:

Let $z = x + iy$, where $x$ and $y$ are real numbers. The point representing $z$ in the complex plane is $(x, y)$.

We first simplify the complex expression $\frac{z-1}{z+1}$ by substituting $z = x+iy$.

$\frac{z-1}{z+1} = \frac{(x+iy)-1}{(x+iy)+1} = \frac{(x-1)+iy}{(x+1)+iy}$

To express this fraction in the form $A+iB$, we multiply the numerator and denominator by the conjugate of the denominator. The denominator is $(x+1)+iy$, and its conjugate is $(x+1)-iy$.

$\frac{(x-1)+iy}{(x+1)+iy} = \frac{(x-1)+iy}{(x+1)+iy} \times \frac{(x+1)-iy}{(x+1)-iy}$

Calculate the numerator:

Numerator $= ((x-1)+iy)((x+1)-iy) = (x-1)(x+1) - (x-1)(iy) + iy(x+1) - (iy)^2$

$= (x^2 - 1) - ixy + iy + ixy + iy - i^2y^2$

$= (x^2 - 1) + i(-xy + y + xy + y) - (-1)y^2$

$= (x^2 - 1 + y^2) + i(2y)$

Calculate the denominator:

Denominator $= ((x+1)+iy)((x+1)-iy) = (x+1)^2 - (iy)^2$

$= (x+1)^2 - i^2y^2 = (x^2 + 2x + 1) - (-1)y^2$

$= x^2 + 2x + 1 + y^2$

So, the complex fraction is:

$\frac{z-1}{z+1} = \frac{(x^2 + y^2 - 1) + i(2y)}{x^2 + y^2 + 2x + 1}$

$= \frac{x^2 + y^2 - 1}{x^2 + y^2 + 2x + 1} + i \frac{2y}{x^2 + y^2 + 2x + 1}$

Let $A = \frac{x^2 + y^2 - 1}{x^2 + y^2 + 2x + 1}$ and $B = \frac{2y}{x^2 + y^2 + 2x + 1}$. The expression is in the form $A+iB$.

We are given that $\arg \left( \frac{z-1}{z+1} \right) = \frac{\pi}{4}$. The argument of a complex number $A+iB$ is $\frac{\pi}{4}$ if and only if the complex number lies on the ray originating from the origin that makes an angle of $\frac{\pi}{4}$ with the positive real axis (excluding the origin itself). For this to be true, the real part $A$ and the imaginary part $B$ must be equal and positive. Thus, $A = B$ and $A > 0$, $B > 0$.

From the condition $A = B$, we have:

$\frac{x^2 + y^2 - 1}{x^2 + y^2 + 2x + 1} = \frac{2y}{x^2 + y^2 + 2x + 1}$

The denominator $x^2 + y^2 + 2x + 1 = (x+1)^2 + y^2$ is zero only if $x=-1$ and $y=0$, which corresponds to $z = -1$. The expression $\frac{z-1}{z+1}$ is undefined at $z=-1$. Assuming $z \neq -1$, the denominator is non-zero, and since it is a square sum, it is positive. We can multiply both sides by the denominator:

$x^2 + y^2 - 1 = 2y$

Rearrange the equation:

$x^2 + y^2 - 2y - 1 = 0$

To show this represents a circle, we complete the square for the $y$ terms:

$x^2 + (y^2 - 2y) - 1 = 0$

$x^2 + (y^2 - 2y + (-1)^2) - (-1)^2 - 1 = 0$

$x^2 + (y-1)^2 - 1 - 1 = 0$

$x^2 + (y-1)^2 = 2$

This is the standard form of the equation of a circle $(x-h)^2 + (y-k)^2 = r^2$, with center $(h, k) = (0, 1)$ and radius $r = \sqrt{2}$.

The conditions $A > 0$ and $B > 0$ imply $\frac{x^2 + y^2 - 1}{x^2 + y^2 + 2x + 1} > 0$ and $\frac{2y}{x^2 + y^2 + 2x + 1} > 0$. Since the denominator is positive for $z \neq -1$, these conditions are equivalent to $x^2 + y^2 - 1 > 0$ and $y > 0$. For points on the circle $x^2 + (y-1)^2 = 2$, we have $x^2 + y^2 - 2y + 1 = 2$, which means $x^2 + y^2 = 2y+1$. Substituting this into $x^2 + y^2 - 1 > 0$, we get $(2y+1) - 1 > 0$, which simplifies to $2y > 0$, or $y > 0$. Thus, for points on the derived circle (where $z \neq -1$), the conditions $A>0$ and $B>0$ are equivalent to $y > 0$. This means the locus is the part of the circle $x^2 + (y-1)^2 = 2$ for which $y > 0$, excluding the points where $y=0$ (i.e., $(1,0)$ and $(-1,0)$ where the expression is undefined). This describes an arc of the circle.

Since the equation derived from the given condition is the equation of a circle, the locus of the point representing $z$ satisfying the condition lies on this circle.

Hence, shown that the complex number z satisfying the condition lies on a circle.

We are asked to solve the equation $|z| = z + 1 + 2i$ for the complex number $z$.

Let the complex number $z$ be represented as $z = x + iy$, where $x$ and $y$ are real numbers.

The modulus of $z$ is given by $|z| = \sqrt{x^2 + y^2}$.

Substitute $z = x + iy$ into the given equation:

$\sqrt{x^2 + y^2} = (x + iy) + 1 + 2i$

Group the real and imaginary parts on the right side:

$\sqrt{x^2 + y^2} = (x + 1) + i(y + 2)$

Since the left side of the equation, $\sqrt{x^2 + y^2}$, is a real number, the imaginary part of the right side must be equal to zero. Also, the real parts must be equal.

Equating the imaginary parts:

$y + 2 = 0$

Equating the real parts:

$\sqrt{x^2 + y^2} = x + 1$

Substitute the value of $y$ from (i) into this equation:

$\sqrt{x^2 + (-2)^2} = x + 1$

$\sqrt{x^2 + 4} = x + 1$

For the square root to be equal to $x+1$, $x+1$ must be non-negative. So, $x+1 \ge 0$, which implies $x \ge -1$.

To solve for $x$, square both sides of the equation:

$(\sqrt{x^2 + 4})^2 = (x + 1)^2$

$x^2 + 4 = x^2 + 2x + 1$

Subtract $x^2$ from both sides:

$4 = 2x + 1$

Subtract 1 from both sides:

$4 - 1 = 2x$

$3 = 2x$

Divide by 2:

$x = \frac{3}{2}$

We must check if this value of $x$ satisfies the condition $x \ge -1$. Since $\frac{3}{2} = 1.5$, which is greater than or equal to $-1$, the condition is satisfied.

Now we have the values for $x$ and $y$: $x = \frac{3}{2}$ and $y = -2$.

Substitute these values back into $z = x + iy$:

$z = \frac{3}{2} + i(-2)$

$z = \frac{3}{2} - 2i$

Let's verify this solution in the original equation $|z| = z + 1 + 2i$.

LHS: $|z| = \left|\frac{3}{2} - 2i\right| = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{9 + 16}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$

RHS: $z + 1 + 2i = \left(\frac{3}{2} - 2i\right) + 1 + 2i = \left(\frac{3}{2} + 1\right) + (-2i + 2i) = \left(\frac{3}{2} + \frac{2}{2}\right) + 0i = \frac{5}{2}$

Since LHS = RHS, the solution is correct.

The final answer is $\mathbf{z = \frac{3}{2} - 2i}$.

We are asked to find the complex number $z$ that satisfies the equation $|z+1| = z + 2 (1 + i)$.

Let $z = x + iy$, where $x$ and $y$ are real numbers.

Substitute $z = x+iy$ into the equation:

$|(x+iy) + 1| = (x+iy) + 2(1+i)$

$|(x+1) + iy| = x + iy + 2 + 2i$

$|(x+1) + iy| = (x+2) + i(y+2)$

The left side, $|(x+1) + iy|$, is the modulus of the complex number $(x+1) + iy$, which is a real number.

$|(x+1) + iy| = \sqrt{(x+1)^2 + y^2}$

So, the equation becomes:

$\sqrt{(x+1)^2 + y^2} = (x+2) + i(y+2)$

For a complex number to be equal to a real number (which is the left side), its imaginary part must be zero.

Equating the imaginary parts of both sides:

$y+2 = 0$

Equating the real parts of both sides:

$\sqrt{(x+1)^2 + y^2} = x+2$

Substitute the value of $y$ from (i) into this equation:

$\sqrt{(x+1)^2 + (-2)^2} = x+2$

$\sqrt{(x+1)^2 + 4} = x+2$

For the square root to be equal to $x+2$, the right side must be non-negative:

This implies $x \ge -2$.

Square both sides of the equation $\sqrt{(x+1)^2 + 4} = x+2$ to solve for $x$:

$(\sqrt{(x+1)^2 + 4})^2 = (x+2)^2$

$(x+1)^2 + 4 = (x+2)^2$

Expand both sides:

$(x^2 + 2x + 1) + 4 = x^2 + 4x + 4$

$x^2 + 2x + 5 = x^2 + 4x + 4$

Subtract $x^2$ from both sides:

$2x + 5 = 4x + 4$

Rearrange the terms to solve for $x$:

$5 - 4 = 4x - 2x$

$1 = 2x$

$x = \frac{1}{2}$

Check if this value of $x$ satisfies the condition (ii): $x \ge -2$. Since $\frac{1}{2} \ge -2$, the condition is satisfied.

We have found $x = \frac{1}{2}$ and $y = -2$. Substitute these values back into $z = x + iy$:

$z = \frac{1}{2} + i(-2)$

$z = \frac{1}{2} - 2i$

The final answer is $\mathbf{z = \frac{1}{2} - 2i}$.

We are given the condition $\arg (z - 1) = \arg (z + 3i)$, where $z = x + iy$. We need to find the ratio $(x - 1) : y$.

Substitute $z = x + iy$ into the arguments:

$z - 1 = (x + iy) - 1 = (x - 1) + iy$

$z + 3i = (x + iy) + 3i = x + i(y + 3)$

The argument of a complex number $a + ib$ is given by $\arg(a+ib) = \tan^{-1}\left(\frac{b}{a}\right)$, considering the quadrant of the complex number.

So, $\arg(z - 1) = \arg((x - 1) + iy) = \tan^{-1}\left(\frac{y}{x - 1}\right)$, provided $x-1 \neq 0$.

And $\arg(z + 3i) = \arg(x + i(y + 3)) = \tan^{-1}\left(\frac{y + 3}{x}\right)$, provided $x \neq 0$.

Given $\arg (z - 1) = \arg (z + 3i)$, we have:

$\tan^{-1}\left(\frac{y}{x - 1}\right) = \tan^{-1}\left(\frac{y + 3}{x}\right)$

Taking the tangent of both sides (and considering the implications on the range of the angles):

$\frac{y}{x - 1} = \frac{y + 3}{x}$

For this equality to hold, the denominators must be non-zero, i.e., $x - 1 \neq 0$ (so $x \neq 1$) and $x \neq 0$. Also, if $y=0$, then the first argument is $\arg(x-1)$, and if $y+3=0$ (i.e., $y=-3$), then the second argument is $\arg(x)$. If $y=0$ and $y+3=0$, this is impossible. If $y=0$ and $x \neq 1$, $\arg(x-1)$ is $0$ if $x-1>0$ and $\pi$ if $x-1<0$. If $y=-3$ and $x \neq 0$, $\arg(x)$ is $0$ if $x>0$ and $\pi$ if $x<0$. For the arguments to be equal and non-zero, neither $y=0$ nor $y=-3$ can hold unless $x-1$ and $x$ have the same sign, leading to both arguments being 0 or both being $\pi$. If $y=0$ and $y+3=0$ were allowed in the $\tan^{-1}$ formula, it would imply $0 = 3/x$, which means $3=0$, a contradiction. Thus, we can proceed assuming $y \neq 0$ and $y \neq -3$.

Cross-multiply:

$y \cdot x = (y + 3) \cdot (x - 1)$

$xy = y(x - 1) + 3(x - 1)$

$xy = xy - y + 3x - 3$

Subtract $xy$ from both sides:

$0 = -y + 3x - 3$

Rearrange the terms to find the relationship between $x$ and $y$:

$y = 3x - 3$

$y = 3(x - 1)$

We are asked to find the ratio $(x - 1) : y$. From the equation $y = 3(x - 1)$, we can rearrange it to find the ratio:

If $y \neq 0$ and $x - 1 \neq 0$, we can divide both sides by $y$ and by 3:

$\frac{y}{y} = \frac{3(x - 1)}{y}$

$1 = 3 \frac{x - 1}{y}$

$\frac{1}{3} = \frac{x - 1}{y}$

So the ratio $\frac{x-1}{y} = \frac{1}{3}$.

The ratio $(x - 1) : y$ is $1 : 3$.

Geometrically, the condition $\arg(z-1) = \arg(z+3i)$ means that the angle made by the line segment from $(1, 0)$ to $(x, y)$ with the positive real axis is the same as the angle made by the line segment from $(0, -3)$ to $(x, y)$ with the positive real axis. This occurs if the points $(1, 0)$, $(0, -3)$, and $(x, y)$ are collinear. The equation of the line passing through $(1, 0)$ and $(0, -3)$ is $\frac{y - 0}{x - 1} = \frac{-3 - 0}{0 - 1} = \frac{-3}{-1} = 3$. So, $\frac{y}{x-1} = 3$, which gives $y = 3(x-1)$, the same equation we found. The locus is the line $y = 3x - 3$, excluding the points $z=1$ and $z=-3i$ where the arguments are undefined.

The ratio $(x-1) : y$ for any point on this line (except where $y=0$ or $x-1=0$) is $\frac{x-1}{y} = \frac{x-1}{3(x-1)} = \frac{1}{3}$.

The final answer is $\mathbf{1 : 3}$.

Given:

The condition $\left| \frac{z-2}{z-3} \right| = 2$.

To Show:

The locus of $z$ satisfying the condition is a circle, and find its center and radius.

Solution:

The given condition is $\left| \frac{z-2}{z-3} \right| = 2$.

Using the property $|z_1 / z_2| = |z_1| / |z_2|$, we can write:

$\frac{|z-2|}{|z-3|} = 2$

Multiply both sides by $|z-3|$, assuming $z \neq 3$ (so $|z-3| \neq 0$):

$|z-2| = 2 |z-3|$

Let $z = x + iy$, where $x$ and $y$ are real numbers. Substitute this into the equation.

$| (x+iy) - 2 | = 2 | (x+iy) - 3 |$

$| (x-2) + iy | = 2 | (x-3) + iy |$

The modulus of a complex number $a+ib$ is $\sqrt{a^2+b^2}$. Apply this to both sides:

$\sqrt{(x-2)^2 + y^2} = 2 \sqrt{(x-3)^2 + y^2}$

Square both sides of the equation to eliminate the square roots:

$(\sqrt{(x-2)^2 + y^2})^2 = (2 \sqrt{(x-3)^2 + y^2})^2$

$(x-2)^2 + y^2 = 4 ((x-3)^2 + y^2)$

Expand the squared terms:

$(x^2 - 4x + 4) + y^2 = 4 (x^2 - 6x + 9 + y^2)$

$x^2 - 4x + 4 + y^2 = 4x^2 - 24x + 36 + 4y^2$

Move all terms to one side to obtain the general form of a conic section equation:

$0 = 4x^2 - x^2 + 4y^2 - y^2 - 24x + 4x + 36 - 4$

$0 = 3x^2 + 3y^2 - 20x + 32$

Divide the entire equation by 3:

$x^2 + y^2 - \frac{20}{3}x + \frac{32}{3} = 0$

This equation is in the general form of a circle $x^2 + y^2 + Dx + Ey + F = 0$, where $D = -\frac{20}{3}$, $E = 0$, and $F = \frac{32}{3}$.

To find the center and radius, we complete the square for the $x$ and $y$ terms.

$(x^2 - \frac{20}{3}x) + y^2 + \frac{32}{3} = 0$

Complete the square for the $x$ terms by adding and subtracting $(\frac{-\frac{20}{3}}{2})^2 = (-\frac{10}{3})^2 = \frac{100}{9}$:

$(x^2 - \frac{20}{3}x + \frac{100}{9}) - \frac{100}{9} + y^2 + \frac{32}{3} = 0$

Rewrite the perfect square trinomial $(x^2 - \frac{20}{3}x + \frac{100}{9})$ as $(x - \frac{10}{3})^2$:

$(x - \frac{10}{3})^2 + y^2 - \frac{100}{9} + \frac{32 \times 3}{3 \times 3} = 0$

$(x - \frac{10}{3})^2 + y^2 - \frac{100}{9} + \frac{96}{9} = 0$

$(x - \frac{10}{3})^2 + y^2 - \frac{4}{9} = 0$

Move the constant term to the right side:

$(x - \frac{10}{3})^2 + y^2 = \frac{4}{9}$

This equation is in the standard form of a circle $(x-h)^2 + (y-k)^2 = r^2$, where:

The center $(h, k)$ is $(\frac{10}{3}, 0)$.

The radius squared $r^2$ is $\frac{4}{9}$. So the radius $r = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

Since the derived equation is the equation of a circle, the locus of the point representing $z$ is a circle.

The point $z=3$ was excluded initially because $|z-3|$ is in the denominator. The equation of the derived circle is $(x - \frac{10}{3})^2 + y^2 = \frac{4}{9}$. If $z=3$, then $x=3, y=0$. Plugging into the circle equation: $(3 - \frac{10}{3})^2 + 0^2 = (\frac{9}{3} - \frac{10}{3})^2 = (-\frac{1}{3})^2 = \frac{1}{9}$. Since $\frac{1}{9} \neq \frac{4}{9}$, the point $(3,0)$ does not lie on this circle, which is consistent with the initial assumption $z \neq 3$.

Hence, shown that $\mathbf{\left| \frac{z-2}{z-3} \right| = 2}$ represents a circle.

The centre of the circle is $\mathbf{\left(\frac{10}{3}, 0\right)}$.

The radius of the circle is $\mathbf{\frac{2}{3}}$.

We are given that the complex number $\frac{z-1}{z+1}$ is purely imaginary, where $z \neq -1$. We need to find the value of $|z|$.

A complex number is purely imaginary if its real part is zero and its imaginary part is non-zero. Also, a complex number $w$ is purely imaginary if and only if $w = - \overline{w}$.

Let $w = \frac{z-1}{z+1}$. Since $w$ is purely imaginary, we have $w = - \overline{w}$.

$\frac{z-1}{z+1} = - \overline{\left(\frac{z-1}{z+1}\right)}$

Using the property $\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}$, we have:

$\frac{z-1}{z+1} = - \frac{\overline{z-1}}{\overline{z+1}}$

Using the property $\overline{z_1 \pm z_2} = \overline{z_1} \pm \overline{z_2}$, we have $\overline{z-1} = \overline{z} - \overline{1} = \overline{z} - 1$ and $\overline{z+1} = \overline{z} + \overline{1} = \overline{z} + 1$.

So, the equation becomes:

$\frac{z-1}{z+1} = - \frac{\overline{z}-1}{\overline{z}+1}$

Multiply both sides by $(z+1)(\overline{z}+1)$, assuming $z \neq -1$ and $\overline{z} \neq -1$ (which is equivalent to $z \neq -1$):

$(z-1)(\overline{z}+1) = - (\overline{z}-1)(z+1)$

Expand both sides:

$z\overline{z} + z - \overline{z} - 1 = - (z\overline{z} + \overline{z} - z - 1)$

$z\overline{z} + z - \overline{z} - 1 = - z\overline{z} - \overline{z} + z + 1$

Move all terms to one side:

$z\overline{z} + z - \overline{z} - 1 + z\overline{z} + \overline{z} - z - 1 = 0$

Combine like terms:

$(z\overline{z} + z\overline{z}) + (z - z) + (-\overline{z} + \overline{z}) + (-1 - 1) = 0$

$2z\overline{z} + 0 + 0 - 2 = 0$

$2z\overline{z} - 2 = 0$

Add 2 to both sides:

$2z\overline{z} = 2$

Divide by 2:

$z\overline{z} = 1$

Recall that $z\overline{z} = |z|^2$.

So, $|z|^2 = 1$.

Since $|z|$ is a non-negative real number, taking the square root of both sides gives:

$|z| = \sqrt{1}$

$|z| = 1$

This result holds provided the imaginary part of $\frac{z-1}{z+1}$ is non-zero. If $\frac{z-1}{z+1}$ is $0 + 0i$, then $z-1=0$, so $z=1$. If $z=1$, then $|z|=1$. The expression $\frac{z-1}{z+1} = \frac{1-1}{1+1} = \frac{0}{2} = 0$, which has a real part of 0 and an imaginary part of 0. A purely imaginary number is generally considered to have a non-zero imaginary part. If we strictly require a non-zero imaginary part, then $z=1$ is excluded from the locus. However, $|z|=1$ is the general condition derived from the real part being zero. The locus of points $z$ such that $|z|=1$ is a circle centered at the origin with radius 1.

The condition that $\frac{z-1}{z+1}$ is purely imaginary corresponds to the locus of points $z$ such that $z$ lies on the circle with diameter connecting the points $(1, 0)$ and $(-1, 0)$, excluding these two points. The equation of this circle is $x^2 + y^2 = 1$, which is $|z|=1$. The points $(1,0)$ and $(-1,0)$ correspond to $z=1$ and $z=-1$. $z=-1$ is already excluded. If $z=1$, $\frac{z-1}{z+1} = 0$, which has an imaginary part of 0. Thus, the locus is the circle $|z|=1$ excluding the point $z=1$. However, the question asks for the value of $|z|$, which is constant for all points satisfying the condition.

The final answer is $\mathbf{1}$.

Given:

1. $|z_1| = |z_2|$

2. $\arg(z_1) + \arg(z_2) = \pi$

To Show:

$z_1 = - \overline{z_2}$

Proof:

Let $z_1$ and $z_2$ be represented in polar form.

$z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)$

$z_2 = r_2 (\cos \theta_2 + i \sin \theta_2)$

From the given condition $|z_1| = |z_2|$, we have $r_1 = r_2$. Let's denote this common modulus by $r$, i.e., $r_1 = r_2 = r$.

From the given condition $\arg(z_1) + \arg(z_2) = \pi$, we have $\theta_1 + \theta_2 = \pi$.

This implies $\theta_1 = \pi - \theta_2$.

Now, let's find the conjugate of $z_2$, which is $\overline{z_2}$.

$\overline{z_2} = r_2 (\cos \theta_2 - i \sin \theta_2)$

Using the trigonometric identities $\cos(-\theta) = \cos \theta$ and $\sin(-\theta) = -\sin \theta$, we can write:

$\overline{z_2} = r_2 (\cos (-\theta_2) + i \sin (-\theta_2))$

Now, consider $- \overline{z_2}$. We can write $-1$ in polar form as $\cos \pi + i \sin \pi$.

$- \overline{z_2} = (-1) \times \overline{z_2}$

$- \overline{z_2} = (\cos \pi + i \sin \pi) \times r_2 (\cos (-\theta_2) + i \sin (-\theta_2))$

Using the property that the product of two complex numbers in polar form is the product of their moduli and the sum of their arguments:

$- \overline{z_2} = r_2 \times 1 \times (\cos (\pi + (-\theta_2)) + i \sin (\pi + (-\theta_2)))$

$- \overline{z_2} = r_2 (\cos (\pi - \theta_2) + i \sin (\pi - \theta_2))$

We know that $r_2 = r$. So,

$- \overline{z_2} = r (\cos (\pi - \theta_2) + i \sin (\pi - \theta_2))$

From the given condition, $\theta_1 = \pi - \theta_2$. Substitute this into the expression for $-\overline{z_2}$:

$- \overline{z_2} = r (\cos \theta_1 + i \sin \theta_1)$

Recall that $z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)$ and $r_1 = r$. So,

$z_1 = r (\cos \theta_1 + i \sin \theta_1)$

Comparing the expressions for $- \overline{z_2}$ and $z_1$, we see that they are the same:

This shows that $z_1 = - \overline{z_2}$.

Hence, shown that $\mathbf{z_1 = - \overline{z_2}}$.

Given:

A complex number $z_1$ such that $|z_1| = 1$ and $z_1 \neq -1$.

Another complex number $z_2 = \frac{z_1 - 1}{z_1 + 1}$.

To Prove:

The real part of $z_2$ is zero, i.e., $\text{Re}(z_2) = 0$.

Solution:

Let $z_1$ be a complex number represented as $z_1 = x + iy$, where $x$ and $y$ are real numbers.

Given that $|z_1| = 1$. The magnitude of $z_1$ is $\sqrt{x^2 + y^2}$.

So, we have $\sqrt{x^2 + y^2} = 1$. Squaring both sides gives:

$x^2 + y^2 = 1$

We are given $z_2 = \frac{z_1 - 1}{z_1 + 1}$. Substitute $z_1 = x + iy$ into this expression:

$z_2 = \frac{(x + iy) - 1}{(x + iy) + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$

To find the real part of $z_2$, we need to express it in the form $A + iB$. We do this by multiplying the numerator and the denominator by the conjugate of the denominator, which is $(x + 1) - iy$:

$z_2 = \frac{((x - 1) + iy)((x + 1) - iy)}{((x + 1) + iy)((x + 1) - iy)}$

Let's calculate the denominator and the numerator separately.

The denominator is:

$((x + 1) + iy)((x + 1) - iy) = (x + 1)^2 - (iy)^2 = (x + 1)^2 - i^2 y^2 = (x + 1)^2 + y^2$

$= (x^2 + 2x + 1) + y^2 = (x^2 + y^2) + 2x + 1$

Using the given condition $x^2 + y^2 = 1$, the denominator becomes:

Denominator = $1 + 2x + 1 = 2 + 2x = 2(1 + x)$

Given $z_1 \neq -1$, which means $x + iy \neq -1$. If $y=0$, then $x \neq -1$. If $y \neq 0$, $x+1+iy \neq 0$, so $x+1$ could be zero if $y \neq 0$. However, with $x^2+y^2=1$, $x=-1$ implies $y=0$. So $z_1 = -1$ implies $(x,y)=(-1,0)$. Thus, $z_1 \neq -1$ means $(x,y) \neq (-1,0)$, which ensures $x+1$ and $y$ are not simultaneously zero. Specifically, $x \neq -1$, so the denominator $2(1+x)$ is non-zero.

The numerator is:

$((x - 1) + iy)((x + 1) - iy) = (x - 1)(x + 1) - iy(x - 1) + iy(x + 1) - (iy)^2$

$= (x^2 - 1) - ixy + iy + ixy + iy + y^2$

$= (x^2 - 1 + y^2) + i(-xy + y + xy + y)$

$= (x^2 + y^2 - 1) + i(2y)$

Using the given condition $x^2 + y^2 = 1$, the real part of the numerator becomes $x^2 + y^2 - 1 = 1 - 1 = 0$.

So, the numerator simplifies to $0 + i(2y) = i(2y)$.

Now substitute the simplified numerator and denominator back into the expression for $z_2$:

$z_2 = \frac{i(2y)}{2(1 + x)} = i \frac{2y}{2(1 + x)} = i \frac{y}{1 + x}$

We can write $z_2$ in the standard form $A + iB$ as $z_2 = 0 + i \left( \frac{y}{1 + x} \right)$.

The real part of $z_2$, denoted by $\text{Re}(z_2)$, is the term without the imaginary unit $i$.

$\text{Re}(z_2) = 0$

Thus, the real part of $z_2$ is zero.

Alternate Solution:

We want to show that $\text{Re}(z_2) = 0$. We know that for any complex number $z$, $\text{Re}(z) = \frac{z + \overline{z}}{2}$.

Let's find the conjugate of $z_2$. Given $z_2 = \frac{z_1 - 1}{z_1 + 1}$.

$\overline{z_2} = \overline{\left(\frac{z_1 - 1}{z_1 + 1}\right)} = \frac{\overline{z_1 - 1}}{\overline{z_1 + 1}}$

Using the property $\overline{z_a \pm z_b} = \overline{z_a} \pm \overline{z_b}$ and $\overline{c} = c$ for a real number $c$:

$\overline{z_2} = \frac{\overline{z_1} - \overline{1}}{\overline{z_1} + \overline{1}} = \frac{\overline{z_1} - 1}{\overline{z_1} + 1}$

Given that $|z_1| = 1$, we have $z_1 \overline{z_1} = |z_1|^2 = 1^2 = 1$. Since $z_1 \neq -1$, $z_1$ cannot be zero (as $|z_1|=1$). Thus, we can write $\overline{z_1} = \frac{1}{z_1}$.

Substitute $\overline{z_1} = \frac{1}{z_1}$ into the expression for $\overline{z_2}$:

$\overline{z_2} = \frac{\frac{1}{z_1} - 1}{\frac{1}{z_1} + 1}$

Multiply the numerator and the denominator by $z_1$ to simplify (since $|z_1|=1$, $z_1 \neq 0$):

$\overline{z_2} = \frac{z_1 \left(\frac{1}{z_1} - 1\right)}{z_1 \left(\frac{1}{z_1} + 1\right)} = \frac{z_1 \cdot \frac{1}{z_1} - z_1 \cdot 1}{z_1 \cdot \frac{1}{z_1} + z_1 \cdot 1} = \frac{1 - z_1}{1 + z_1}$

Now, calculate $z_2 + \overline{z_2}$:

$z_2 + \overline{z_2} = \frac{z_1 - 1}{z_1 + 1} + \frac{1 - z_1}{1 + z_1}$

Notice that $\frac{1 - z_1}{1 + z_1} = \frac{-(z_1 - 1)}{z_1 + 1}$.

$z_2 + \overline{z_2} = \frac{z_1 - 1}{z_1 + 1} + \frac{-(z_1 - 1)}{z_1 + 1} = \frac{(z_1 - 1) - (z_1 - 1)}{z_1 + 1} = \frac{z_1 - 1 - z_1 + 1}{z_1 + 1} = \frac{0}{z_1 + 1}$

Given $z_1 \neq -1$, the denominator $z_1 + 1$ is non-zero. Therefore, $\frac{0}{z_1 + 1} = 0$.

$z_2 + \overline{z_2} = 0$

Now, we can find the real part of $z_2$:

$\text{Re}(z_2) = \frac{z_2 + \overline{z_2}}{2} = \frac{0}{2} = 0$

Thus, the real part of $z_2$ is zero.

Given:

$z_1, z_2$ are conjugate complex numbers, which means $z_2 = \overline{z_1}$.

$z_3, z_4$ are conjugate complex numbers, which means $z_4 = \overline{z_3}$.

To Find:

The value of $\text{arg}\left( \frac{z_{1}}{z_{4}} \right) + \text{arg}\left( \frac{z_{2}}{z_{3}} \right)$.

Solution:

We are given that $z_2 = \overline{z_1}$ and $z_4 = \overline{z_3}$.

We need to find the value of $\text{arg}\left( \frac{z_{1}}{z_{4}} \right) + \text{arg}\left( \frac{z_{2}}{z_{3}} \right)$.

Using the property that $\text{arg}(a) + \text{arg}(b) = \text{arg}(ab)$, we can write the expression as:

$\text{arg}\left( \frac{z_{1}}{z_{4}} \right) + \text{arg}\left( \frac{z_{2}}{z_{3}} \right) = \text{arg}\left( \left( \frac{z_{1}}{z_{4}} \right) \left( \frac{z_{2}}{z_{3}} \right) \right)$

Now, substitute the given relationships $z_2 = \overline{z_1}$ and $z_4 = \overline{z_3}$ into the product:

$\left( \frac{z_{1}}{z_{4}} \right) \left( \frac{z_{2}}{z_{3}} \right) = \frac{z_{1}}{\overline{z_{3}}} \cdot \frac{\overline{z_{1}}}{z_{3}}$

Rearrange the terms:

$\frac{z_{1} \overline{z_{1}}}{\overline{z_{3}} z_{3}}$

Using the property of complex numbers $z \overline{z} = |z|^2$, we have:

$\frac{z_{1} \overline{z_{1}}}{\overline{z_{3}} z_{3}} = \frac{|z_1|^2}{|z_3|^2}$

For the arguments $\text{arg}\left( \frac{z_1}{z_4} \right)$ and $\text{arg}\left( \frac{z_2}{z_3} \right)$ to be defined, the complex numbers $\frac{z_1}{z_4}$ and $\frac{z_2}{z_3}$ must be non-zero. This requires $z_1 \neq 0$, $z_2 \neq 0$, $z_3 \neq 0$, and $z_4 \neq 0$. Since $z_2 = \overline{z_1}$ and $z_4 = \overline{z_3}$, this is equivalent to requiring $z_1 \neq 0$ and $z_3 \neq 0$.

If $z_1 \neq 0$ and $z_3 \neq 0$, then $|z_1| > 0$ and $|z_3| > 0$. Consequently, $|z_1|^2 > 0$ and $|z_3|^2 > 0$.

Therefore, the expression $\frac{|z_1|^2}{|z_3|^2}$ is a positive real number.

The argument of any positive real number is 0.

So, $\text{arg}\left( \frac{|z_1|^2}{|z_3|^2} \right) = 0$.

Hence,

$\text{arg}\left( \frac{z_{1}}{z_{4}} \right) + \text{arg}\left( \frac{z_{2}}{z_{3}} \right) = \text{arg}\left( \frac{|z_1|^2}{|z_3|^2} \right) = \mathbf{0}$

Alternate Solution:

Let $\text{arg}(z_1) = \theta_1$ and $\text{arg}(z_3) = \theta_3$.

Since $z_2 = \overline{z_1}$, we know that $\text{arg}(z_2) = \text{arg}(\overline{z_1}) = -\text{arg}(z_1) = -\theta_1$.

Since $z_4 = \overline{z_3}$, we know that $\text{arg}(z_4) = \text{arg}(\overline{z_3}) = -\text{arg}(z_3) = -\theta_3$.

Using the property that $\text{arg}\left(\frac{z_a}{z_b}\right) = \text{arg}(z_a) - \text{arg}(z_b)$, we have:

$\text{arg}\left( \frac{z_{1}}{z_{4}} \right) = \text{arg}(z_1) - \text{arg}(z_4) = \theta_1 - (-\theta_3) = \theta_1 + \theta_3$

$\text{arg}\left( \frac{z_{2}}{z_{3}} \right) = \text{arg}(z_2) - \text{arg}(z_3) = -\theta_1 - \theta_3 = -(\theta_1 + \theta_3)$

Now, add these two arguments:

$\text{arg}\left( \frac{z_{1}}{z_{4}} \right) + \text{arg}\left( \frac{z_{2}}{z_{3}} \right) = (\theta_1 + \theta_3) + (-(\theta_1 + \theta_3))$

$= \theta_1 + \theta_3 - \theta_1 - \theta_3 = 0$

Assuming $z_1, z_2, z_3, z_4$ are non-zero such that the arguments are defined, the result is 0.

Given:

A set of $n$ complex numbers $z_1, z_2, \dots, z_n$ such that $|z_k| = 1$ for each $k = 1, 2, \dots, n$.

To Prove:

$|z_1 + z_2 + z_3 + \dots + z_n | = \left| \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}+...+\frac{1}{z_{n}} \right|$.

Proof:

We are given that $|z_k| = 1$ for $k = 1, 2, \dots, n$.

By the definition of magnitude of a complex number, $|z|^2 = z \overline{z}$.

So, for each $z_k$, we have $|z_k|^2 = z_k \overline{z_k}$.

Since $|z_k| = 1$, we have $1^2 = z_k \overline{z_k}$, which means $z_k \overline{z_k} = 1$.

Since $|z_k|=1$, $z_k$ cannot be zero. Therefore, we can divide by $z_k$ to find an expression for $\overline{z_k}$:

$\overline{z_k} = \frac{1}{z_k}$

This relationship holds for every $k = 1, 2, \dots, n$.

Now, consider the expression on the right side of the equation we need to prove:

$\left| \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}+...+\frac{1}{z_{n}} \right|$

Using the relationship $\frac{1}{z_k} = \overline{z_k}$, we can replace each term inside the magnitude:

$\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}+...+\frac{1}{z_{n}} = \overline{z_1} + \overline{z_2} + \overline{z_3} + \dots + \overline{z_n}$

We know that the sum of the conjugates of complex numbers is equal to the conjugate of their sum. That is, for any complex numbers $w_1, w_2, \dots, w_n$, we have $\overline{w_1} + \overline{w_2} + \dots + \overline{w_n} = \overline{w_1 + w_2 + \dots + w_n}$.

Applying this property to our expression:

$\overline{z_1} + \overline{z_2} + \overline{z_3} + \dots + \overline{z_n} = \overline{z_1 + z_2 + z_3 + \dots + z_n}$

So, the expression inside the magnitude on the right side becomes $\overline{z_1 + z_2 + z_3 + \dots + z_n}$.

The right side of the equation we need to prove is thus:

$\left| \overline{z_1 + z_2 + z_3 + \dots + z_n} \right|$

Let $Z = z_1 + z_2 + z_3 + \dots + z_n$. The expression is $|\overline{Z}|$.

We know that the magnitude of a complex number is equal to the magnitude of its conjugate, i.e., $|z| = |\overline{z}|$.

Applying this property, we have $|\overline{Z}| = |Z|$.

Substituting back $Z = z_1 + z_2 + z_3 + \dots + z_n$, we get:

$\left| \overline{z_1 + z_2 + z_3 + \dots + z_n} \right| = |z_1 + z_2 + z_3 + \dots + z_n|$

Thus, we have shown that:

$\left| \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}+...+\frac{1}{z_{n}} \right| = |z_1 + z_2 + z_3 + \dots + z_n|$

This completes the proof.

Given:

For complex numbers $z_1$ and $z_2$, $\text{arg}(z_1) - \text{arg}(z_2) = 0$.

To Prove:

$|z_1 - z_2| = |z_1| - |z_2|$.

Solution:

The given condition is $\text{arg}(z_1) - \text{arg}(z_2) = 0$.

We know that for any two non-zero complex numbers $z_a$ and $z_b$, $\text{arg}\left(\frac{z_a}{z_b}\right) = \text{arg}(z_a) - \text{arg}(z_b)$.

Thus, the given condition implies $\text{arg}\left(\frac{z_1}{z_2}\right) = 0$, provided $z_2 \neq 0$.

If $z_2 = 0$, then the arguments are undefined unless $z_1 = 0$ as well. If $z_1 = z_2 = 0$, then $|0-0| = |0|-|0|$, which is $0 = 0$, so the equality holds. Let's assume $z_2 \neq 0$ and consequently $z_1 \neq 0$ (since if $z_1=0$, $\text{arg}(0)$ is undefined, or if we take a common argument like principal argument, $\text{arg}(0)$ is not 0, while $\text{arg}(z_2)$ would be defined).

The condition $\text{arg}\left(\frac{z_1}{z_2}\right) = 0$ means that the complex number $\frac{z_1}{z_2}$ is a positive real number.

Let $\frac{z_1}{z_2} = k$, where $k$ is a real number and $k > 0$.

From this, we can write $z_1 = k z_2$.

Now, let's consider the left side of the equation we need to prove, $|z_1 - z_2|$.

Substitute $z_1 = k z_2$ into the expression:

$|z_1 - z_2| = |k z_2 - z_2|$

$|z_1 - z_2| = |(k - 1) z_2|$

Using the property of magnitudes $|ab| = |a||b|$, we get:

$|z_1 - z_2| = |k - 1| |z_2|$

Now, let's consider the right side of the equation we need to prove, $|z_1| - |z_2|$.

Substitute $z_1 = k z_2$ into the expression:

$|z_1| - |z_2| = |k z_2| - |z_2|$

Using the property $|ab| = |a||b|$, we get:

$|z_1| - |z_2| = |k| |z_2| - |z_2|$

Since $k$ is a positive real number ($k > 0$), we have $|k| = k$.

So, $|z_1| - |z_2| = k |z_2| - |z_2|$

Factor out $|z_2|$:

$|z_1| - |z_2| = (k - 1) |z_2|$

We are asked to show that $|z_1 - z_2| = |z_1| - |z_2|$.

Substituting the expressions we derived:

$|k - 1| |z_2| = (k - 1) |z_2|$

Since we assumed $z_2 \neq 0$, we have $|z_2| \neq 0$. We can divide both sides by $|z_2|$.

We need to show that $|k - 1| = k - 1$.

The equality $|A| = A$ holds if and only if $A \geq 0$.

Therefore, $|k - 1| = k - 1$ holds if and only if $k - 1 \geq 0$, which means $k \geq 1$.

Recall that $k = \frac{|z_1|}{|z_2|}$ (from $|z_1| = |k z_2| = |k||z_2| = k|z_2|$, assuming $z_2 \neq 0$).

So, the condition $k \geq 1$ is equivalent to $\frac{|z_1|}{|z_2|} \geq 1$, which implies $|z_1| \geq |z_2|$ (since $|z_2| > 0$).

Thus, the equality $|z_1 - z_2| = |z_1| - |z_2|$ holds if and only if $\text{arg}(z_1) - \text{arg}(z_2) = 0$ AND $|z_1| \geq |z_2|$.

Given the wording of the question, it implies that the condition $\text{arg}(z_1) - \text{arg}(z_2) = 0$ is sufficient for the equality $|z_1 - z_2| = |z_1| - |z_2|$ to hold exactly in this form. Our derivation shows this is true precisely when $|z_1| \geq |z_2|$.

Given:

The system of equations involving a complex number $z$:

1. $\text{Re}(z^2) = 0$

2. $|z| = 2$

To Solve:

Find the complex number(s) $z$ that satisfy both equations.

Solution:

Let the complex number $z$ be represented in the Cartesian form as $z = x + iy$, where $x$ and $y$ are real numbers.

Consider the first equation, $\text{Re}(z^2) = 0$.

First, calculate $z^2$:

$z^2 = (x + iy)^2$

$z^2 = x^2 + 2ixy + (iy)^2$

$z^2 = x^2 + 2ixy - y^2$

$z^2 = (x^2 - y^2) + i(2xy)$

The real part of $z^2$ is $\text{Re}(z^2) = x^2 - y^2$.

The first equation gives us:

$x^2 - y^2 = 0$

This can be factored as $(x - y)(x + y) = 0$.

This implies that either $x - y = 0$ or $x + y = 0$.

So, we have two possibilities for the relationship between $x$ and $y$:

$y = x$ or $y = -x$.

Consider the second equation, $|z| = 2$.

The magnitude of $z = x + iy$ is $|z| = \sqrt{x^2 + y^2}$.

The second equation gives us:

$\sqrt{x^2 + y^2} = 2$

Squaring both sides, we get:

$x^2 + y^2 = 4$

Now we need to solve the system of equations:

1. $x^2 - y^2 = 0$

2. $x^2 + y^2 = 4$

We consider the two cases derived from the first equation.

Case 1: $y = x$

Substitute $y = x$ into the second equation ($x^2 + y^2 = 4$):

$x^2 + (x)^2 = 4$

$x^2 + x^2 = 4$

$2x^2 = 4$

$x^2 = 2$

$x = \pm \sqrt{2}$

If $x = \sqrt{2}$, then $y = x = \sqrt{2}$. This gives the complex number $z = \sqrt{2} + i\sqrt{2}$.

If $x = -\sqrt{2}$, then $y = x = -\sqrt{2}$. This gives the complex number $z = -\sqrt{2} - i\sqrt{2}$.

Case 2: $y = -x$

Substitute $y = -x$ into the second equation ($x^2 + y^2 = 4$):

$x^2 + (-x)^2 = 4$

$x^2 + x^2 = 4$

$2x^2 = 4$

$x^2 = 2$

$x = \pm \sqrt{2}$

If $x = \sqrt{2}$, then $y = -x = -\sqrt{2}$. This gives the complex number $z = \sqrt{2} - i\sqrt{2}$.

If $x = -\sqrt{2}$, then $y = -x = -(-\sqrt{2}) = \sqrt{2}$. This gives the complex number $z = -\sqrt{2} + i\sqrt{2}$.

The solutions to the system of equations are the complex numbers found in these two cases.

The solutions are $\sqrt{2} + i\sqrt{2}$, $-\sqrt{2} - i\sqrt{2}$, $\sqrt{2} - i\sqrt{2}$, and $-\sqrt{2} + i\sqrt{2}$.

We can also write these solutions using the polar form. Since $|z|=2$ and $x = \pm \sqrt{2}, y = \pm \sqrt{2}$, the arguments correspond to angles whose tangent is $\pm 1$. These are $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ (or $\frac{\pi}{4}, \frac{3\pi}{4}, -\frac{3\pi}{4}, -\frac{\pi}{4}$).

$z = 2e^{i \pi/4} = 2(\cos(\pi/4) + i \sin(\pi/4)) = 2(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}) = \sqrt{2} + i\sqrt{2}$.

$z = 2e^{i 3\pi/4} = 2(\cos(3\pi/4) + i \sin(3\pi/4)) = 2(-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}) = -\sqrt{2} + i\sqrt{2}$.

$z = 2e^{i 5\pi/4} = 2(\cos(5\pi/4) + i \sin(5\pi/4)) = 2(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}) = -\sqrt{2} - i\sqrt{2}$.

$z = 2e^{i 7\pi/4} = 2(\cos(7\pi/4) + i \sin(7\pi/4)) = 2(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}) = \sqrt{2} - i\sqrt{2}$.

The solutions are $\sqrt{2} + i\sqrt{2}$, $-\sqrt{2} + i\sqrt{2}$, $-\sqrt{2} - i\sqrt{2}$, and $\sqrt{2} - i\sqrt{2}$.

Given:

The equation involving a complex number $z$: $z + \sqrt{2}\;|(z + 1)| + i = 0$.

To Find:

The complex number $z$ satisfying the given equation.

Solution:

Let the complex number be $z = x + iy$, where $x$ and $y$ are real numbers.

Substitute $z = x + iy$ into the given equation:

$(x + iy) + \sqrt{2}\;|((x + iy) + 1)| + i = 0$

Group the real and imaginary terms inside the magnitude:

$(x + iy) + \sqrt{2}\;|(x + 1) + iy| + i = 0$

The magnitude of a complex number $a + bi$ is $\sqrt{a^2 + b^2}$. Here, $a = x + 1$ and $b = y$.

So the magnitude term is $|(x + 1) + iy| = \sqrt{(x + 1)^2 + y^2}$.

Substitute this back into the equation:

$(x + iy) + \sqrt{2}\; \sqrt{(x + 1)^2 + y^2} + i = 0$

Rearrange the terms to group the real and imaginary parts of the entire equation:

$(x + \sqrt{2}\; \sqrt{(x + 1)^2 + y^2}) + i(y + 1) = 0$

For a complex number to be equal to zero, both its real part and its imaginary part must be zero.

Equating the imaginary part to zero:

$y + 1 = 0$

Solving for $y$, we get:

$y = -1$

Equating the real part to zero:

$x + \sqrt{2}\; \sqrt{(x + 1)^2 + y^2} = 0$

Substitute the value $y = -1$ into this equation:

$x + \sqrt{2}\; \sqrt{(x + 1)^2 + (-1)^2} = 0$

$x + \sqrt{2}\; \sqrt{(x + 1)^2 + 1} = 0$

Isolate the square root term by moving $x$ to the right side:

$\sqrt{2}\; \sqrt{(x + 1)^2 + 1} = -x$

The left side of this equation, $\sqrt{2}\; \sqrt{(x + 1)^2 + 1}$, represents a non-negative real number (since it's $\sqrt{2}$ times a magnitude). Therefore, the right side, $-x$, must also be non-negative.

This implies $-x \geq 0$, which means $x \leq 0$.

Now, square both sides of the equation $\sqrt{2}\; \sqrt{(x + 1)^2 + 1} = -x$ to eliminate the square root:

$(\sqrt{2}\; \sqrt{(x + 1)^2 + 1})^2 = (-x)^2$

$(\sqrt{2})^2 \left( \sqrt{(x + 1)^2 + 1} \right)^2 = x^2$

$2 \left( (x + 1)^2 + 1 \right) = x^2$

Expand the term $(x + 1)^2$:

$2 \left( (x^2 + 2x + 1) + 1 \right) = x^2$

$2 \left( x^2 + 2x + 2 \right) = x^2$

Distribute the 2 on the left side:

$2x^2 + 4x + 4 = x^2$

Move all terms to one side to form a standard quadratic equation:

$2x^2 - x^2 + 4x + 4 = 0$

$x^2 + 4x + 4 = 0$

This quadratic equation is a perfect square trinomial, which can be factored as:

$(x + 2)^2 = 0$

Solving for $x$:

$x + 2 = 0$

$x = -2$

We must check if this value of $x$ satisfies the condition $x \leq 0$ that we derived earlier. Since $-2 \leq 0$, the value $x = -2$ is valid.

We have found the values for $x$ and $y$: $x = -2$ and $y = -1$.

Therefore, the complex number $z$ that satisfies the given equation is $z = x + iy = -2 + i(-1)$.

$z = -2 - i$

The complex number is $z = -2 - i$.

Given:

The complex number $z = \frac{1\;-\;i}{\cos\frac{\pi}{3} \;+\; i\sin\frac{\pi}{3}}$.

To Find:

The polar form of the complex number $z$.

Solution:

Let $z = \frac{N}{D}$, where $N = 1 - i$ is the numerator and $D = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}$ is the denominator.

First, we find the polar form of the numerator $N = 1 - i$.

The magnitude of $N$ is $|N| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

To find the argument of $N$, let $\theta_N = \text{arg}(1 - i)$. We have $\tan(\theta_N) = \frac{-1}{1} = -1$. Since the real part is positive (1) and the imaginary part is negative (-1), the complex number $1-i$ lies in the fourth quadrant. The principal argument is in the interval $(-\pi, \pi]$.

Thus, $\theta_N = -\frac{\pi}{4}$.

The polar form of the numerator is $N = \sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right) \right)$.

Next, we consider the denominator $D = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}$.

This complex number is already in polar form.

The magnitude of $D$ is $|D| = \sqrt{\cos^2\left(\frac{\pi}{3}\right) + \sin^2\left(\frac{\pi}{3}\right)} = \sqrt{1} = 1$.

The argument of $D$ is $\theta_D = \text{arg}\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = \frac{\pi}{3}$.

Now, we find the magnitude and argument of the quotient $z = \frac{N}{D}$.

The magnitude of $z$ is $|z| = \frac{|N|}{|D|} = \frac{\sqrt{2}}{1} = \sqrt{2}$.

The argument of $z$ is $\text{arg}(z) = \text{arg}(N) - \text{arg}(D)$.

$\text{arg}(z) = -\frac{\pi}{4} - \frac{\pi}{3}$

To subtract these fractions, we find a common denominator, which is 12:

$-\frac{\pi}{4} - \frac{\pi}{3} = -\frac{3\pi}{12} - \frac{4\pi}{12} = -\frac{7\pi}{12}$

The argument of $z$ is $-\frac{7\pi}{12}$. This value is within the principal argument range $(-\pi, \pi]$.

The polar form of a complex number with magnitude $r$ and argument $\theta$ is $r(\cos\theta + i\sin\theta)$.

Using the magnitude $|z| = \sqrt{2}$ and argument $\text{arg}(z) = -\frac{7\pi}{12}$, the polar form of $z$ is:

$z = \sqrt{2} \left( \cos\left(-\frac{7\pi}{12}\right) + i\sin\left(-\frac{7\pi}{12}\right) \right)$

Using the identities $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$, this can also be written as:

$z = \sqrt{2} \left( \cos\left(\frac{7\pi}{12}\right) - i\sin\left(\frac{7\pi}{12}\right) \right)$

However, the standard polar form uses the calculated argument directly.

The polar form of the complex number is $\sqrt{2} \left( \cos\left(-\frac{7\pi}{12}\right) + i\sin\left(-\frac{7\pi}{12}\right) \right)$.

Given:

Two complex numbers $z$ and $w$ satisfying the conditions:

1. $|zw| = 1$

2. $\text{arg}(z) - \text{arg}(w) = \frac{\pi}{2}$

To Prove:

$\overline{z}w = -i$.

Proof:

Let the complex numbers $z$ and $w$ be represented in their polar forms.

Let $z = |z| (\cos(\text{arg}(z)) + i\sin(\text{arg}(z)))$ and $w = |w| (\cos(\text{arg}(w)) + i\sin(\text{arg}(w)))$.

Using Euler's formula, we can write $z = |z| e^{i \text{arg}(z)}$ and $w = |w| e^{i \text{arg}(w)}$.

From the first given condition, $|zw| = 1$.

We know that $|zw| = |z||w|$. So, we have:

$|z||w| = 1$

From the second given condition, $\text{arg}(z) - \text{arg}(w) = \frac{\pi}{2}$.

Let $\theta_z = \text{arg}(z)$ and $\theta_w = \text{arg}(w)$. So, $\theta_z - \theta_w = \frac{\pi}{2}$.

We need to evaluate the expression $\overline{z}w$.

The conjugate of $z$ in polar form is $\overline{z} = |z| e^{-i \text{arg}(z)} = |z| e^{-i \theta_z}$.

Now, multiply $\overline{z}$ and $w$:

$\overline{z}w = (|z| e^{-i \theta_z}) (|w| e^{i \theta_w})$

Using the properties of exponents, we can combine the terms:

$\overline{z}w = |z||w| e^{i (\theta_w - \theta_z)}$

From the given conditions, we know that $|z||w| = 1$ and $\theta_w - \theta_z = -(\theta_z - \theta_w) = -\frac{\pi}{2}$.

Substitute these values into the expression for $\overline{z}w$:

$\overline{z}w = 1 \cdot e^{i \left(-\frac{\pi}{2}\right)}$

$\overline{z}w = e^{-i \frac{\pi}{2}}$

Now, we convert the exponential form $e^{-i \frac{\pi}{2}}$ back to the Cartesian form using Euler's formula $e^{i\phi} = \cos\phi + i\sin\phi$:

$e^{-i \frac{\pi}{2}} = \cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)$

We know that $\cos\left(-\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0$ and $\sin\left(-\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1$.

So, $e^{-i \frac{\pi}{2}} = 0 + i(-1) = -i$.

Therefore, we have shown that $\overline{z}w = -i$.

This completes the proof.

(i) For any two complex numbers $z_1$, $z_2$ and any real numbers a, b, $|az_1 - bz_2|^2 + |bz_1 + az_2|^2 = \left(a^2 + b^2\right)\left(|z_1|^2 + |z_2|^2\right)$.

(ii) The value of $\sqrt{-25} \times \sqrt{-9}$ is $-15$.

(iii) The number $\frac{(1-i)^{3}}{1-i^{3}}$ is equal to $-2$.

(iv) The sum of the series $i + i^2 + i^3 + …$ upto 1000 terms is $0$.

(v) Multiplicative inverse of $1 + i$ is $\frac{1}{2} - \frac{1}{2}i$.

(vi) If $z_1$ and $z_2$ are complex numbers such that $z_1 + z_2$ is a real number, then $z_2 = \overline{z_1} + k$ where $k$ is a real number. (Note: A common simplified answer sometimes given is $\overline{z_1}$, although this is only a specific case).

(vii) arg(z) + arg $\overline{z}$ ($\overline{z}$ ≠ 0) is $0$.

(viii) If $|z + 4| ≤ 3$, then the greatest and least values of $|z +1|$ are $6$ and $0$.

(ix) If $\left| \frac{z-2}{z+2} \right|=\frac{\pi}{6}$, then the locus of z is a circle.

(x) If $|z| = 4$ and arg (z) = $\frac{5\pi}{6}$, then z = $-2\sqrt{3} + 2i$.

(i) The order relation is defined on the set of complex numbers. False

(ii) Multiplication of a non zero complex number by $– i$ rotates the point about origin through a right angle in the anti-clockwise direction. False

(iii) For any complex number z the minimum value of |z| + |z - 1| is 1. True

(iv) The locus represented by |z - 1| = |z - $i$| is a line perpendicular to the join of (1, 0) and (0, 1). True

(v) If z is a complex number such that z ≠ 0 and Re (z) = 0, then Im (z²) = 0. True

(vi) The inequality |z - 4| < |z - 2| represents the region given by x > 3. True

(vii) Let z₁ and z₂ be two complex numbers such that |z₁ + z₂| = |z₁| + |z₂| , then arg (z₁ – z₂ ) = 0. False

(viii) 2 is not a complex number. False

To match the statements, we will analyse each part of Column A and find its corresponding result or description in Column B.

(a) The polar form of $i + \sqrt{3}$

Let the complex number be $z = \sqrt{3} + i$. We need to find its polar form $z = r(\cos \theta + i \sin \theta)$, where $r$ is the modulus and $\theta$ is the argument.

The modulus is $r = |z| = \sqrt{(\text{Re}(z))^2 + (\text{Im}(z))^2}$.

$r = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.

To find the argument $\theta$, we note that the point $(\sqrt{3}, 1)$ lies in the first quadrant. The reference angle $\alpha$ is given by $\tan \alpha = \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right|$.

$\tan \alpha = \left| \frac{1}{\sqrt{3}} \right| = \frac{1}{\sqrt{3}}$.

This gives $\alpha = \frac{\pi}{6}$. Since the point is in the first quadrant, the argument $\theta = \alpha$.

$\theta = \frac{\pi}{6}$.

The polar form is $z = 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right)$.

This matches statement (v) in Column B.

(b) The amplitude of $-1 + \sqrt{-3}$

Let the complex number be $z = -1 + \sqrt{-3} = -1 + i\sqrt{3}$. We need to find its amplitude (principal argument).

The point $(-1, \sqrt{3})$ lies in the second quadrant.

The reference angle $\alpha$ is given by $\tan \alpha = \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right|$.

$\tan \alpha = \left| \frac{\sqrt{3}}{-1} \right| = \sqrt{3}$.

This gives $\alpha = \frac{\pi}{3}$. Since the point is in the second quadrant, the principal argument $\theta = \pi - \alpha$.

$\theta = \pi - \frac{\pi}{3} = \frac{3\pi - \pi}{3} = \frac{2\pi}{3}$.

The amplitude is $\frac{2\pi}{3}$.

This matches statement (iii) in Column B.

(c) If $|z + 2| = |z - 2|$, then locus of $z$ is

Let $z = x + iy$. The given condition is $|z - (-2)| = |z - 2|$. This means the distance of $z$ from the point $(-2, 0)$ is equal to its distance from the point $(2, 0)$.

The locus of a point that is equidistant from two fixed points is the perpendicular bisector of the segment joining these two points.

The two fixed points are $A = (-2, 0)$ and $B = (2, 0)$. The segment joining them is on the x-axis.

The midpoint of the segment AB is $\left( \frac{-2+2}{2}, \frac{0+0}{2} \right) = (0, 0)$.

The segment AB is horizontal. The perpendicular bisector is a vertical line passing through the midpoint $(0, 0)$. The equation of this line is $x = 0$ (the y-axis).

Alternatively, using the algebraic method:

$|x + iy + 2| = |x + iy - 2|$

$|(x+2) + iy| = |(x-2) + iy|$

$\sqrt{(x+2)^2 + y^2} = \sqrt{(x-2)^2 + y^2}$

Squaring both sides:

$(x+2)^2 + y^2 = (x-2)^2 + y^2$

$x^2 + 4x + 4 + y^2 = x^2 - 4x + 4 + y^2$

Subtract $x^2 + y^2 + 4$ from both sides:

$4x = -4x$

$8x = 0$

$x = 0$.

The locus is the line $x=0$, which is the perpendicular bisector of the segment joining $(-2, 0)$ and $(2, 0)$.

This matches statement (i) in Column B.

(d) If $|z + 2i| = |z - 2i|$, then locus of $z$ is

Let $z = x + iy$. The given condition is $|z - (-2i)| = |z - 2i|$. This means the distance of $z$ from the point $(0, -2)$ is equal to its distance from the point $(0, 2)$.

The locus of a point that is equidistant from two fixed points is the perpendicular bisector of the segment joining these two points.

The two fixed points are $C = (0, -2)$ and $D = (0, 2)$. The segment joining them is on the y-axis.

The midpoint of the segment CD is $\left( \frac{0+0}{2}, \frac{-2+2}{2} \right) = (0, 0)$.

The segment CD is vertical. The perpendicular bisector is a horizontal line passing through the midpoint $(0, 0)$. The equation of this line is $y = 0$ (the x-axis).

Alternatively, using the algebraic method:

$|x + iy + 2i| = |x + iy - 2i|$

$|x + i(y+2)| = |x + i(y-2)|$

$\sqrt{x^2 + (y+2)^2} = \sqrt{x^2 + (y-2)^2}$

Squaring both sides:

$x^2 + (y+2)^2 = x^2 + (y-2)^2$

$x^2 + y^2 + 4y + 4 = x^2 + y^2 - 4y + 4$

Subtract $x^2 + y^2 + 4$ from both sides:

$4y = -4y$

$8y = 0$

$y = 0$.

The locus is the line $y=0$, which is the perpendicular bisector of the segment joining $(0, -2)$ and $(0, 2)$.

This matches statement (iv) in Column B.

(e) Region represented by $|z + 4i| \ge 3$ is

Let $z = x + iy$. The inequality is $|z - (-4i)| \ge 3$. This means the distance of $z$ from the point $(0, -4)$ is greater than or equal to 3.

The equation $|z - (-4i)| = 3$ represents a circle with centre at $(0, -4)$ and radius 3.

The inequality $|z - (-4i)| \ge 3$ represents all points outside or on this circle.

Alternatively, using the algebraic method:

$|x + iy + 4i| \ge 3$

$|x + i(y+4)| \ge 3$

$\sqrt{x^2 + (y+4)^2} \ge 3$

Squaring both sides:

$x^2 + (y+4)^2 \ge 3^2$

$x^2 + (y - (-4))^2 \ge 3^2$.

This is the equation of a circle with centre $(0, -4)$ and radius $3$. The inequality $\ge$ indicates the region outside or on the circle.

This matches statement (ii) in Column B.

(f) Region represented by $|z + 4| \le 3$ is

Let $z = x + iy$. The inequality is $|z - (-4)| \le 3$. This means the distance of $z$ from the point $(-4, 0)$ is less than or equal to 3.

The equation $|z - (-4)| = 3$ represents a circle with centre at $(-4, 0)$ and radius 3.

The inequality $|z - (-4)| \le 3$ represents all points inside or on this circle.

Alternatively, using the algebraic method:

$|x + iy + 4| \le 3$

$|(x+4) + iy| \le 3$

$\sqrt{(x+4)^2 + y^2} \le 3$

Squaring both sides:

$(x+4)^2 + y^2 \le 3^2$

$(x - (-4))^2 + (y - 0)^2 \le 3^2$.

This is the equation of a circle with centre $(-4, 0)$ and radius $3$. The inequality $\le$ indicates the region inside or on the circle.

This matches statement (vi) in Column B.

(g) Conjugate of $\frac{1 + 2i}{1 - i}$ lies in

First, simplify the complex number $\frac{1 + 2i}{1 - i}$ by multiplying the numerator and denominator by the conjugate of the denominator.

$\frac{1 + 2i}{1 - i} = \frac{1 + 2i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(1)(1) + (1)(i) + (2i)(1) + (2i)(i)}{1^2 - i^2}$

$= \frac{1 + i + 2i + 2i^2}{1 - (-1)} = \frac{1 + 3i - 2}{1 + 1} = \frac{-1 + 3i}{2} = -\frac{1}{2} + \frac{3}{2}i$.

The conjugate of this complex number is obtained by changing the sign of the imaginary part.

Conjugate = $\overline{-\frac{1}{2} + \frac{3}{2}i} = -\frac{1}{2} - \frac{3}{2}i$.

The real part of the conjugate is $-\frac{1}{2}$ (negative) and the imaginary part is $-\frac{3}{2}$ (negative).

A complex number with a negative real part and a negative imaginary part lies in the third quadrant.

This matches statement (viii) in Column B.

(h) Reciprocal of $1 - i$ lies in

The reciprocal of $1 - i$ is $\frac{1}{1 - i}$. We simplify this complex number by multiplying the numerator and denominator by the conjugate of the denominator.

$\frac{1}{1 - i} = \frac{1}{1 - i} \times \frac{1 + i}{1 + i} = \frac{1 + i}{1^2 - i^2} = \frac{1 + i}{1 - (-1)} = \frac{1 + i}{2} = \frac{1}{2} + \frac{1}{2}i$.

The real part of the reciprocal is $\frac{1}{2}$ (positive) and the imaginary part is $\frac{1}{2}$ (positive).

A complex number with a positive real part and a positive imaginary part lies in the first quadrant.

This matches statement (vii) in Column B.

Summary of matches:

(a) $\to$ (v)

(b) $\to$ (iii)

(c) $\to$ (i)

(d) $\to$ (iv)

(e) $\to$ (ii)

(f) $\to$ (vi)

(g) $\to$ (viii)

(h) $\to$ (vii)

Let the given complex number be $z = \frac{2-i}{(1-2i)^{2}}$.

We first need to simplify the denominator $(1-2i)^{2}$.

Simplify the denominator:

$(1-2i)^{2} = 1^2 - 2(1)(2i) + (2i)^2$

$(1-2i)^{2} = 1 - 4i + 4i^2$

Since $i^2 = -1$, we have:

$(1-2i)^{2} = 1 - 4i + 4(-1)$

$(1-2i)^{2} = 1 - 4i - 4$

$(1-2i)^{2} = -3 - 4i$

Now, substitute the simplified denominator back into the expression for $z$:

$z = \frac{2-i}{-3-4i}$

To express $z$ in the form $a + bi$, we multiply the numerator and denominator by the conjugate of the denominator.

The conjugate of $-3 - 4i$ is $-3 + 4i$.

$z = \frac{2-i}{-3-4i} \times \frac{-3+4i}{-3+4i}$

Multiply the numerators:

$(2-i)(-3+4i) = (2)(-3) + (2)(4i) + (-i)(-3) + (-i)(4i)$

$= -6 + 8i + 3i - 4i^2$

$= -6 + 11i - 4(-1)$

$= -6 + 11i + 4$

$= -2 + 11i$

Multiply the denominators:

$(-3-4i)(-3+4i) = (-3)^2 - (4i)^2$

$= 9 - 16i^2$

$= 9 - 16(-1)$

$= 9 + 16$

$= 25$

So, the simplified form of $z$ is:

$z = \frac{-2 + 11i}{25}$

$z = -\frac{2}{25} + \frac{11}{25}i$

Find the conjugate of $z$:

The conjugate of a complex number $a + bi$ is $\overline{z} = a - bi$.

For $z = -\frac{2}{25} + \frac{11}{25}i$, the real part is $a = -\frac{2}{25}$ and the imaginary part is $b = \frac{11}{25}$.

The conjugate $\overline{z}$ is obtained by changing the sign of the imaginary part.

$\overline{z} = -\frac{2}{25} - \frac{11}{25}i$

The conjugate of $\frac{2-i}{(1-2i)^{2}}$ is $-\frac{2}{25} - \frac{11}{25}i$.

No, it is not necessary that $z_1 = z_2$ if $|z_1| = |z_2|$.

The modulus of a complex number $z$, denoted by $|z|$, represents the distance of the point representing $z$ from the origin $(0,0)$ in the complex plane. For a complex number $z = x + iy$, its modulus is given by $|z| = \sqrt{x^2 + y^2}$.

The condition $|z_1| = |z_2|$ means that the complex numbers $z_1$ and $z_2$ are at the same distance from the origin. Geometrically, this implies that both $z_1$ and $z_2$ lie on the same circle centered at the origin with radius equal to the common modulus value.

However, simply lying on the same circle centered at the origin does not mean the points themselves are the same. Different points can exist on the same circle.

Consider a counterexample to illustrate this.

Let $z_1 = 1$ and $z_2 = -1$.

We calculate the modulus of $z_1$:

$|z_1| = |1| = 1$

We calculate the modulus of $z_2$:

$|z_2| = |-1| = 1$

In this example, we have $|z_1| = |z_2| = 1$.

However, the complex numbers themselves are $z_1 = 1$ and $z_2 = -1$. Clearly, $z_1 \ne z_2$.

Another counterexample could be $z_1 = i$ and $z_2 = -i$.

$|z_1| = |i| = \sqrt{0^2 + 1^2} = \sqrt{1} = 1$.

$|z_2| = |-i| = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1$.

Again, $|z_1| = |z_2| = 1$, but $z_1 \ne z_2$.

Since we can find examples where $|z_1| = |z_2|$ but $z_1 \ne z_2$, the statement is not necessarily true.

Given: $\frac{(a^{2}+1)^{2}}{2a-i} = x + iy$, where $a$, $x$, and $y$ are real numbers.

To Find: The value of $x^{2} + y^{2}$.

Solution:

We are given the equation: $\frac{(a^{2}+1)^{2}}{2a-i} = x + iy$

We need to find the value of $x^2 + y^2$. Recall that for a complex number $z = x + iy$, the square of its modulus is $|z|^2 = x^2 + y^2$. So, $x^2 + y^2 = |x + iy|^2$.

Taking the modulus of both sides of the given equation, we have:

$\left| \frac{(a^{2}+1)^{2}}{2a-i} \right| = |x + iy|$

Using the property of modulus that $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$ for $z_2 \ne 0$, we can write:

$\frac{|(a^{2}+1)^{2}|}{|2a-i|} = |x + iy|$

Now, we use the property $|z^n| = |z|^n$ for the numerator:

$\frac{|a^{2}+1|^2}{|2a-i|} = |x + iy|$

Let's calculate the modulus of the numerator and the denominator on the left side.

For the numerator, $a^2+1$: Since $a$ is a real number, $a^2 \ge 0$, so $a^2+1 \ge 1$. Thus, $a^2+1$ is a positive real number. The modulus of a positive real number $r$ is $|r| = r$.

$|a^{2}+1| = a^{2}+1$.

So, $|a^{2}+1|^2 = (a^{2}+1)^2$.

For the denominator, $2a-i$: This is a complex number of the form $A + Bi$, where $A = 2a$ and $B = -1$. The modulus is $\sqrt{A^2 + B^2}$.

$|2a-i| = \sqrt{(2a)^2 + (-1)^2} = \sqrt{4a^2 + 1}$.

Substitute these values back into the equation:

$\frac{(a^{2}+1)^2}{\sqrt{4a^2 + 1}} = |x + iy|$

To find $x^2 + y^2 = |x + iy|^2$, we square both sides of the equation:

$\left( \frac{(a^{2}+1)^2}{\sqrt{4a^2 + 1}} \right)^2 = (|x + iy|)^2$

$\frac{((a^{2}+1)^2)^2}{(\sqrt{4a^2 + 1})^2} = x^2 + y^2$

$\frac{(a^{2}+1)^{4}}{4a^2 + 1} = x^2 + y^2$

Thus, the value of $x^2 + y^2$ is $\frac{(a^{2}+1)^4}{4a^2 + 1}$.

Given:

Modulus of $z$, $|z| = 4$

Argument of $z$, $\text{arg}(z) = \frac{5\pi}{6}$

To Find: The complex number $z$.

Solution:

A complex number $z$ can be expressed in polar form as $z = r(\cos \theta + i \sin \theta)$, where $r = |z|$ is the modulus and $\theta = \text{arg}(z)$ is the argument.

We are given $r = 4$ and $\theta = \frac{5\pi}{6}$.

Substitute these values into the polar form formula:

$z = 4 \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right)$

Now, we need to evaluate $\cos \frac{5\pi}{6}$ and $\sin \frac{5\pi}{6}$. The angle $\frac{5\pi}{6}$ is in the second quadrant, as $0 < \frac{5\pi}{6} < \pi$.

Using the properties of trigonometric functions:

$\cos \frac{5\pi}{6} = \cos \left( \pi - \frac{\pi}{6} \right) = -\cos \frac{\pi}{6}$

We know that $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$, so $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$.

$\sin \frac{5\pi}{6} = \sin \left( \pi - \frac{\pi}{6} \right) = \sin \frac{\pi}{6}$

We know that $\sin \frac{\pi}{6} = \frac{1}{2}$, so $\sin \frac{5\pi}{6} = \frac{1}{2}$.

Substitute these values back into the expression for $z$:

$z = 4 \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right)$

Distribute the $4$ into the parentheses:

$z = 4 \times \left(-\frac{\sqrt{3}}{2}\right) + 4 \times \left(i \frac{1}{2}\right)$

$z = -2\sqrt{3} + 2i$

This is the complex number $z$ in the standard form $x + iy$, where $x = -2\sqrt{3}$ and $y = 2$.

Thus, the complex number $z$ is $-2\sqrt{3} + 2i$.

To Find: The value of $\left|(1+i) \frac{(2+i)}{(3+i)} \right|$.

Solution:

We need to find the modulus of the given complex expression. We can use the properties of the modulus of complex numbers:

1. $|z_1 z_2| = |z_1| |z_2|$ (Modulus of a product)

2. $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$ (Modulus of a quotient, where $z_2 \ne 0$)

Using these properties, we can write the given expression as:

$\left|(1+i) \frac{(2+i)}{(3+i)} \right| = |1+i| \times \left|\frac{2+i}{3+i}\right|$

$= |1+i| \times \frac{|2+i|}{|3+i|}$

Now, we calculate the modulus of each complex number involved:

For a complex number $z = x + iy$, the modulus is $|z| = \sqrt{x^2 + y^2}$.

For $1+i$: The real part is $x=1$ and the imaginary part is $y=1$.

$|1+i| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$.

For $2+i$: The real part is $x=2$ and the imaginary part is $y=1$.

$|2+i| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$.

For $3+i$: The real part is $x=3$ and the imaginary part is $y=1$.

$|3+i| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$.

Substitute these values back into the expression for the modulus:

$\left|(1+i) \frac{(2+i)}{(3+i)} \right| = \sqrt{2} \times \frac{\sqrt{5}}{\sqrt{10}}$

We can simplify this expression:

$= \frac{\sqrt{2} \times \sqrt{5}}{\sqrt{10}}$

Using the property $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ and $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$:

$= \frac{\sqrt{2 \times 5}}{\sqrt{10}} = \frac{\sqrt{10}}{\sqrt{10}}$

$= 1$

Alternatively, we could first simplify the complex number inside the modulus and then find its modulus.

Let $z = (1+i) \frac{(2+i)}{(3+i)}$.

$z = \frac{(1+i)(2+i)}{3+i} = \frac{2 + i + 2i + i^2}{3+i} = \frac{2 + 3i - 1}{3+i} = \frac{1 + 3i}{3+i}$

Multiply the numerator and denominator by the conjugate of the denominator ($3-i$):

$z = \frac{1 + 3i}{3+i} \times \frac{3-i}{3-i} = \frac{(1)(3) + (1)(-i) + (3i)(3) + (3i)(-i)}{3^2 - i^2}$

$z = \frac{3 - i + 9i - 3i^2}{9 - (-1)} = \frac{3 + 8i - 3(-1)}{9 + 1} = \frac{3 + 8i + 3}{10} = \frac{6 + 8i}{10}$

$z = \frac{6}{10} + \frac{8}{10}i = \frac{3}{5} + \frac{4}{5}i$

Now, find the modulus of this simplified complex number:

$|z| = \left|\frac{3}{5} + \frac{4}{5}i\right| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2}$

$|z| = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{9+16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1$

Both methods yield the same result.

The value of $\left|(1+i) \frac{(2+i)}{(3+i)} \right|$ is $1$.

Given: The complex number $(1+i\sqrt{3})^{2}$.

To Find: The principal argument of $(1+i\sqrt{3})^{2}$.

Solution:

First, let's simplify the complex number $(1+i\sqrt{3})^{2}$.

$(1+i\sqrt{3})^{2} = 1^2 + 2(1)(i\sqrt{3}) + (i\sqrt{3})^2$

Using the property $(a+b)^2 = a^2 + 2ab + b^2$ and $i^2 = -1$:

$(1+i\sqrt{3})^{2} = 1 + 2i\sqrt{3} + i^2 (\sqrt{3})^2$

$= 1 + 2i\sqrt{3} + (-1)(3)$

$= 1 + 2i\sqrt{3} - 3$

$= -2 + 2i\sqrt{3}$

Let $z = -2 + 2i\sqrt{3}$. This is in the form $x + iy$, where $x = -2$ and $y = 2\sqrt{3}$.

To find the principal argument of $z$, we consider the quadrant in which the point $(x, y) = (-2, 2\sqrt{3})$ lies in the complex plane.

Since $x = -2$ is negative and $y = 2\sqrt{3}$ is positive, the point $(-2, 2\sqrt{3})$ lies in the second quadrant.

The principal argument $\theta$ for a complex number $x+iy$ is the angle $\theta$ such that $x = r \cos \theta$ and $y = r \sin \theta$, where $r = |z| = \sqrt{x^2+y^2}$ and $-\pi < \theta \le \pi$.

The reference angle $\alpha$ is given by $\tan \alpha = \left| \frac{y}{x} \right|$.

$\tan \alpha = \left| \frac{2\sqrt{3}}{-2} \right| = |-\sqrt{3}| = \sqrt{3}$.

The angle $\alpha$ such that $\tan \alpha = \sqrt{3}$ and $0 < \alpha < \pi/2$ is $\alpha = \frac{\pi}{3}$.

Since the complex number lies in the second quadrant, the principal argument $\theta$ is given by $\theta = \pi - \alpha$.

$\theta = \pi - \frac{\pi}{3} = \frac{3\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3}$.

The value $\frac{2\pi}{3}$ is in the range $(-\pi, \pi]$, so it is the principal argument.

Alternatively, using polar form properties:

Let $z_0 = 1 + i\sqrt{3}$.

Modulus: $|z_0| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2$.

Argument: The point $(1, \sqrt{3})$ is in the first quadrant. Let $\theta_0 = \text{arg}(z_0)$.

$\tan \theta_0 = \frac{\sqrt{3}}{1} = \sqrt{3}$. Since it's in the first quadrant, $\theta_0 = \frac{\pi}{3}$.

So, $1+i\sqrt{3}$ in polar form is $2 \left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right)$.

Now we need to find the argument of $z = (1+i\sqrt{3})^2$. Using De Moivre's Theorem, if $z_0 = r(\cos \theta_0 + i \sin \theta_0)$, then $z_0^n = r^n (\cos(n\theta_0) + i \sin(n\theta_0))$.

Here $r=2$, $\theta_0 = \frac{\pi}{3}$, and $n=2$.

$z = (1+i\sqrt{3})^2 = 2^2 \left(\cos\left(2 \times \frac{\pi}{3}\right) + i \sin\left(2 \times \frac{\pi}{3}\right)\right)$

$z = 4 \left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right)$

The argument obtained directly from this polar form is $\frac{2\pi}{3}$. This value is within the range $(-\pi, \pi]$, so it is the principal argument.

The principal argument of $(1+i\sqrt{3})^{2}$ is $\frac{2\pi}{3}$.

Given: The condition $\left| \frac{z-5i}{z+5i} \right|=1$.

To Find: Where $z$ lies in the complex plane.

Solution:

The given condition is $\left| \frac{z-5i}{z+5i} \right|=1$.

We can use the property of modulus that for complex numbers $z_1$ and $z_2$ ($z_2 \ne 0$), $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$.

Applying this property, the given equation becomes:

Assuming $z+5i \ne 0$, we can multiply both sides by $|z+5i|$:

This equation has a geometric interpretation. $|z-z_0|$ represents the distance between the complex numbers $z$ and $z_0$ in the complex plane.

The equation $|z-5i| = |z-(-5i)|$ means that the distance of the complex number $z$ from the point representing $5i$ is equal to its distance from the point representing $-5i$.

In the complex plane, $5i$ corresponds to the point $(0, 5)$ on the imaginary axis, and $-5i$ corresponds to the point $(0, -5)$ on the imaginary axis.

The locus of points that are equidistant from two fixed points is the perpendicular bisector of the line segment joining these two points.

The two fixed points here are $(0, 5)$ and $(0, -5)$. The line segment joining these points lies along the y-axis.

The midpoint of the segment joining $(0, 5)$ and $(0, -5)$ is $\left( \frac{0+0}{2}, \frac{5+(-5)}{2} \right) = (0, 0)$.

The segment is vertical. The line perpendicular to a vertical line is a horizontal line. The perpendicular bisector must pass through the midpoint $(0, 0)$.

The equation of a horizontal line passing through $(0, 0)$ is $y=0$.

Alternatively, we can solve this algebraically by letting $z = x+iy$, where $x$ and $y$ are real numbers.

Substitute $z = x+iy$ into the equation $|z-5i| = |z+5i|$:

$|x+iy-5i| = |x+iy+5i|$

$|x + i(y-5)| = |x + i(y+5)|$

Using the definition of modulus, $|a+bi| = \sqrt{a^2+b^2}$:

$\sqrt{x^2 + (y-5)^2} = \sqrt{x^2 + (y+5)^2}$

Square both sides of the equation to eliminate the square roots:

$x^2 + (y-5)^2 = x^2 + (y+5)^2$

Expand the squared terms:

$x^2 + (y^2 - 10y + 25) = x^2 + (y^2 + 10y + 25)$

$x^2 + y^2 - 10y + 25 = x^2 + y^2 + 10y + 25$

Subtract $x^2 + y^2 + 25$ from both sides of the equation:

$-10y = 10y$

Add $10y$ to both sides:

$0 = 20y$

Divide by 20:

$y = 0$

The condition $y=0$ means that the imaginary part of the complex number $z = x+iy$ must be zero. Complex numbers with an imaginary part of zero are real numbers.

In the complex plane, the set of points where the imaginary part is zero is the x-axis, which is also called the real axis.

The condition $\left| \frac{z-5i}{z+5i} \right|=1$ implies that $|z-5i| = |z+5i|$. This means $z$ is equidistant from the points $5i$ and $-5i$. The locus of such points is the perpendicular bisector of the segment joining $5i$ and $-5i$. This perpendicular bisector is the real axis.

Thus, $z$ lies on the real axis.

Given:

Two complex numbers: $z_1 = \sin x + i \cos 2x$ and $z_2 = \cos x - i \sin 2x$.

To Check:

For which value of $x$ are $z_1$ and $z_2$ conjugate to each other?

Solution:

Two complex numbers $z_1 = a + bi$ and $z_2 = c + di$ are conjugate to each other if $z_1 = \overline{z_2}$ or $z_2 = \overline{z_1}$.

Let's assume $z_1 = \overline{z_2}$.

The conjugate of $z_2 = \cos x - i \sin 2x$ is $\overline{z_2} = \overline{\cos x - i \sin 2x} = \cos x + i \sin 2x$.

Setting $z_1 = \overline{z_2}$, we get:

For two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal.

Equating the real parts:

Equating the imaginary parts:

Solve equation (1): $\sin x = \cos x$.

If $\cos x = 0$, then $\sin x = \pm 1$, so $\sin x = \cos x$ is not possible. Thus, $\cos x \ne 0$. We can divide by $\cos x$:

$\frac{\sin x}{\cos x} = 1$

$\tan x = 1$

The general solution for $\tan x = 1$ is $x = n\pi + \frac{\pi}{4}$, where $n$ is an integer ($n \in \mathbb{Z}$).

Solve equation (2): $\cos 2x = \sin 2x$.

If $\cos 2x = 0$, then $\sin 2x = \pm 1$, so $\cos 2x = \sin 2x$ is not possible. Thus, $\cos 2x \ne 0$. We can divide by $\cos 2x$:

$\frac{\sin 2x}{\cos 2x} = 1$

$\tan 2x = 1$

The general solution for $\tan \theta = 1$ is $\theta = m\pi + \frac{\pi}{4}$, where $m$ is an integer ($m \in \mathbb{Z}$).

So, $2x = m\pi + \frac{\pi}{4}$.

Dividing by 2, we get $x = \frac{m\pi}{2} + \frac{\pi}{8}$, where $m$ is an integer ($m \in \mathbb{Z}$).

For the two complex numbers to be conjugates, both conditions must hold simultaneously.

So, we must have $x = n\pi + \frac{\pi}{4}$ and $x = \frac{m\pi}{2} + \frac{\pi}{8}$ for some integers $n$ and $m$.

Set the two expressions for $x$ equal:

$n\pi + \frac{\pi}{4} = \frac{m\pi}{2} + \frac{\pi}{8}$

Divide by $\pi$ (since $\pi \ne 0$):

$n + \frac{1}{4} = \frac{m}{2} + \frac{1}{8}$

Multiply the entire equation by 8 to clear the fractions:

$8 \left( n + \frac{1}{4} \right) = 8 \left( \frac{m}{2} + \frac{1}{8} \right)$

$8n + 8 \times \frac{1}{4} = 8 \times \frac{m}{2} + 8 \times \frac{1}{8}$

$8n + 2 = 4m + 1$

Rearrange the terms to isolate integer variables:

$8n - 4m = 1 - 2$

$8n - 4m = -1$

Factor out 4 from the left side:

$4(2n - m) = -1$

The left side of the equation is an integer multiple of 4, since $n$ and $m$ are integers, so $2n-m$ is an integer. An integer multiple of 4 can only be ..., -8, -4, 0, 4, 8, ...

The right side of the equation is $-1$.

An integer multiple of 4 cannot equal $-1$. Therefore, there are no integers $n$ and $m$ that can satisfy this equation.

This implies that there is no value of $x$ for which both $\sin x = \cos x$ and $\cos 2x = \sin 2x$ are simultaneously true.

Thus, the two given complex numbers are not conjugate to each other for any value of $x$.

The correct option is (D) No value of x.

Given: The complex expression $\frac{1-i \;sin\; \alpha}{1+\;2i\;sin \ \alpha}$, where $\alpha$ is a real value.

To Find: The real value(s) of $\alpha$ for which the expression is purely real.

Solution:

Let the given complex number be $z = \frac{1-i \;sin\; \alpha}{1+\;2i\;sin \ \alpha}$.

A complex number is purely real if its imaginary part is zero.

To find the real and imaginary parts of $z$, we multiply the numerator and the denominator by the conjugate of the denominator.

The denominator is $1+2i \sin \alpha$. Its conjugate is $1-2i \sin \alpha$.

$z = \frac{1-i \sin \alpha}{1+2i \sin \alpha} \times \frac{1-2i \sin \alpha}{1-2i \sin \alpha}$

Calculate the numerator:

Numerator $= (1 - i \sin \alpha)(1 - 2i \sin \alpha)$

$= 1(1) + 1(-2i \sin \alpha) + (-i \sin \alpha)(1) + (-i \sin \alpha)(-2i \sin \alpha)$

$= 1 - 2i \sin \alpha - i \sin \alpha + 2i^2 \sin^2 \alpha$

Using $i^2 = -1$:

$= 1 - 3i \sin \alpha + 2(-1) \sin^2 \alpha$

$= 1 - 2 \sin^2 \alpha - 3i \sin \alpha$

Calculate the denominator:

Denominator $= (1 + 2i \sin \alpha)(1 - 2i \sin \alpha)$

Using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$:

$= 1^2 - (2i \sin \alpha)^2$

$= 1 - (4i^2 \sin^2 \alpha)$

Using $i^2 = -1$:

$= 1 - (4(-1) \sin^2 \alpha)$

$= 1 - (-4 \sin^2 \alpha)$

$= 1 + 4 \sin^2 \alpha$

So, the complex number $z$ is:

$z = \frac{1 - 2 \sin^2 \alpha - 3i \sin \alpha}{1 + 4 \sin^2 \alpha}$

We can write this in the form $x+iy$:

$z = \frac{1 - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} - i \frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha}$

For $z$ to be purely real, the imaginary part must be zero.

Imaginary part $= -\frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha}$

Set the imaginary part equal to zero:

$-\frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha} = 0$

For a fraction to be zero, the numerator must be zero, provided the denominator is non-zero.

The denominator is $1 + 4 \sin^2 \alpha$. Since $\sin^2 \alpha \ge 0$ for real $\alpha$, $4 \sin^2 \alpha \ge 0$. Therefore, $1 + 4 \sin^2 \alpha \ge 1$. The denominator is always non-zero for real $\alpha$.

So, we only need the numerator to be zero:

$3 \sin \alpha = 0$

$\sin \alpha = 0$

The general solution for $\sin \alpha = 0$ is when $\alpha$ is an integer multiple of $\pi$.

$\alpha = n\pi$, where $n$ is any integer ($n \in \mathbb{Z}$).

Now, we compare this solution with the given options:

(A) $(n + 1) \frac{\pi}{2}$: This gives angles like $\frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, ...$ (for $n=0, 1, 2, 3, ...$) or $... -\frac{3\pi}{2}, -\pi, -\frac{\pi}{2}, 0, ...$ (for $n=-3, -2, -1, 0, ...$). For $\alpha = \frac{\pi}{2}$ or $\frac{3\pi}{2}$, $\sin \alpha = \pm 1 \ne 0$. For $\alpha = \pi$ or $2\pi$, $\sin \alpha = 0$. This option does not exclusively produce angles where $\sin \alpha = 0$.

(B) $(2n + 1) \frac{\pi}{2}$: This gives angles like $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, ...$ (for $n=0, 1, 2, ...$) or $... -\frac{3\pi}{2}, -\frac{\pi}{2}, ...$ (for $n=-2, -1, ...$). For these angles, $\sin \alpha = \pm 1 \ne 0$.

(C) $n\pi$: This gives angles like $..., -2\pi, -\pi, 0, \pi, 2\pi, ...$ (for $n \in \mathbb{Z}$). For all these angles, $\sin \alpha = 0$. This matches our solution.

(D) None of these: This is incorrect as option (C) matches our solution.

Assuming $n$ in options (A), (B), and (C) represents an integer, the set of values for $\alpha$ that make the expression purely real is given by $\alpha = n\pi$.

The correct answer is (C) x = nπ.

Given: $z = x + iy$ lies in the third quadrant.

To Find: The condition on $x$ and $y$ such that $\frac{\overline{z}}{z}$ also lies in the third quadrant.

Solution:

If $z = x + iy$ lies in the third quadrant, then its real part $x$ and imaginary part $y$ are both negative. So, we have $x < 0$ and $y < 0$.

Now, let's find the expression for $\frac{\overline{z}}{z}$.

The conjugate of $z = x + iy$ is $\overline{z} = x - iy$.

$\frac{\overline{z}}{z} = \frac{x - iy}{x + iy}$

To express this in the standard form $A + Bi$, we multiply the numerator and denominator by the conjugate of the denominator:

$\frac{\overline{z}}{z} = \frac{x - iy}{x + iy} \times \frac{x - iy}{x - iy}$

$= \frac{(x - iy)^2}{(x + iy)(x - iy)}$

Using the identities $(a-b)^2 = a^2 - 2ab + b^2$ and $(a+b)(a-b) = a^2 - b^2$:

Numerator $= x^2 - 2i xy + (iy)^2 = x^2 - 2ixy + i^2 y^2 = x^2 - 2ixy - y^2$ (since $i^2 = -1$)

Denominator $= x^2 - (iy)^2 = x^2 - i^2 y^2 = x^2 - (-1)y^2 = x^2 + y^2$

So, $\frac{\overline{z}}{z} = \frac{x^2 - y^2 - 2ixy}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2} - i \frac{2xy}{x^2 + y^2}$.

Let $Z = \frac{\overline{z}}{z}$. Then $Z = A + Bi$, where $A = \frac{x^2 - y^2}{x^2 + y^2}$ is the real part and $B = -\frac{2xy}{x^2 + y^2}$ is the imaginary part.

For $Z$ to lie in the third quadrant, its real part $A$ must be negative, and its imaginary part $B$ must be negative.

Condition 1: Real part $A < 0$

$\frac{x^2 - y^2}{x^2 + y^2} < 0$

Since $z$ is in the third quadrant, $x$ and $y$ are non-zero, so $x^2 + y^2 = |z|^2 > 0$. Thus, the inequality holds if the numerator is negative:

$x^2 - y^2 < 0$

Condition 2: Imaginary part $B < 0$

$-\frac{2xy}{x^2 + y^2} < 0$

Since $x^2 + y^2 > 0$, this inequality holds if the numerator is positive:

$-2xy < 0$

$2xy > 0$

We are given that $z$ is in the third quadrant, which means $x < 0$ and $y < 0$.

Check condition (ii): If $x < 0$ and $y < 0$, then their product $xy$ is positive. So, condition (ii) $xy > 0$ is already satisfied when $z$ is in the third quadrant.

Now, consider condition (i): $x^2 < y^2$.

Taking the square root of both sides gives $|x| < |y|$.

Since $x < 0$, $|x| = -x$.

Since $y < 0$, $|y| = -y.

Substituting these into the inequality $|x| < |y|$:

$-x < -y$

Multiplying both sides by $-1$ and reversing the inequality sign:

$x > y$

So, for $z$ in the third quadrant ($x < 0$ and $y < 0$), the condition $x^2 < y^2$ is equivalent to $y < x$.

Combining the conditions for $z$ to be in the third quadrant ($x < 0$ and $y < 0$) and the condition for $\frac{\overline{z}}{z}$ to be in the third quadrant ($y < x$):

We need $x < 0$, $y < 0$, and $y < x$.

This can be written as $y < x < 0$.

Comparing this with the given options:

(A) $x > y > 0$: $x$ and $y$ are positive (z is in the first quadrant).

(B) $x < y < 0$: $x$ and $y$ are negative, but $x < y$ (does not satisfy $y < x$).

(C) $y < x < 0$: $x$ and $y$ are negative, and $y < x$. This matches our condition.

(D) $y > x > 0$: $x$ and $y$ are positive (z is in the first quadrant).

The correct answer is (C) y < x < 0.

Given: The expression $(z + 3) (\overline{z} + 3)$.

To Find: Which of the given options is equivalent to the expression.

Solution:

Let the given expression be $E = (z + 3) (\overline{z} + 3)$.

We use the property of complex conjugates that for any complex numbers $z_1$ and $z_2$, the conjugate of their sum is the sum of their conjugates: $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$.

Let's consider the term $(\overline{z} + 3)$. This is the conjugate of the complex number $(z+3)$, because:

$\overline{z + 3} = \overline{z} + \overline{3}$

Since $3$ is a real number, its conjugate is itself, i.e., $\overline{3} = 3$.

So, $\overline{z + 3} = \overline{z} + 3$.

Now, substitute this back into the given expression:

$E = (z + 3) \overline{(z + 3)}$

We use the property of the modulus of a complex number that for any complex number $w$, $w \overline{w} = |w|^2$.

Let $w = z + 3$. Then the expression becomes $w \overline{w}$.

So, the value of $(z + 3) (\overline{z} + 3)$ is equivalent to $|z + 3|^2$.

Let's verify this algebraically. Let $z = x+iy$, where $x$ and $y$ are real numbers.

Then $\overline{z} = x - iy$.

The expression is $(x+iy + 3)(x-iy + 3)$.

Rearrange the terms:

$((x+3) + iy)((x+3) - iy)$

This is in the form $(A+B)(A-B)$, where $A = x+3$ and $B = iy$.

Using the difference of squares formula, $(A+B)(A-B) = A^2 - B^2$:

$= (x+3)^2 - (iy)^2$

$= (x+3)^2 - i^2 y^2$

Since $i^2 = -1$:

$= (x+3)^2 - (-1)y^2$

$= (x+3)^2 + y^2$

Now, let's look at option (A): $|z + 3|^2$.

Let $w = z + 3 = (x+iy) + 3 = (x+3) + iy$.

The modulus of $w$ is $|w| = \sqrt{(\text{Re}(w))^2 + (\text{Im}(w))^2}$.

$|z+3| = |(x+3) + iy| = \sqrt{(x+3)^2 + y^2}$.

Squaring the modulus:

$|z+3|^2 = \left(\sqrt{(x+3)^2 + y^2}\right)^2 = (x+3)^2 + y^2$.

Since both expressions are equal to $(x+3)^2 + y^2$, they are equivalent.

Comparing our result with the given options:

(A) $|z + 3|^2$: Matches our result.

(B) $|z - 3|$: This is $\sqrt{(x-3)^2 + y^2}$, which is generally not equal to $(x+3)^2 + y^2$.

(C) $z^2 + 3$: This is $(x+iy)^2 + 3 = (x^2 - y^2 + 2ixy) + 3 = (x^2 - y^2 + 3) + 2ixy$. This is generally a complex number, while the given expression is always a real number $|z+3|^2 \ge 0$. So they are not equivalent.

(D) None of these: This is incorrect because option (A) is equivalent.

The correct answer is (A) |z + 3|².

Given: The equation $\left( \frac{1+i}{1-i} \right)^{x} = 1$, where $x$ is an integer.

To Find: The general value of $x$ that satisfies the equation.

Solution:

First, simplify the complex number inside the parentheses, $\frac{1+i}{1-i}$.

Multiply the numerator and denominator by the conjugate of the denominator ($1+i$):

$\frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2}$

Calculate the numerator:

$(1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i$

Calculate the denominator:

$1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$

So, $\frac{1+i}{1-i} = \frac{2i}{2} = i$.

Now, substitute this back into the given equation:

We need to find the integer values of $x$ for which $i^x = 1$.

Recall the powers of $i$:

$i^1 = i$

$i^2 = -1$

$i^3 = i^2 \times i = -1 \times i = -i$

$i^4 = i^2 \times i^2 = (-1) \times (-1) = 1$

The powers of $i$ repeat with a cycle of 4: $i, -1, -i, 1, i, -1, -i, 1, ...$

The value $i^x = 1$ occurs when the exponent $x$ is a multiple of 4.

So, $x$ must be of the form $4n$, where $n$ is an integer ($n \in \mathbb{Z}$).

The question specifies $n \in N$, which usually means the set of natural numbers. There is a convention conflict for whether $0$ is included in $N$. If $N = \{1, 2, 3, ...\}$, the options only cover positive values. If $N = \{0, 1, 2, 3, ...\}$, the options cover non-negative values. Let's check which option fits the general solution $x = 4n$ where $n$ is an integer.

Our solution is $x = 4n$ for $n \in \mathbb{Z}$.

Let's check the options:

(A) $x = 2n + 1$: Odd numbers (e.g., for $n=1$, $x=3$, $i^3 = -i \ne 1$).

(B) $x = 4n$: Multiples of 4 (e.g., for $n=1$, $x=4$, $i^4 = 1$; for $n=0$, $x=0$, $i^0 = 1$; for $n=-1$, $x=-4$, $i^{-4} = (i^4)^{-1} = 1^{-1} = 1$). This matches our general solution for $n \in \mathbb{Z}$.

(C) $x = 2n$: Even numbers (e.g., for $n=1$, $x=2$, $i^2 = -1 \ne 1$).

(D) $x = 4n + 1$: Numbers of the form 1, 5, 9, ... (e.g., for $n=1$, $x=5$, $i^5 = i \ne 1$).

If the option's $n$ is restricted to $N = \{1, 2, 3, ...\}$, then option (B) gives $x = 4, 8, 12, ...$. This is a subset of our solution. However, $x$ can also be 0 (if $n=0$) or negative multiples of 4 (if $n < 0$).

Given the options, option (B) $x = 4n$ is the correct form for the values of $x$ that satisfy $i^x = 1$. The wording "where $n \in N$" in the options might be interpreted as restricting the possible forms of $x$ to positive integers (or non-negative integers if $0 \in N$). However, the problem $i^x=1$ has integer solutions for $x$. Option (B) correctly describes all multiples of 4, provided $n$ is understood as any integer for this specific question context, or it represents the general form where $n$ can be any integer, which is then used to describe the set of solutions.

Assuming the options list the *form* of $x$ rather than restricting $n$ to strictly natural numbers for the solution set, $x=4n$ is the correct form for any integer $n$. If $N$ specifically means $\{1, 2, 3, ...\}$, then the options only cover positive values of $x$. In that case, $x$ must be a positive multiple of 4. Option (B) for $n \in \{1, 2, 3, ...\}$ gives $x \in \{4, 8, 12, ...\}$. However, $i^0 = 1$, so $x=0$ is also a solution. If $N = \{0, 1, 2, 3, ...\}$, then option (B) gives $x \in \{0, 4, 8, 12, ...\}$. This covers the non-negative solutions.

Without a clear definition of $N$ in the problem, the most mathematically accurate representation of the solutions is $x = 4n$ for $n \in \mathbb{Z}$. Among the options provided, $x = 4n$ (option B) is the correct form, and the constraint "where $n \in N$" from the question statement applies to the $n$ in the options. This implies we choose the option that gives the correct condition on $x$. $i^x=1$ if and only if $x$ is a multiple of 4. The set of multiples of 4 can be represented as $\{4n | n \in \mathbb{Z}\}$. Option (B) provides this form. The context of multiple choice questions often implies that one of the options must be the correct general form. Thus, $x = 4n$ is the correct form of the solution.

The correct answer is (B) x = 4n.

Given: The equation $\left(\frac{3-4ix}{3+4ix} \right) = \alpha - i \beta$, where $x$, $\alpha$, and $\beta$ are real numbers.

To Find: The value of $\alpha^2 + \beta^2$.

Solution:

We are given that $\alpha - i \beta = \frac{3-4ix}{3+4ix}$.

We need to find the value of $\alpha^2 + \beta^2$. For a complex number $z = a + bi$, the square of its modulus is $|z|^2 = a^2 + b^2$. In this case, the complex number is $\alpha - i \beta$, and its square of the modulus is $|\alpha - i \beta|^2 = \alpha^2 + (-\beta)^2 = \alpha^2 + \beta^2$.

So, we need to find the square of the modulus of the right side of the equation:

Using the property of modulus that $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$ for $z_2 \ne 0$, we have:

$\left| \frac{3-4ix}{3+4ix} \right| = \frac{|3-4ix|}{|3+4ix|}$

So, $\left| \frac{3-4ix}{3+4ix} \right|^2 = \left( \frac{|3-4ix|}{|3+4ix|} \right)^2 = \frac{|3-4ix|^2}{|3+4ix|^2}$.

Now, we calculate the square of the modulus for the numerator and the denominator.

For a complex number $a+bi$, $|a+bi|^2 = a^2 + b^2$.

The numerator is $3 - 4ix$. The real part is $3$ and the imaginary part is $-4x$.

$|3 - 4ix|^2 = (3)^2 + (-4x)^2 = 9 + 16x^2$.

The denominator is $3 + 4ix$. The real part is $3$ and the imaginary part is $4x$.

$|3 + 4ix|^2 = (3)^2 + (4x)^2 = 9 + 16x^2$.

Note that since $x$ is a real number, $4x$ is also a real number. The denominator $3+4ix$ is zero only if $3=0$ and $4x=0$, which is impossible. So the denominator is never zero for any real value of $x$.

Now substitute the calculated modulus squares back into the expression for $\alpha^2 + \beta^2$:

$\alpha^2 + \beta^2 = \frac{9 + 16x^2}{9 + 16x^2}$

Since the numerator and the denominator are the same and non-zero (as $9+16x^2 \ge 9$), the fraction simplifies to 1.

Thus, the value of $\alpha^2 + \beta^2$ is 1, regardless of the real value of $x$.

Comparing our result with the given options:

(A) 1: Matches our result.

(B) – 1: Does not match.

(C) 2: Does not match.

(D) – 2: Does not match.

The correct answer is (A) 1.

We are asked to identify the correct statement for any two complex numbers $z_1$ and $z_2$ from the given options.

Let's examine each option:

(A) $|z_1z_2| = |z_1| |z_2|$

This statement claims that the modulus of the product of two complex numbers is equal to the product of their moduli. This is a fundamental property of the modulus of complex numbers and is always true for any complex numbers $z_1$ and $z_2$.

Let $z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)$ and $z_2 = r_2 (\cos \theta_2 + i \sin \theta_2)$ be the polar forms of $z_1$ and $z_2$, where $r_1 = |z_1|$, $\theta_1 = \text{arg}(z_1)$, $r_2 = |z_2|$, and $\theta_2 = \text{arg}(z_2)$.

The product is $z_1 z_2 = r_1 r_2 (\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2))$.

The modulus of the product is $|z_1 z_2| = r_1 r_2$.

Since $r_1 = |z_1|$ and $r_2 = |z_2|$, we have $|z_1 z_2| = |z_1| |z_2|$.

This statement is correct.

(B) $\text{arg} (z_1z_2) = \text{arg} (z_1) \cdot \text{arg} (z_2)$

This statement claims that the argument of the product of two complex numbers is equal to the product of their arguments. This is incorrect.

The correct property is that the argument of the product is the sum of the arguments (modulo $2\pi$). $\text{arg}(z_1 z_2) = \text{arg}(z_1) + \text{arg}(z_2) + 2k\pi$, for some integer $k$, provided $z_1, z_2 \ne 0$.

For example, let $z_1 = i$ and $z_2 = i$. $z_1 z_2 = i \times i = i^2 = -1$. $\text{arg}(z_1) = \text{arg}(i) = \frac{\pi}{2}$. $\text{arg}(z_2) = \text{arg}(i) = \frac{\pi}{2}$. $\text{arg}(z_1 z_2) = \text{arg}(-1) = \pi$. Product of arguments = $\text{arg}(z_1) \cdot \text{arg}(z_2) = \frac{\pi}{2} \times \frac{\pi}{2} = \frac{\pi^2}{4}$. Clearly, $\pi \ne \frac{\pi^2}{4}$.

This statement is incorrect.

(C) $|z_1 + z_2| = |z_1| + |z_2|$

This statement claims that the modulus of the sum of two complex numbers is equal to the sum of their moduli. This is the triangle inequality stated as an equality. This is not true for any two complex numbers.

The triangle inequality states that $|z_1 + z_2| \le |z_1| + |z_2|$. The equality holds if and only if $z_1$ and $z_2$ have the same direction (i.e., $\text{arg}(z_1) = \text{arg}(z_2)$ or one of them is zero).

For example, let $z_1 = 1$ and $z_2 = i$. $|z_1 + z_2| = |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}$. $|z_1| + |z_2| = |1| + |i| = 1 + 1 = 2$. $\sqrt{2} \ne 2$.

This statement is incorrect.

(D) $|z_1 + z_2| \ge |z_1| - |z_2|$

This statement is a variation of the triangle inequality (specifically, the reverse triangle inequality is $|z_1+z_2| \ge ||z_1| - |z_2||$). Let's show that $|z_1 + z_2| \ge |z_1| - |z_2|$ is always true.

We know the triangle inequality: $|a+b| \le |a|+|b|$. Let $a = z_1 + z_2$ and $b = -z_2$. Then $|(z_1 + z_2) + (-z_2)| \le |z_1 + z_2| + |-z_2|$. $|z_1| \le |z_1 + z_2| + |z_2|$. Rearranging the inequality, we get: $|z_1| - |z_2| \le |z_1 + z_2|$.

This statement is always true for any complex numbers $z_1$ and $z_2$. Thus, option (D) is correct.

Both option (A) and option (D) are correct statements for any two complex numbers. However, in multiple-choice questions, there is typically only one intended correct answer. Option (A) is a direct property relating the modulus of a product to the product of moduli, whereas (D) is an inequality derived from the triangle inequality for sums. Given the options, (A) is the most fundamental and commonly listed property among the equalities and inequalities provided.

Assuming this is a standard question expecting a direct property match, option (A) is the most likely intended answer.

The correct answer is (A) |z₁z₂| = |z₁| |z₂|.

Given:

The complex number representing a point: $z = 2 - i$.

Rotation is about the origin.

Angle of rotation is $\frac{\pi}{2}$ (90 degrees) in the clockwise direction.

To Find:

The new position of the point after rotation, represented by a new complex number $z'$.

Solution:

Rotation of a complex number $z$ about the origin by an angle $\theta$ is given by multiplying $z$ by $e^{i\theta}$, i.e., $z' = z e^{i\theta}$.

The angle of rotation is $\frac{\pi}{2}$. Clockwise rotation corresponds to a negative angle in the counter-clockwise convention. So, the angle $\theta = -\frac{\pi}{2}$.

The complex number $e^{i\theta}$ is $e^{i(-\pi/2)} = \cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right)$.

We know that $\cos(-\theta) = \cos \theta$ and $\sin(-\theta) = -\sin \theta$.

So, $\cos\left(-\frac{\pi}{2}\right) = \cos \frac{\pi}{2} = 0$.

And $\sin\left(-\frac{\pi}{2}\right) = -\sin \frac{\pi}{2} = -1$.

Thus, $e^{i(-\pi/2)} = 0 + i(-1) = -i$.

The new complex number $z'$ is given by $z' = z \times (-i)$.

Substitute the value of $z = 2 - i$:

$z' = (2 - i) \times (-i)$

$z' = 2(-i) - i(-i)$

$z' = -2i - i^2$

Using $i^2 = -1$:

$z' = -2i - (-1)$

$z' = -2i + 1$

$z' = 1 - 2i$

The new position of the point is represented by the complex number $1 - 2i$. This corresponds to the point $(1, -2)$ in the complex plane.

Let's check the options:

(A) $1 + 2i$: Corresponds to the point $(1, 2)$.

(B) $-1 - 2i$: Corresponds to the point $(-1, -2)$.

(C) $2 + i$: Corresponds to the point $(2, 1)$.

(D) $-1 + 2i$: Corresponds to the point $(-1, 2)$.

None of the options match our calculated result $1 - 2i$. Let's re-check the problem statement and calculation.

Original point $(2, -1)$. Rotating $(x,y)$ by $-90^\circ$ (clockwise) gives $(y, -x)$. So, $(2, -1)$ rotated by $-90^\circ$ is $(-1, -(2)) = (-1, -2)$. The complex number for $(-1, -2)$ is $-1 - 2i$.

Let's re-calculate $z' = z \times (-i)$ carefully:

$z = 2 - i$.

$z' = (2 - i)(-i)$

$z' = (2)(-i) + (-i)(-i)$

$z' = -2i + i^2$

$z' = -2i - 1$

$z' = -1 - 2i$

Yes, the calculation is correct. The new position is $-1 - 2i$.

Let's re-examine the options again, specifically the value corresponding to option (B).

(B) $-1 - 2i$: This matches our calculated result.

The correct answer is (B) –1 – 2i.

Given: A complex number $z = x + iy$, where $x$ and $y$ are real numbers ($x, y \in \mathbb{R}$).

To Find: The condition under which $z$ is a non-real complex number.

Solution:

A complex number $z = x + iy$ is classified as:

1. Purely Real: If the imaginary part is zero, i.e., $y = 0$. In this case, $z = x$, which is a real number.

2. Purely Imaginary: If the real part is zero, i.e., $x = 0$, and the imaginary part is non-zero, i.e., $y \ne 0$. In this case, $z = iy$, where $y \ne 0$.

3. Non-real Complex Number (or simply Complex Number): If the imaginary part is non-zero, i.e., $y \ne 0$. This includes purely imaginary numbers (where $x=0$ and $y \ne 0$) and complex numbers with both non-zero real and imaginary parts (where $x \ne 0$ and $y \ne 0$).

The term "non-real complex number" is used to distinguish complex numbers that are not purely real. A complex number $x+iy$ is not purely real if its imaginary part $y$ is not equal to zero.

So, $x+iy$ is a non-real complex number if and only if $y \ne 0$.

Let's examine the given options:

(A) $x = 0$: If $x=0$, $z = iy$. If $y \ne 0$, this is purely imaginary (which is a non-real complex number). If $y=0$, $z=0$, which is real. So $x=0$ alone does not guarantee that $z$ is non-real.

(B) $y = 0$: If $y=0$, $z = x$. This is always a real number. So this is the condition for the complex number to be real, not non-real.

(C) $x \ne 0$: If $x \ne 0$, $z = x+iy$. If $y=0$, $z=x$, which is real. If $y \ne 0$, $z=x+iy$ is a non-real complex number. So $x \ne 0$ alone does not guarantee that $z$ is non-real.

(D) $y \ne 0$: If $y \ne 0$, the imaginary part of $z$ is non-zero. This means $z = x+iy$ is not a real number. Therefore, it is a non-real complex number.

The correct answer is (D) y $\neq$ 0.

Given: The equality of two complex numbers $a + ib = c + id$, where $a, b, c, d$ are real numbers.

To Find: Which of the given options is correct based on the given equality.

Solution:

The equality of two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ is defined by the condition that their real parts are equal and their imaginary parts are equal.

Given $a + ib = c + id$, where $a, b, c, d \in \mathbb{R}$.

By the definition of equality of complex numbers:

Now, let's examine each option using these conditions.

(A) $a^2 + c^2 = 0$

Since $a=c$, this equation becomes $a^2 + a^2 = 0$, which simplifies to $2a^2 = 0$. This implies $a^2 = 0$, so $a = 0$. Since $a=c$, we also have $c=0$. This condition means that the real parts of the complex numbers must be zero. This is not necessarily true if $a+ib = c+id$. For example, if $z_1 = 1 + 2i$ and $z_2 = 1 + 2i$, then $a=1, b=2, c=1, d=2$. Here $a=c=1 \ne 0$. Thus, $a^2+c^2 = 1^2+1^2 = 2 \ne 0$. So option (A) is incorrect.

(B) $b^2 + c^2 = 0$

Since $b$ and $c$ are real numbers, $b^2 \ge 0$ and $c^2 \ge 0$. Their sum $b^2 + c^2 = 0$ if and only if both $b^2 = 0$ and $c^2 = 0$. This implies $b = 0$ and $c = 0$. Since $a=c$ and $b=d$, this means $a=0$ and $d=0$. This condition requires that the imaginary part of the first number is zero and the real part of the second number is zero. This is not necessarily true if $a+ib = c+id$. For example, if $z_1 = 1 + 2i$ and $z_2 = 1 + 2i$, then $a=1, b=2, c=1, d=2$. Here $b=2 \ne 0$ and $c=1 \ne 0$. Thus, $b^2+c^2 = 2^2+1^2 = 5 \ne 0$. So option (B) is incorrect.

(C) $b^2 + d^2 = 0$

Since $b=d$, this equation becomes $b^2 + b^2 = 0$, which simplifies to $2b^2 = 0$. This implies $b^2 = 0$, so $b = 0$. Since $b=d$, we also have $d=0$. This condition means that the imaginary parts of the complex numbers must be zero. This is not necessarily true if $a+ib = c+id$. For example, if $z_1 = 1 + 2i$ and $z_2 = 1 + 2i$, then $a=1, b=2, c=1, d=2$. Here $b=d=2 \ne 0$. Thus, $b^2+d^2 = 2^2+2^2 = 8 \ne 0$. So option (C) is incorrect.

(D) $a^2 + b^2 = c^2 + d^2$

We have the conditions $a=c$ and $b=d$. Substitute these into the equation $a^2 + b^2 = c^2 + d^2$. The left side is $a^2 + b^2$. The right side is $c^2 + d^2$. Since $c=a$ and $d=b$, the right side becomes $a^2 + b^2$. So, the equation becomes $a^2 + b^2 = a^2 + b^2$, which is always true for any real numbers $a$ and $b$. This means that if $a+ib = c+id$, then the equality $a^2 + b^2 = c^2 + d^2$ must hold.

Alternatively, recall that for a complex number $z = x+iy$, its modulus is $|z| = \sqrt{x^2+y^2}$, and $|z|^2 = x^2+y^2$. The given equality $a+ib = c+id$ means that the two complex numbers are the same. Equal complex numbers have the same modulus. $|a+ib| = |c+id|$ Squaring both sides gives: $|a+ib|^2 = |c+id|^2$ $(a^2 + b^2) = (c^2 + d^2)$.

This equality $a^2 + b^2 = c^2 + d^2$ is always true whenever $a+ib = c+id$.

The correct answer is (D) a² + b² = c² + d².

Given: The condition $\left| \frac{i + z}{i - z} \right| = 1$ for a complex number $z$.

To Find: The locus of $z$ that satisfies the given condition.

Solution:

The given condition is $\left| \frac{i + z}{i - z} \right| = 1$.

We can use the property of modulus that $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$ for $z_2 \ne 0$.

Applying this property, the given equation becomes:

Provided $i - z \ne 0$, we can multiply both sides by $|i - z|$:

This equation has a geometric interpretation. $|z_1 - z_2|$ represents the distance between the complex numbers $z_1$ and $z_2$ in the complex plane.

The equation can be rewritten as $|z - (-i)| = |z - i|$. This means that the distance of the complex number $z$ from the point representing $-i$ is equal to its distance from the point representing $i$.

In the complex plane, $-i$ corresponds to the point $(0, -1)$ on the imaginary axis, and $i$ corresponds to the point $(0, 1)$ on the imaginary axis.

The locus of points that are equidistant from two fixed points is the perpendicular bisector of the line segment joining these two points.

The two fixed points here are $(0, -1)$ and $(0, 1)$. The line segment joining these points lies along the y-axis.

The midpoint of the segment joining $(0, -1)$ and $(0, 1)$ is $\left( \frac{0+0}{2}, \frac{-1+1}{2} \right) = (0, 0)$.

The segment is vertical. The line perpendicular to a vertical line is a horizontal line. The perpendicular bisector must pass through the midpoint $(0, 0)$.

The equation of a horizontal line passing through $(0, 0)$ is $y=0$.

In the complex plane, the set of points where the imaginary part is zero is the x-axis, which is also called the real axis.

We also need to consider the condition $i - z \ne 0$, which means $z \ne i$. If $z=i$, the original expression is undefined. The point $z=i$ is $(0, 1)$, which is one of the fixed points and is on the perpendicular bisector $y=0$. However, the original expression requires the denominator to be non-zero. The locus we found $y=0$ includes the origin $(0,0)$, which is fine. The point $z=i$ is not on the line $y=0$. So, the condition $z \ne i$ is automatically satisfied by the points on the locus $y=0$.

Alternatively, we can solve this algebraically by letting $z = x+iy$, where $x$ and $y$ are real numbers.

Substitute $z = x+iy$ into the equation $|i + z| = |i - z|$:

$|i + (x+iy)| = |i - (x+iy)|$

$|x + i(y+1)| = |-x + i(1-y)|$

Using the definition of modulus, $|a+bi| = \sqrt{a^2+b^2}$:

$\sqrt{x^2 + (y+1)^2} = \sqrt{(-x)^2 + (1-y)^2}$

Square both sides of the equation to eliminate the square roots:

$x^2 + (y+1)^2 = (-x)^2 + (1-y)^2$

$x^2 + (y^2 + 2y + 1) = x^2 + (1 - 2y + y^2)$

$x^2 + y^2 + 2y + 1 = x^2 + y^2 - 2y + 1$

Subtract $x^2 + y^2 + 1$ from both sides of the equation:

$2y = -2y$

Add $2y$ to both sides:

$4y = 0$

$y = 0$

The condition $y=0$ means that the imaginary part of the complex number $z = x+iy$ must be zero. Complex numbers with an imaginary part of zero are real numbers, which lie on the real axis (the x-axis) in the complex plane.

Comparing this with the given options:

(A) circle $x^2 + y^2 = 1$: This is the unit circle centered at the origin.

(B) the x-axis: This is the line $y=0$. This matches our result.

(C) the y-axis: This is the line $x=0$.

(D) the line $x + y = 1$.

The complex numbers $z$ satisfying the condition lie on the x-axis.

The correct answer is (B) the x-axis.

Given: $z$ is a complex number.

To Find: The correct relationship between $|z^2|$ and $|z|^2$.

Solution:

We use the property of the modulus of complex numbers that for any two complex numbers $z_1$ and $z_2$, $|z_1 z_2| = |z_1| |z_2|$.

Consider $z^2 = z \times z$.

Using the property of the modulus of a product, we have:

So,

Thus, for any complex number $z$, the modulus of $z^2$ is equal to the square of the modulus of $z$.

Let's verify this with an example. Let $z = 3 + 4i$.

First, calculate $z^2$:

$z^2 = (3 + 4i)^2 = 3^2 + 2(3)(4i) + (4i)^2 = 9 + 24i + 16i^2 = 9 + 24i - 16 = -7 + 24i$.

Now, calculate $|z^2|$:

$|z^2| = |-7 + 24i| = \sqrt{(-7)^2 + (24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.

Next, calculate $|z|$:

$|z| = |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

Now, calculate $|z|^2$:

$|z|^2 = (5)^2 = 25$.

We see that $|z^2| = 25$ and $|z|^2 = 25$, so $|z^2| = |z|^2$ holds in this example.

Comparing our result with the given options:

(A) $|z^2| > |z|^2$: Incorrect.

(B) $|z^2| = |z|^2$: Correct.

(C) $|z^2| < |z|^2$: Incorrect.

(D) $|z^2| \ge |z|^2$: This is true, but option (B) is a more specific and precise equality that holds for all complex numbers. If (B) is correct, then (D) is also technically correct, but (B) is the primary property.

In multiple choice questions, the best or most specific correct answer is usually expected. The equality $|z^2| = |z|^2$ is always true.

The correct answer is (B) |z²| = |z|².

Given: The equation $|z_1 + z_2| = |z_1| + |z_2|$, where $z_1$ and $z_2$ are complex numbers.

To Find: The condition under which this equality holds.

Solution:

The inequality $|z_1 + z_2| \le |z_1| + |z_2|$ is known as the triangle inequality for complex numbers. The equality $|z_1 + z_2| = |z_1| + |z_2|$ holds if and only if the complex numbers $z_1$ and $z_2$ lie on the same ray from the origin in the complex plane. This occurs in two cases:

Case 1: One or both of the complex numbers are zero. If $z_1 = 0$, then $|0 + z_2| = |z_2|$ and $|0| + |z_2| = 0 + |z_2| = |z_2|$. So the equality holds. Similarly, if $z_2 = 0$, the equality holds.

Case 2: Both complex numbers are non-zero, i.e., $z_1 \ne 0$ and $z_2 \ne 0$. In this case, the equality $|z_1 + z_2| = |z_1| + |z_2|$ holds if and only if $z_1$ and $z_2$ have the same direction. Having the same direction means that the argument of $z_1$ is equal to the argument of $z_2$ (possibly differing by a multiple of $2\pi$). This condition is expressed as $\text{arg}(z_1) = \text{arg}(z_2)$. Alternatively, this means that the ratio $\frac{z_1}{z_2}$ (or $\frac{z_2}{z_1}$) is a positive real number. For instance, $z_2 = k z_1$ for some real number $k \ge 0$. If $k=0$, then $z_2=0$, which is Case 1. If $k>0$, then $z_1$ and $z_2$ are in the same direction, and $\text{arg}(z_1) = \text{arg}(z_2)$.

Let's check the given options:

(A) $z_2 = \overline{z_1}$: This condition implies $|z_2| = |\overline{z_1}| = |z_1|$. If $z_1 = 1+i$, $z_2 = 1-i$. $|z_1+z_2| = |2|=2$. $|z_1|+|z_2| = \sqrt{2}+\sqrt{2} = 2\sqrt{2}$. $2 \ne 2\sqrt{2}$. This condition is not sufficient in general.

(B) $z_2 = \frac{1}{z_{1}}$: Assume $z_1 \ne 0$. This condition implies $|z_2| = |\frac{1}{z_1}| = \frac{1}{|z_1|}$. If $z_1=i$, $z_2 = 1/i = -i$. $|z_1+z_2| = |0|=0$. $|z_1|+|z_2| = 1+1=2$. $0 \ne 2$. This condition is not sufficient in general.

(C) $\text{arg} (z_1) = \text{arg} (z_2)$: Assuming $z_1 \ne 0$ and $z_2 \ne 0$, this condition implies that $z_1$ and $z_2$ lie on the same ray from the origin. As discussed above, this is the condition for the equality to hold in the triangle inequality when both numbers are non-zero. If this condition is met, the equality $|z_1 + z_2| = |z_1| + |z_2|$ holds.

(D) $|z_1| = |z_2|$: This condition means $z_1$ and $z_2$ lie on the same circle centered at the origin. If $z_1=1$ and $z_2=i$, then $|z_1|=|z_2|=1$. $|z_1+z_2| = |1+i| = \sqrt{2}$. $|z_1|+|z_2| = 1+1=2$. $\sqrt{2} \ne 2$. This condition is not sufficient in general.

Among the given options, the condition $\text{arg}(z_1) = \text{arg}(z_2)$ is the correct condition for the equality $|z_1 + z_2| = |z_1| + |z_2|$ to be possible (specifically when $z_1$ and $z_2$ are non-zero). If this condition is met, the equality holds.

The correct answer is (C) arg (z₁) = arg (z₂).

Given: The complex expression $\frac{1 \;+\;i\; cos\;θ}{1\;-\;2i\;cos\;θ}$, where $\theta$ is a real value.

To Find: The real value(s) of $\theta$ for which the expression is a real number.

Solution:

Let the given complex number be $z = \frac{1 \;+\;i\; cos\;θ}{1\;-\;2i\;cos\;θ}$.

A complex number is a real number if its imaginary part is zero.

To find the real and imaginary parts of $z$, we multiply the numerator and the denominator by the conjugate of the denominator.

The denominator is $1-2i \cos \theta$. Its conjugate is $1+2i \cos \theta$.

$z = \frac{1+i \cos \theta}{1-2i \cos \theta} \times \frac{1+2i \cos \theta}{1+2i \cos \theta}$

Calculate the numerator:

Numerator $= (1 + i \cos \theta)(1 + 2i \cos \theta)$

$= 1(1) + 1(2i \cos \theta) + (i \cos \theta)(1) + (i \cos \theta)(2i \cos \theta)$

$= 1 + 2i \cos \theta + i \cos \theta + 2i^2 \cos^2 \theta$

Using $i^2 = -1$:

$= 1 + 3i \cos \theta + 2(-1) \cos^2 \theta$

$= 1 - 2 \cos^2 \theta + 3i \cos \theta$

Calculate the denominator:

Denominator $= (1 - 2i \cos \theta)(1 + 2i \cos \theta)$

Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$:

$= 1^2 - (2i \cos \theta)^2$

$= 1 - (4i^2 \cos^2 \theta)$

Using $i^2 = -1$:

$= 1 - (4(-1) \cos^2 \theta)$

$= 1 - (-4 \cos^2 \theta)$

$= 1 + 4 \cos^2 \theta$

So, the complex number $z$ is:

$z = \frac{1 - 2 \cos^2 \theta + 3i \cos \theta}{1 + 4 \cos^2 \theta}$

We can write this in the form $x+iy$:

$z = \frac{1 - 2 \cos^2 \theta}{1 + 4 \cos^2 \theta} + i \frac{3 \cos \theta}{1 + 4 \cos^2 \theta}$

For $z$ to be a real number, the imaginary part must be zero.

Imaginary part $= \frac{3 \cos \theta}{1 + 4 \cos^2 \theta}$

Set the imaginary part equal to zero:

$\frac{3 \cos \theta}{1 + 4 \cos^2 \theta} = 0$

For a fraction to be zero, the numerator must be zero, provided the denominator is non-zero.

The denominator is $1 + 4 \cos^2 \theta$. Since $\cos^2 \theta \ge 0$ for real $\theta$, $4 \cos^2 \theta \ge 0$. Therefore, $1 + 4 \cos^2 \theta \ge 1$. The denominator is always non-zero for real $\theta$.

So, we only need the numerator to be zero:

$3 \cos \theta = 0$

$\cos \theta = 0$

The general solution for $\cos \theta = 0$ is when $\theta$ is an odd multiple of $\frac{\pi}{2}$.

$\theta = (2n + 1) \frac{\pi}{2}$, where $n$ is any integer ($n \in \mathbb{Z}$).

Now, we compare this solution with the given options:

(A) $n\pi + \frac{\pi}{4}$: This gives angles where $\cos \theta = \pm \frac{1}{\sqrt{2}} \ne 0$.

(B) $n\pi + (-1)^n \frac{\pi}{4}$: This gives angles where $|\cos \theta| = \frac{1}{\sqrt{2}} \ne 0$.

(C) $2n\pi \pm \frac{\pi}{2}$: This represents angles of the form $2n\pi + \frac{\pi}{2}$ and $2n\pi - \frac{\pi}{2}$. $2n\pi + \frac{\pi}{2} = (4n+1)\frac{\pi}{2}$. This represents odd multiples of $\frac{\pi}{2}$. $2n\pi - \frac{\pi}{2} = (4n-1)\frac{\pi}{2}$. This also represents odd multiples of $\frac{\pi}{2}$. The union of these two sets covers all odd multiples of $\frac{\pi}{2}$. This matches our solution $\theta = (2k+1)\frac{\pi}{2}$ for any integer $k$. For $\theta$ being an odd multiple of $\frac{\pi}{2}$, $\cos \theta = 0$. Thus, this option is correct.

(D) none of these: This is incorrect as option (C) matches our solution.

The correct answer is (C) 2nπ ± $\frac{\pi}{2}$.

Given: A real number $x < 0$.

To Find: The principal argument of $x$, denoted as $\text{arg}(x)$.

Solution:

A real number $x$ can be represented as a complex number $z = x + 0i$.

If $x < 0$, the complex number $z$ lies on the negative real axis in the complex plane.

The principal argument of a complex number $z = x+iy$ is the angle $\theta$ measured counter-clockwise from the positive real axis to the line segment connecting the origin to the point $(x,y)$, such that $-\pi < \theta \le \pi$.

For a point on the negative real axis (where the real part is negative and the imaginary part is zero), the angle made with the positive real axis is $\pi$ radians (or $180^\circ$).

Since $\pi$ is in the range $(-\pi, \pi]$, the principal argument of $x$ when $x < 0$ is $\pi$.

Let's verify this using the polar form $z = |z| (\cos \theta + i \sin \theta)$.

If $z = x$ and $x < 0$, then $|z| = |x| = -x$ (since $x$ is negative).

So, $x = (-x) (\cos \theta + i \sin \theta)$.

Dividing by $-x$ (which is a positive number):

$\frac{x}{-x} = \cos \theta + i \sin \theta$

$-1 = \cos \theta + i \sin \theta$

For this equality to hold, the real parts must be equal, and the imaginary parts must be equal:

The angle $\theta$ in the interval $(-\pi, \pi]$ that satisfies both $\cos \theta = -1$ and $\sin \theta = 0$ is $\theta = \pi$.

Comparing this with the given options:

(A) 0: This is the argument for positive real numbers ($x > 0$).

(B) $\frac{\pi}{2}$: This is the argument for purely imaginary numbers on the positive imaginary axis ($z = iy, y > 0$).

(C) $\pi$: This is the argument for negative real numbers ($x < 0$). This matches our result.

(D) none of these: This is incorrect as option (C) matches our result.

The correct answer is (C) π.

Given:

The function $f(z) = \frac{7-z}{1-z^{2}}$.

The complex number $z = 1 + 2i$.

To Find:

The value of $|f(z)|$ when $z = 1 + 2i$, and identify which option matches this value in terms of $|z|$.

Solution:

Substitute the value of $z = 1 + 2i$ into the expression for $f(z)$.

$f(z) = f(1 + 2i) = \frac{7 - (1 + 2i)}{1 - (1 + 2i)^2}$

First, simplify the numerator:

Numerator $= 7 - (1 + 2i) = 7 - 1 - 2i = 6 - 2i$.

Next, simplify the denominator. We need to calculate $z^2$ first.

$z^2 = (1 + 2i)^2$

Using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$z^2 = 1^2 + 2(1)(2i) + (2i)^2$

$z^2 = 1 + 4i + 4i^2$

Since $i^2 = -1$:

$z^2 = 1 + 4i + 4(-1)$

$z^2 = 1 + 4i - 4$

$z^2 = -3 + 4i$

Now, calculate the denominator $1 - z^2$:

Denominator $= 1 - (-3 + 4i)$

Denominator $= 1 + 3 - 4i$

Denominator $= 4 - 4i$

So, $f(z)$ becomes:

$f(z) = \frac{6 - 2i}{4 - 4i}$

We can simplify this fraction by dividing both the numerator and the denominator by 2:

$f(z) = \frac{3 - i}{2 - 2i}$

To express $f(z)$ in the standard form $a+bi$, multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $2 - 2i$ is $2 + 2i$.

$f(z) = \frac{3 - i}{2 - 2i} \times \frac{2 + 2i}{2 + 2i}$

Calculate the numerator of the product:

Numerator $= (3 - i)(2 + 2i) = 3(2) + 3(2i) - i(2) - i(2i)$

$= 6 + 6i - 2i - 2i^2$

$= 6 + 4i - 2(-1)$

$= 6 + 4i + 2$

$= 8 + 4i$

Calculate the denominator of the product:

Denominator $= (2 - 2i)(2 + 2i) = 2^2 - (2i)^2$

$= 4 - 4i^2$

$= 4 - 4(-1)$

$= 4 + 4$

$= 8$

So, the simplified form of $f(z)$ is:

$f(z) = \frac{8 + 4i}{8} = \frac{8}{8} + \frac{4i}{8} = 1 + \frac{1}{2}i$.

Now, calculate the modulus of $f(z) = 1 + \frac{1}{2}i$.

For a complex number $a+bi$, the modulus is $|a+bi| = \sqrt{a^2 + b^2}$.

$|f(z)| = \left|1 + \frac{1}{2}i\right| = \sqrt{1^2 + \left(\frac{1}{2}\right)^2}$

$|f(z)| = \sqrt{1 + \frac{1}{4}}$

$|f(z)| = \sqrt{\frac{4}{4} + \frac{1}{4}}$

$|f(z)| = \sqrt{\frac{5}{4}}$

$|f(z)| = \frac{\sqrt{5}}{\sqrt{4}} = \frac{\sqrt{5}}{2}$.

Next, calculate the modulus of the given $z = 1 + 2i$.

$|z| = |1 + 2i| = \sqrt{1^2 + 2^2}$

$|z| = \sqrt{1 + 4}$

$|z| = \sqrt{5}$.

Now, let's check the given options and compare their values with $|f(z)| = \frac{\sqrt{5}}{2}$ using $|z| = \sqrt{5}$.

(A) $\frac{|z|}{2}$: Substitute $|z| = \sqrt{5}$: $\frac{\sqrt{5}}{2}$. This matches $|f(z)|$.

(B) $|z|$: Substitute $|z| = \sqrt{5}$: $\sqrt{5}$. This does not match $|f(z)|$.

(C) $2|z|$: Substitute $|z| = \sqrt{5}$: $2\sqrt{5}$. This does not match $|f(z)|$.

(D) none of these: This is incorrect because option (A) matches the value of $|f(z)|$.

The value of $|f(z)|$ is $\frac{\sqrt{5}}{2}$, which is equal to $\frac{|z|}{2}$.

The correct answer is (A) $\frac{|z|}{2}$.