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Chapter 5 Complex Numbers and Quadratic Equations (Concepts)
Embark on a significant expansion of our numerical landscape as we introduce the fascinating realm of Complex Numbers. Our journey through the number system, from natural numbers to integers, rationals, and finally real numbers, might seem complete. However, the real number system has inherent limitations, most notably revealed when attempting to solve certain algebraic equations. Consider a simple quadratic equation like $x^2 + 1 = 0$. Within the confines of real numbers, there is no value for $x$ whose square is $-1$. To overcome this algebraic barrier and provide solutions to all polynomial equations, mathematicians conceived of extending the number system further. This chapter introduces the fundamental building block for this extension and explores the rich algebra and geometry of these new numbers.
The journey begins by introducing the imaginary unit, denoted by the symbol $i$. This unique entity is defined specifically as the square root of negative one: $\mathbf{i = \sqrt{-1}}$. From this definition, it immediately follows that $\mathbf{i^2 = -1}$. This single definition opens the door to a whole new system. A complex number is then formally defined as any number that can be expressed in the standard form $\mathbf{z = a + ib}$ (or $a+bi$), where $a$ and $b$ are real numbers. In this form:
- $a$ is called the real part of the complex number $z$, denoted as $\text{Re}(z)$.
- $b$ is called the imaginary part of the complex number $z$, denoted as $\text{Im}(z)$.
We then explore the algebra of complex numbers, defining operations in a way that is consistent with real number arithmetic while incorporating the property $i^2 = -1$:
- Addition: $(a+ib) + (c+id) = (a+c) + i(b+d)$ (Add real parts and imaginary parts separately).
- Subtraction: $(a+ib) - (c+id) = (a-c) + i(b-d)$
- Multiplication: $(a+ib)(c+id) = ac + iad + ibc + i^2bd = (ac - bd) + i(ad + bc)$ (Multiply as binomials, then substitute $i^2 = -1$).
- Division: To divide $\frac{a+ib}{c+id}$, we multiply the numerator and denominator by the conjugate of the denominator.
Complex numbers lend themselves beautifully to geometric visualization on the Argand plane (or complex plane). This is similar to the Cartesian plane, but the horizontal axis represents the real part (Real Axis) and the vertical axis represents the imaginary part (Imaginary Axis). Each complex number $a+ib$ corresponds uniquely to the point $(a, b)$ on this plane. This geometric view allows us to introduce the polar representation (or trigonometric form) of a complex number. Instead of using rectangular coordinates $(a, b)$, we use the distance from the origin $r = |z|$ (the modulus) and the angle $\theta$ (theta) made by the line connecting the origin to the point $(a,b)$ with the positive real axis. This angle $\theta$ is called the argument or amplitude of $z$, denoted as $\arg(z)$. The polar form is written as $\mathbf{z = r(\cos \theta + i \sin \theta)}$.
Finally, equipped with complex numbers, we revisit quadratic equations $ax^2 + bx + c = 0$. The limitation encountered when the discriminant $D = b^2 - 4ac < 0$ is now overcome. Since $D$ is negative, we can write $D = - (4ac - b^2)$, where $4ac - b^2$ is positive. Then $\sqrt{D} = \sqrt{-(4ac - b^2)} = \sqrt{-1} \sqrt{4ac - b^2} = i\sqrt{4ac - b^2}$. The quadratic formula still applies, yielding complex roots: $$ \mathbf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}} $$ Thus, within the complex number system $\mathbb{C}$, every quadratic equation has exactly two roots (which may be real and distinct, real and equal, or a complex conjugate pair). We might also briefly touch upon methods for finding the square root of a complex number itself.
Complex Numbers
In the journey of mathematics, the number system has been progressively extended to solve equations that were previously unsolvable within the existing system. We started with natural numbers, extended to integers to solve $x+a=b$, then to rational numbers to solve $ax=b$, and further to real numbers to handle limits and equations like $x^2=2$. However, even within the set of real numbers, there are simple equations like the quadratic equation $x^2 + 1 = 0$ that have no real solutions. This limitation led to the invention of complex numbers.
The Imaginary Unit $i$
The core idea behind extending the real number system to include solutions for equations like $x^2 = -1$ was the introduction of a new entity. This new number, called the imaginary unit, is denoted by the symbol $i$. It is defined by the fundamental property that its square is equal to $-1$.
$i^2 = -1$
... (1)
Based on this definition, we can formally say that $i$ is the square root of -1. While $\sqrt{-1} = i$ is often used, it's important to strictly use the property $i^2 = -1$ in calculations involving powers of $i$ to avoid potential ambiguities with properties of square roots of positive numbers when extended to negative numbers.
$i = \sqrt{-1}$
With the imaginary unit $i$, we can now express the square root of any negative real number. For any positive real number $a$, $\sqrt{-a} = \sqrt{a \times -1} = \sqrt{a} \times \sqrt{-1} = \sqrt{a}i$. For example:
- $\sqrt{-4} = \sqrt{4} \times \sqrt{-1} = 2i$
- $\sqrt{-9} = \sqrt{9} \times \sqrt{-1} = 3i$
- $\sqrt{-7} = \sqrt{7} \times \sqrt{-1} = \sqrt{7}i$
Now, the equation $x^2 + 1 = 0$ can be solved: $x^2 = -1 \implies x = \pm \sqrt{-1} = \pm i$. The solutions are $x=i$ and $x=-i$.
Definition of a Complex Number
By combining real numbers with the imaginary unit $i$, we form a new set of numbers called complex numbers. A complex number is formally defined as a number that can be written in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit ($i^2 = -1$).
Complex numbers are typically denoted by the variable $z$. So, the standard form of a complex number is:
$z = a + bi$
... (2)
In the complex number $z = a + bi$:
- The real number $a$ is called the real part of $z$. It is denoted by $\text{Re}(z)$.
- The real number $b$ is called the imaginary part of $z$. It is denoted by $\text{Im}(z)$.
It is crucial to remember that the imaginary part is the real number $b$, not the term $bi$.
Examples of Complex Numbers and their Parts
Let's identify the real and imaginary parts of some complex numbers:
- $z = 3 + 4i$: $\text{Re}(z) = 3$, $\text{Im}(z) = 4$.
- $z = -2 + i$: $\text{Re}(z) = -2$, $\text{Im}(z) = 1$ (since $i = 1i$).
- $z = 5$: This is a real number. It can be written as $5 + 0i$. $\text{Re}(z) = 5$, $\text{Im}(z) = 0$. A real number is a complex number with an imaginary part equal to zero.
- $z = -6i$: This is a purely imaginary number. It can be written as $0 - 6i$. $\text{Re}(z) = 0$, $\text{Im}(z) = -6$. A complex number with a real part equal to zero is called a purely imaginary number (unless $b=0$ also, in which case it's $0+0i = 0$, which is both real and purely imaginary).
The Set of Complex Numbers
Having defined the real part and the imaginary part, we can now formally define the entire universe of numbers they belong to. The collection of all numbers that can be written in the form $a + bi$, where $a$ and $b$ are any real numbers, is known as the set of complex numbers.
This all-encompassing set is denoted by the symbol $\mathbb{C}$.
Formal Definition
Using set-builder notation, the set of complex numbers is defined as:
$\mathbb{C} = \{ a + bi : a \in \mathbb{R}, b \in \mathbb{R} \}$
... (3)
Let's break down what this notation means:
- The symbol $\mathbb{C}$ is the name for the set.
- The curly braces $\{ ... \}$ mean "the set of all elements..."
- The expression $a + bi$ describes the form of every element in the set.
- The colon ":" is read as "such that".
- The condition $a \in \mathbb{R}, b \in \mathbb{R}$ means that both the real part '$a$' and the imaginary part '$b$' must be real numbers.
In simple terms, the definition says: "$\mathbb{C}$ is the set of all numbers that can be written in the form $a+bi$, provided that $a$ and $b$ are real numbers."
The Family of Numbers: How Complex Numbers Fit In
A key concept to understand is that the set of complex numbers does not exist separately from real numbers; it actually contains them. The complex number system is an extension of the real number system.
Any real number, let's say $a$, can be written as a complex number by giving it an imaginary part of zero.
$a = a + 0i$
Because every real number can be expressed in this form, the set of all real numbers ($\mathbb{R}$) is a subset of the set of all complex numbers ($\mathbb{C}$).
Let's look at some examples:
- The real number 5 is a complex number because it can be written as $5 + 0i$. (Here, $a=5, b=0$).
- The purely imaginary number 3i is a complex number because it can be written as $0 + 3i$. (Here, $a=0, b=3$).
- A number like -2 + 7i is a complex number where both parts are non-zero. (Here, $a=-2, b=7$).
This shows that the set $\mathbb{C}$ is a "super set" that includes all other number systems you have learned about. The hierarchy is as follows:
This relationship is written formally as: $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$.
Equality of Complex Numbers
Similar to how we define equality for ordered pairs or vectors based on their components, the equality of two complex numbers is defined based on their real and imaginary parts. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
Let $z_1 = a + bi$ and $z_2 = c + di$ be two complex numbers, where $a, b, c,$ and $d$ are real numbers.
Then, $z_1$ is equal to $z_2$ if and only if the real part of $z_1$ equals the real part of $z_2$, AND the imaginary part of $z_1$ equals the imaginary part of $z_2$.
$a + bi = c + di \iff a = c \text{ and } b = d$
... (4)
This definition provides a method to solve equations where both sides are expressed as complex numbers in the form $x + iy$. By equating the real and imaginary parts, we obtain a system of two equations involving real numbers, which we can then solve.
Example 1. Find the values of $x$ and $y$ if $(x + 3) + i(y - 2) = 5 + i(4)$, where $x$ and $y$ are real numbers.
Answer:
Given:
The equation $(x + 3) + i(y - 2) = 5 + 4i$.
We are also given that $x \in \mathbb{R}$ and $y \in \mathbb{R}$.
To Find:
The values of $x$ and $y$.
Solution:
The given equation is an equality between two complex numbers written in the standard form $a + bi$.
The left side of the equation is $(x + 3) + i(y - 2)$.
Here, the real part is $(x + 3)$.
The imaginary part is $(y - 2)$.
The right side of the equation is $5 + 4i$.
Here, the real part is 5.
The imaginary part is 4.
According to the definition of equality of complex numbers, two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
Equating the real parts of both sides:
$x + 3 = 5$
Solving for $x$:
$x = 5 - 3$
$x = 2$
Equating the imaginary parts of both sides:
$y - 2 = 4$
Solving for $y$:
$y = 4 + 2$
$y = 6$
Thus, the values of the real numbers $x$ and $y$ that satisfy the given equation are $x=2$ and $y=6$.
The final answer is x = 2, y = 6.
Algebra of Complex Numbers
Just like real numbers, complex numbers can be combined using the basic arithmetic operations: addition, subtraction, multiplication, and division. These operations are defined in a way that extends the properties of real numbers and incorporates the property of the imaginary unit $i$, $i^2 = -1$.
1. Addition of Complex Numbers
Adding complex numbers is straightforward. To find the sum of two complex numbers, we simply add their corresponding real parts together and their corresponding imaginary parts together.
Let $z_1 = a + bi$ and $z_2 = c + di$ be two complex numbers, where $a, b, c,$ and $d$ are real numbers.
Their sum $z_1 + z_2$ is defined as:
$z_1 + z_2 = (a + bi) + (c + di)$
Grouping the real and imaginary terms:
= $(a + c) + (bi + di)$
Factoring out $i$ from the imaginary terms:
$z_1 + z_2 = (a + c) + i(b + d)$
... (1)
The result is a new complex number with real part $(a+c)$ and imaginary part $(b+d)$.
Properties of Addition of Complex Numbers
Complex number addition satisfies the same fundamental properties as real number addition:
(i) Closure Property: The sum of two complex numbers is always a complex number. If $z_1, z_2 \in \mathbb{C}$, then $z_1 + z_2 \in \mathbb{C}$.
(ii) Commutative Law: The order of addition does not matter. For any complex numbers $z_1, z_2$, $z_1 + z_2 = z_2 + z_1$. This holds because addition of real numbers is commutative ($a+c = c+a$ and $b+d = d+b$).
(iii) Associative Law: The grouping of numbers in addition does not matter. For any complex numbers $z_1, z_2, z_3$, $(z_1 + z_2) + z_3 = z_1 + (z_2 + z_3)$. This holds because addition of real numbers is associative.
(iv) Existence of Additive Identity (Zero Element): There exists a unique complex number, $0 + 0i$, denoted simply as 0, which acts as the additive identity. Adding 0 to any complex number leaves it unchanged.
For any $z = a + bi \in \mathbb{C}$, we can show this as follows:
$z + (0 + 0i) = (a + bi) + (0 + 0i)$
By adding the real parts and the imaginary parts separately, we get:
$= (a + 0) + i(b + 0)$
This simplifies to:
$= a + bi$
Which is equal to the original complex number:
$= z$
(v) Existence of Additive Inverse (Negative of a Complex Number): For every complex number $z = a + bi$, there exists a unique complex number, denoted by $-z$, such that their sum is the additive identity (0). The additive inverse of $z=a+bi$ is $-z = -a - bi$.
For any $z = a + bi \in \mathbb{C}$, there exists its additive inverse $-z = -a - bi \in \mathbb{C}$. Their sum is calculated as follows:
$z + (-z) = (a + bi) + (-a - bi)$
By grouping the real parts and the imaginary parts, we get:
$= (a - a) + i(b - b)$
This simplifies to:
$= 0 + 0i$
Which is the additive identity:
$= 0$
2. Subtraction of Complex Numbers
Subtracting one complex number from another is defined as adding the additive inverse of the second complex number. In practice, it's similar to addition: subtract the real parts and subtract the imaginary parts separately.
Let $z_1 = a + bi$ and $z_2 = c + di$.
$z_1 - z_2 = z_1 + (-z_2) = (a + bi) + (-c - di)$
Using the rule for addition:
$z_1 - z_2 = (a - c) + i(b - d)$
... (2)
The result is a new complex number with real part $(a-c)$ and imaginary part $(b-d)$.
3. Multiplication of Complex Numbers
Multiplying complex numbers is similar to multiplying binomials in algebra, with the crucial addition of using the property $i^2 = -1$.
Let $z_1 = a + bi$ and $z_2 = c + di$.
$z_1 z_2 = (a + bi)(c + di)$
Using the distributive property:
= $a(c + di) + bi(c + di)$
= $ac + adi + bci + bdi^2$
Substitute $i^2 = -1$:
= $ac + adi + bci + bd(-1)$
= $ac + adi + bci - bd$
Group the real terms and the imaginary terms:
$z_1 z_2 = (ac - bd) + i(ad + bc)$
... (3)
The result is a new complex number with real part $(ac - bd)$ and imaginary part $(ad + bc)$.
Properties of Multiplication of Complex Numbers
Complex number multiplication also satisfies familiar algebraic properties:
(i) Closure Property: The product of two complex numbers is always a complex number. If $z_1, z_2 \in \mathbb{C}$, then $z_1 z_2 \in \mathbb{C}$.
(ii) Commutative Law: The order of multiplication does not matter. For any complex numbers $z_1, z_2$, $z_1 z_2 = z_2 z_1$. This can be verified using Formula (3) and the commutative property of real number multiplication and addition.
(iii) Associative Law: The grouping of numbers in multiplication does not matter. For any complex numbers $z_1, z_2, z_3$, $(z_1 z_2) z_3 = z_1 (z_2 z_3)$.
(iv) Existence of Multiplicative Identity (Unit Element): There exists a unique complex number, $1 + 0i$, denoted simply as 1, which acts as the multiplicative identity. Multiplying any complex number by 1 leaves it unchanged.
For any $z = a + bi \in \mathbb{C}$, the product with the multiplicative identity $(1 + 0i)$ is:
$z \cdot (1 + 0i) = (a + bi) \cdot (1 + 0i)$
Using the formula for multiplication of complex numbers, $(x+iy)(u+iv) = (xu - yv) + i(xv + yu)$:
$= (a \cdot 1 - b \cdot 0) + i(a \cdot 0 + b \cdot 1)$
Simplifying the terms gives:
$= (a - 0) + i(0 + b)$
$= a + bi$
This is the original complex number:
$= z$
(v) Existence of Multiplicative Inverse (Reciprocal): For every non-zero complex number $z = a + bi$ (i.e., not $0+0i$), there exists a unique complex number, denoted by $z^{-1}$ or $\frac{1}{z}$, such that their product is the multiplicative identity (1). The multiplicative inverse will be discussed further under division.
(vi) Distributive Law: Multiplication distributes over addition. For any complex numbers $z_1, z_2, z_3$, the following properties hold:
- $z_1 (z_2 + z_3) = z_1 z_2 + z_1 z_3$
- $(z_1 + z_2) z_3 = z_1 z_3 + z_2 z_3$
4. Division of Complex Numbers
Dividing one complex number by another non-zero complex number is achieved by using the concept of the complex conjugate of the denominator. To eliminate the imaginary part from the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator.
Let $z_1 = a + bi$ and $z_2 = c + di$ be two complex numbers, with $z_2 \neq 0$ (meaning at least one of $c$ or $d$ is non-zero). We want to compute $\frac{z_1}{z_2} = \frac{a + bi}{c + di}$.
The complex conjugate of the denominator $z_2 = c + di$ is $\overline{z_2} = c - di$.
Multiply the numerator and denominator by the conjugate of the denominator:
$\frac{z_1}{z_2} = \frac{a + bi}{c + di} = \frac{(a + bi) \times (c - di)}{(c + di) \times (c - di)}$
Simplify the denominator using the property $(x+iy)(x-iy) = x^2+y^2$:
Denominator: $(c + di)(c - di) = c^2 + d^2$.
Since $z_2 \neq 0$, at least one of $c$ or $d$ is non-zero, which guarantees that $c^2 + d^2 > 0$.
Simplify the numerator using the multiplication rule:
Numerator = $(a + bi)(c - di)$
$= (ac - b(-d)) + i(a(-d) + bc)$
$= (ac + bd) + i(bc - ad)$
Combine the simplified numerator and denominator:
$\frac{z_1}{z_2} = \frac{(ac + bd) + i(bc - ad)}{c^2 + d^2}$
Write the result in the standard form $A + Bi$ by separating the real and imaginary parts:
$\frac{z_1}{z_2} = \frac{ac + bd}{c^2 + d^2} + i \frac{bc - ad}{c^2 + d^2}$
... (4)
This is a complex number in standard form, where $A = \frac{ac + bd}{c^2 + d^2}$ and $B = \frac{bc - ad}{c^2 + d^2}$ are real numbers.
Multiplicative Inverse (Reciprocal)
The multiplicative inverse of a non-zero complex number $z = a + bi$ is $\frac{1}{z}$. We can find this using the division method by setting the numerator to $1 + 0i$.
$\frac{1}{z} = \frac{1}{a + bi} = \frac{1 \cdot (a - bi)}{(a + bi)(a - bi)} = \frac{a - bi}{a^2 + b^2}$
$z^{-1} = \frac{1}{a + bi} = \frac{a}{a^2 + b^2} - i \frac{b}{a^2 + b^2}$
... (5)
This exists for any $a+bi \neq 0$, which means $a$ and $b$ are not both zero, so $a^2 + b^2 > 0$.
Example 1. Perform the following operations:
(i) $(3 + 2i) + (1 - 4i)$
(ii) $(5 - 3i) - (2 + 6i)$
(iii) $(4 + i)(2 - 3i)$
(iv) $\frac{2 + 3i}{1 - i}$
Answer:
Solution:
(i) Addition:
(3 + 2i) + (1 - 4i)
Add the real parts (3 and 1) and the imaginary parts (2 and -4):
= $(3 + 1) + i(2 - 4)$
= $4 + i(-2)$
= $4 - 2i$
(ii) Subtraction:
(5 - 3i) - (2 + 6i)
Subtract the real parts (5 and 2) and the imaginary parts (-3 and 6):
= $(5 - 2) + i(-3 - 6)$
= $3 + i(-9)$
= $3 - 9i$
(iii) Multiplication:
(4 + i)(2 - 3i)
Multiply like binomials, using $i^2 = -1$:
= $4(2) + 4(-3i) + i(2) + i(-3i)$
= $8 - 12i + 2i - 3i^2$
Substitute $i^2 = -1$ and combine like terms:
= $8 - 10i - 3(-1)$
= $8 - 10i + 3$
= $11 - 10i$
(iv) Division:
$\frac{2 + 3i}{1 - i}$
Multiply the numerator and denominator by the complex conjugate of the denominator. The denominator is $1 - i$, and its conjugate is $1 + i$.
$= \frac{(2 + 3i)(1 + i)}{(1 - i)(1 + i)}$
Calculate the numerator and denominator separately:
Numerator Calculation:
We need to multiply $(2 + 3i)$ by $(1 + i)$.
Step 1: Apply the distributive property (FOIL method).
$ (2 + 3i)(1 + i) = 2(1) + 2(i) + 3i(1) + 3i(i) $
Step 2: Simplify each term.
$ = 2 + 2i + 3i + 3i^2 $
Step 3: Substitute $i^2 = -1$.
$ = 2 + 2i + 3i + 3(-1) $
$ = 2 + 2i + 3i - 3 $
Step 4: Group the real and imaginary parts.
$ = (2 - 3) + (2i + 3i) $
Step 5: Combine like terms to get the final result.
$ = -1 + 5i $
Denominator Calculation:
We need to multiply the complex conjugates $(1 - i)$ and $(1 + i)$.
Step 1: Apply the distributive property (FOIL method) or the difference of squares formula $(a-b)(a+b) = a^2 - b^2$.
$ (1 - i)(1 + i) = 1(1) + 1(i) - i(1) - i(i) $
Step 2: Simplify each term.
$ = 1 + i - i - i^2 $
Step 3: The imaginary terms cancel each other out.
$ = 1 - i^2 $
Step 4: Substitute $i^2 = -1$.
$ = 1 - (-1) $
Step 5: Simplify to get the final result.
$ = 1 + 1 = 2 $
Substitute these back into the fraction:
= $\frac{-1 + 5i}{2}$
Write the result in standard form $a + bi$:
= $-\frac{1}{2} + \frac{5}{2}i$
The final answers are $4 - 2i$, $3 - 9i$, $11 - 10i$, and $-\frac{1}{2} + \frac{5}{2}i$.
Powers of $i$
The imaginary unit $i$ is defined by the property $i^2 = -1$. By repeatedly multiplying by $i$, we can find the integer powers of $i$. A remarkable and useful property is that these powers follow a repeating pattern.
Calculating Powers of $i$:
Let's calculate the first few positive integer powers of $i$:
By definition:
$i^1 = i$
$i^2 = -1$
(By definition)
Using the definition of $i^2$, we find higher powers:
$i^3 = i^2 \cdot i^1 = (-1) \cdot i = -i$
$i^4 = i^2 \cdot i^2 = (-1) \cdot (-1) = 1$
Now let's calculate the next few powers:
$i^5 = i^4 \cdot i^1 = 1 \cdot i = i$
$i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1$
$i^7 = i^4 \cdot i^3 = 1 \cdot (-i) = -i$
$i^8 = i^4 \cdot i^4 = 1 \cdot 1 = 1$
We can see that the values of the powers of $i$ repeat in a cycle of four: $i, -1, -i, 1$.
General Form for Integer Powers of $i$:
The cyclic nature of powers of $i$ means that the value of $i^n$ for any integer $n$ depends only on the remainder when $n$ is divided by 4. We can use the division algorithm to write any integer $n$ in the form $n = 4q + r$, where $q$ is the quotient (an integer) and $r$ is the remainder, with $0 \le r < 4$. The possible remainders are 0, 1, 2, and 3.
Using the exponent rule $a^{m+k} = a^m a^k$ and $(a^m)^k = a^{mk}$:
$i^n = i^{4q + r} = i^{4q} \cdot i^r = (i^4)^q \cdot i^r$
Since $i^4 = 1$, and $1$ raised to any integer power $q$ is still $1$ ($1^q = 1$):
$i^n = 1^q \cdot i^r = 1 \cdot i^r = i^r$
... (1)
So, to find the value of $i^n$, simply divide $n$ by 4 and find the remainder $r$. The value of $i^n$ is the same as the value of $i^r$.
Summary based on Remainder:
- If $n = 4q$ (remainder $r=0$), $i^n = i^0$. By convention, any non-zero number raised to the power of 0 is 1, so $i^0 = 1$.
$i^{4q} = 1$
- If $n = 4q + 1$ (remainder $r=1$), $i^n = i^1$.
$i^{4q+1} = i$
- If $n = 4q + 2$ (remainder $r=2$), $i^n = i^2$.
$i^{4q+2} = -1$
- If $n = 4q + 3$ (remainder $r=3$), $i^n = i^3$.
$i^{4q+3} = -i$
Summary Table for Powers of $i$:
Form of Exponent $n$ | Remainder $r$ when $n \div 4$ | Value of $i^n = i^r$ |
---|---|---|
$4q$ | 0 | $i^0 = 1$ |
$4q+1$ | 1 | $i^1 = i$ |
$4q+2$ | 2 | $i^2 = -1$ |
$4q+3$ | 3 | $i^3 = -i$ |
where $q$ is any integer ($\in \mathbb{Z}$). This table summarizes the repeating cycle of powers of $i$.
Example 1. Evaluate $i^{25}$.
Answer:
Given:
The expression $i^{25}$.
To Evaluate:
The value of $i^{25}$.
Solution:
We need to find the remainder when the exponent, 25, is divided by 4.
Using division:
$25 \div 4$
$25 = 4 \times 6 + 1$. The quotient is 6 and the remainder is 1.
$25 = 4 \times 6 + 1$
According to the general form $i^n = i^r$ where $r$ is the remainder, the value of $i^{25}$ is the same as $i$ raised to the power of the remainder 1.
$i^{25} = i^1$
$i^{25} = i$
The final answer is $\textbf{i}$.
Example 2. Evaluate $i^{-19}$.
Answer:
Given:
The expression $i^{-19}$.
To Evaluate:
The value of $i^{-19}$.
Solution:
We need to evaluate $i$ raised to a negative integer power. We can still use the property $i^n = i^r$, where $r$ is the remainder when $n$ is divided by 4. We apply the division algorithm to $-19$.
Divide -19 by 4:
$-19 \div 4$
We look for an integer quotient $q$ and a remainder $r$ ($0 \le r < 4$) such that $-19 = 4q + r$. We know $4 \times (-5) = -20$. Then $-19 = -20 + 1$. So, $-19 = 4 \times (-5) + 1$. Here, the quotient $q = -5$ and the remainder $r = 1$.
$-19 = 4 \times (-5) + 1$
The remainder is 1. According to the general form $i^n = i^r$, the value of $i^{-19}$ is the same as $i$ raised to the power of the remainder 1.
$i^{-19} = i^1$
$i^{-19} = i$
Alternate Solution:
We can also solve this by writing $i^{-19}$ as $\frac{1}{i^{19}}$.
$i^{-19} = \frac{1}{i^{19}}$
First, evaluate $i^{19}$ by finding the remainder when 19 is divided by 4. $19 = 4 \times 4 + 3$. The remainder is 3. So, $i^{19} = i^3 = -i$.
$i^{-19} = \frac{1}{-i}$
To express this in the standard $a+bi$ form, multiply the numerator and the denominator by $i$:
$i^{-19} = \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2}$
Substitute $i^2 = -1$:
$i^{-19} = \frac{i}{-(-1)} = \frac{i}{1} = i$
Both methods yield the same result.
The final answer is $\textbf{i}$.
Identities of Complex Numbers
The set of complex numbers $\mathbb{C}$, together with the operations of addition and multiplication defined in the previous section, forms a mathematical structure called a field. A field is a set on which addition and multiplication are defined and satisfy certain fundamental properties or identities. These identities are direct extensions of the properties of real numbers.
Common Identities of Complex Numbers:
Let $z_1, z_2,$ and $z_3$ be any complex numbers. The following identities hold:
1. Identities related to Addition:
These properties describe how complex numbers behave under addition:
- Commutative Law of Addition: The order of addition can be swapped without changing the result.
$z_1 + z_2 = z_2 + z_1$
- Associative Law of Addition: When adding three or more complex numbers, the grouping of numbers does not affect the sum.
$(z_1 + z_2) + z_3 = z_1 + (z_2 + z_3)$
- Existence of Additive Identity: There is a unique complex number, $0 + 0i$ (which is the real number 0), that when added to any complex number $z$, results in $z$. This is the additive identity.
$z + 0 = 0 + z = z$
- Existence of Additive Inverse: For every complex number $z = a + bi$, there exists a unique complex number, $-z = -a - bi$, such that their sum is the additive identity (0). This is the additive inverse.
$z + (-z) = (-z) + z = 0$
2. Identities related to Multiplication:
These properties describe how complex numbers behave under multiplication:
- Commutative Law of Multiplication: The order of multiplication can be swapped without changing the result.
$z_1 z_2 = z_2 z_1$
- Associative Law of Multiplication: When multiplying three or more complex numbers, the grouping of numbers does not affect the product.
$(z_1 z_2) z_3 = z_1 (z_2 z_3)$
- Existence of Multiplicative Identity: There is a unique complex number, $1 + 0i$ (which is the real number 1), that when multiplied by any complex number $z$, results in $z$. This is the multiplicative identity.
$z \cdot 1 = 1 \cdot z = z$
- Existence of Multiplicative Inverse: For every non-zero complex number $z$, there exists a unique complex number, $z^{-1}$ (or $\frac{1}{z}$), such that their product is the multiplicative identity (1). This is the multiplicative inverse or reciprocal, which exists for all $z \in \mathbb{C}$ such that $z \neq 0$.
$z \cdot z^{-1} = z^{-1} \cdot z = 1$ for $z \neq 0$
3. Distributive Law:
The distributive law is a fundamental property that connects the operations of multiplication and addition. It describes how to multiply a sum of complex numbers.
For any three complex numbers $z_1$, $z_2$, and $z_3$:
(i) Left Distributive Law
This rule applies when we multiply a sum from the left side.
$z_1 (z_2 + z_3) = z_1 z_2 + z_1 z_3$
This means you can either add $z_2$ and $z_3$ first and then multiply the result by $z_1$, or you can "distribute" $z_1$ to each term inside the parentheses (multiplying $z_1$ by $z_2$ and $z_1$ by $z_3$ separately) and then add those products together. The final answer will be the same.
(ii) Right Distributive Law
Similarly, the law works when multiplying a sum from the right side.
$(z_1 + z_2) z_3 = z_1 z_3 + z_2 z_3$
This shows that multiplying the sum $(z_1 + z_2)$ by $z_3$ is equivalent to distributing $z_3$ to both $z_1$ and $z_2$ individually and then adding the results.
Proof of Distributive Law (Example):
Let $z_1 = a+bi$, $z_2 = c+di$, and $z_3 = e+fi$.
LHS: $z_1 (z_2 + z_3)$
First, compute the sum inside the parentheses:
$z_2 + z_3 = (c + di) + (e + fi) = (c + e) + i(d + f)$
Now, multiply $z_1$ by this sum. Using the multiplication rule $(x+yi)(u+vi) \ $$ = (xu-yv) + \ $$ i(xv+yu)$, where $x=a, y=b, \ $$ u=(c+e), v=(d+f)$:
$z_1 (z_2 + z_3) = (a + bi)((c + e) + i(d + f))$
$= (a(c+e) - b(d+f)) + i(a(d+f) + b(c+e))$
$= (ac + ae - bd - bf) + i(ad + af + bc + be)$
... (A)
RHS: $z_1 z_2 + z_1 z_3$
First, compute the products $z_1 z_2$ and $z_1 z_3$ using the multiplication rule $(ac-bd) + i(ad+bc)$:
$z_1 z_2 = (a + bi)(c + di) = (ac - bd) + i(ad + bc)$
$z_1 z_3 = (a + bi)(e + fi) = (ae - bf) + i(af + be)$
Now, add these two complex numbers:
$z_1 z_2 + z_1 z_3 = ((ac - bd) + i(ad + bc)) + ((ae - bf) + i(af + be))$
Add the real parts and imaginary parts separately:
$= (ac - bd + ae - bf) + i(ad + bc + af + be)$
Rearrange the terms in the real and imaginary parts:
$= (ac + ae - bd - bf) + i(ad + af + bc + be)$
... (B)
Comparing expression (A) for LHS and (B) for RHS, we see that they are identical.
Thus, the distributive law $z_1 (z_2 + z_3) = z_1 z_2 + z_1 z_3$ is proven for complex numbers.
4. Other useful algebraic identities:
Since complex numbers satisfy the field axioms (commutative, associative, distributive laws, and existence of identities and inverses), standard algebraic identities that hold for real numbers also hold for complex numbers. You can derive these using the fundamental definitions and properties, or simply treat complex numbers like algebraic expressions involving $i$ and substitute $i^2 = -1$ where needed.
Let $z_1$ and $z_2$ be complex numbers.
$(z_1 + z_2)^2 = z_1^2 + 2z_1 z_2 + z_2^2$
... (1)
$(z_1 - z_2)^2 = z_1^2 - 2z_1 z_2 + z_2^2$
... (2)
$(z_1 + z_2)^3 = z_1^3 + 3z_1^2 z_2 + 3z_1 z_2^2 + z_2^3$
... (3)
$(z_1 - z_2)^3 = z_1^3 - 3z_1^2 z_2 + 3z_1 z_2^2 - z_2^3$
... (4)
$(z_1 + z_2)(z_1 - z_2) = z_1^2 - z_2^2$
... (5)
$(z_1 + z_2)(z_1^2 - z_1 z_2 + z_2^2) = z_1^3 + z_2^3$
... (6)
$(z_1 - z_2)(z_1^2 + z_1 z_2 + z_2^2) = z_1^3 - z_2^3$
... (7)
These identities are valid for all complex numbers $z_1, z_2$ where the terms are defined. The fundamental properties (commutative, associative, distributive, identities, inverses) are the reason why algebraic manipulations with complex numbers using these identities are valid, just like with real numbers.
The Argand Plane (The Complex Plane)
We know that real numbers can be visualized as points on a one-dimensional number line. However, a complex number like $z = a + bi$ has two independent components: a real part ($a$) and an imaginary part ($b$). A single line is not enough to represent both parts at the same time.
To visualize complex numbers, we need a two-dimensional space. This special plane is called the complex plane or the Argand plane, named after the mathematician Jean-Robert Argand who helped popularize this geometric representation.
Representing Complex Numbers on the Plane
The main idea is simple: we create a two-dimensional coordinate system where every complex number $z = a + bi$ corresponds to a unique point $(a, b)$.
The Argand plane is constructed as follows:
- The Real Axis: The horizontal axis (the familiar x-axis) is used to plot the real part, $a$. For this reason, it is called the real axis. Any real number, like $4 = 4 + 0i$, lies on this axis at the point $(4, 0)$.
- The Imaginary Axis: The vertical axis (the familiar y-axis) is used to plot the imaginary part, $b$. It is called the imaginary axis. Any purely imaginary number, like $3i = 0 + 3i$, lies on this axis at the point $(0, 3)$.
- The Origin: The point where the axes cross, $(0, 0)$, represents the complex number $0 + 0i$, which is simply 0.
This setup creates a perfect one-to-one mapping. Every complex number corresponds to a unique point on the plane, and every point on the plane corresponds to a unique complex number.
Vector Representation
Besides representing $z = a + bi$ as a point, it is often more useful to think of it as a position vector. This is an arrow that starts at the origin $(0, 0)$ and ends at the point $(a, b)$. This vector viewpoint is the key to understanding complex number operations geometrically.
Geometric Interpretation of Operations
The Argand plane allows us to visualize the effects of complex number algebra in a very intuitive way.
1. Addition
Algebraically, we add complex numbers by adding their real and imaginary parts separately: $(a + bi) + (c + di) = (a+c) + i(b+d)$.
Geometrically, this is identical to vector addition. To find the sum $z_1 + z_2$, you can use the Parallelogram Law. Draw the vectors for $z_1$ and $z_2$. These two vectors form two adjacent sides of a parallelogram. The vector representing the sum, $z_1 + z_2$, is the diagonal of this parallelogram that starts at the origin.
2. Subtraction
Algebraically, subtraction is defined as: $(a + bi) \ $$ - (c + di) = \ $$ (a-c) + \ $$ i(b-d)$.
Geometrically, subtracting $z_2$ from $z_1$ is the same as adding the negative of $z_2$ to $z_1$. The vector for $-z_2 = -c - di$ has the same length as the vector for $z_2$ but points in the exact opposite direction. You can then use the parallelogram law to find $z_1 + (-z_2)$.
3. Multiplication by $i$ (Rotation)
The geometric meaning of general multiplication is a combination of rotation and scaling. A simple but powerful case is multiplication by the imaginary unit, $i$.
Multiplying any complex number $z$ by $i$ results in a 90° counter-clockwise rotation of its vector about the origin. For example, if $z = 3 + i$, its point is $(3, 1)$. Multiplying by $i$ gives: $i \cdot z = i(3+i) = 3i + i^2 = 3i - 1 = -1 + 3i$. The new point is $(-1, 3)$, which is exactly where the original point lands after a 90° rotation.
4. The Complex Conjugate
The conjugate of $z = a + bi$ is $\overline{z} = a - bi$.
Geometrically, the point for $\overline{z}$ is $(a, -b)$. This point is the reflection or "mirror image" of the point for $z$ across the real (horizontal) axis.
5. The Modulus
The modulus of $z = a + bi$, written as $|z|$, is the distance from the origin $(0, 0)$ to the point $(a, b)$. This is simply the length (or magnitude) of the position vector for $z$.
Using the Pythagorean theorem on the right-angled triangle formed by the real part ($a$), the imaginary part ($b$), and the vector itself, we get the formula for the length of the hypotenuse:
$$|z| = \sqrt{a^2 + b^2}$$
The modulus is always a non-negative real number, as it represents a physical distance.
Polar Representation of Complex Numbers
We have seen that a complex number $z = a + bi$ can be represented by Cartesian coordinates $(a, b)$ on the Argand plane. While this is useful, another powerful way to describe the position of a point is by using its distance from the origin and the angle it makes with a reference direction. This is the basis of the polar form or trigonometric form of a complex number.
From Cartesian to Polar Coordinates
Consider a non-zero complex number $z = a + bi$, represented by the point $P(a, b)$ in the Argand plane. Instead of using the horizontal distance 'a' and vertical distance 'b', we can define the point P using:
- $r$: The distance of the point $P$ from the origin $O(0, 0)$.
- $\theta$: The angle that the line segment OP makes with the positive direction of the real axis. This angle is measured in radians (positive for counter-clockwise rotation, negative for clockwise).
Looking at the right-angled triangle in the diagram, we can use basic trigonometry to relate the Cartesian coordinates $(a, b)$ to the polar coordinates $(r, \theta)$:
$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{a}{r} \implies a = r \cos \theta$
$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{b}{r} \implies b = r \sin \theta$
Modulus and Argument
The two components of the polar form have special names:
1. Modulus ($r$):
The distance $r$ is called the modulus of the complex number $z$ and is denoted by $|z|$. It represents the magnitude or length of the vector for $z$. Using the Pythagorean theorem:
$r = |z| = \sqrt{a^2 + b^2}$
(Modulus)
The modulus is always a non-negative real number.
2. Argument ($\theta$):
The angle $\theta$ is called the argument or amplitude of $z$, denoted by $\arg(z)$. It is determined by the equations:
$\cos \theta = \frac{a}{r}$ and $\sin \theta = \frac{b}{r}$
The argument $\theta$ is not unique because you can add any integer multiple of $2\pi$ (a full circle) and get the same point. For a unique value, we define the principal argument, denoted $\text{Arg}(z)$, which is the unique angle $\theta$ in the interval $(-\pi, \pi]$.
How to Find the Principal Argument ($\theta$)
To find the principal argument of $z = a + bi$:
Step 1: Find the reference angle $\alpha$ using the absolute values of $a$ and $b$.
$\alpha = \tan^{-1}\left|\frac{b}{a}\right|$, where $\alpha \in [0, \frac{\pi}{2})$
Step 2: Adjust $\alpha$ based on the Quadrant
Determine which quadrant the point $(a, b)$ lies in. Then, use the following rules to find the principal argument $\theta$.
Quadrant I ($a > 0, b > 0$): The angle is just the reference angle.
$\theta = \alpha$
Quadrant II ($a < 0, b > 0$): Start at $\pi$ (180°) and subtract the reference angle.
$\theta = \pi - \alpha$
Quadrant III ($a < 0, b < 0$): Start at $-\pi$ (-180°) and add the reference angle.
$\theta = -(\pi - \alpha) = \alpha - \pi$
Quadrant IV ($a > 0, b < 0$): The angle is the negative of the reference angle.
$\theta = -\alpha$
For points on the axes, the argument is found by inspection:
- Positive Real Axis ($z=a, a>0$): $\theta = 0$
- Positive Imaginary Axis ($z=bi, b>0$): $\theta = \frac{\pi}{2}$
- Negative Real Axis ($z=a, a < 0$): $\theta = \pi$
- Negative Imaginary Axis ($z=bi, b < 0$): $\theta = -\frac{\pi}{2}$
The Polar Form
To get the polar form of $z = a + bi$, we substitute $a = r \cos \theta$ and $b = r \sin \theta$:
$z = (r \cos \theta) + i (r \sin \theta)$
Factoring out $r$, we get the standard polar form:
$z = r(\cos \theta + i \sin \theta)$
... (1)
This form is very useful because it separates the magnitude ($r$) from the direction ($\theta$) of the complex number. It simplifies multiplication and division of complex numbers immensely.
The term $\cos \theta + i \sin \theta$ is sometimes abbreviated as $\text{cis } \theta$. Furthermore, through Euler's formula, it is equal to $e^{i\theta}$, leading to the exponential form $z = re^{i\theta}$, which is extremely important in advanced mathematics, physics, and engineering.
Example 1. Represent the complex number $z = 1 + i\sqrt{3}$ in polar form.
Answer:
Given:
The complex number $z = 1 + i\sqrt{3}$. Comparing this with the Cartesian form $z = a + bi$, we have $a = 1$ and $b = \sqrt{3}$.
To Represent:
The complex number $z$ in polar form $r(\cos \theta + i \sin \theta)$.
Solution:
We need to find the modulus $r$ and the principal argument $\theta$.
Step 1: Find the modulus $r$
Using the formula $r = \sqrt{a^2 + b^2}$:
$r = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4}$
$r = 2$
Step 2: Find the argument $\theta$
Since $a = 1$ (positive) and $b = \sqrt{3}$ (positive), the point $(1, \sqrt{3})$ lies in the first quadrant.
First, find the reference angle $\alpha$:
$\alpha = \tan^{-1}\left|\frac{b}{a}\right| = \tan^{-1}\left|\frac{\sqrt{3}}{1}\right| = \tan^{-1}(\sqrt{3})$
We know that $\tan(\frac{\pi}{3}) = \sqrt{3}$, so $\alpha = \frac{\pi}{3}$.
For the first quadrant, the principal argument $\theta$ is equal to the reference angle $\alpha$.
$\theta = \alpha = \frac{\pi}{3}$
Step 3: Write in Polar Form
Substitute the values of $r$ and $\theta$ into the polar form $z = r(\cos \theta + i \sin \theta)$:
$z = 2\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right)$
The final answer is $2\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right)$.
Modulus of a Complex Number and Its Properties
In the Argand plane, a complex number $z = a + bi$ is represented by the point $(a, b)$. The modulus of the complex number $z$ is defined as the distance of this point from the origin $(0, 0)$. It is a non-negative real number and represents the magnitude or length of the complex number when considered as a vector from the origin to the point $(a, b)$.
Definition of Modulus
For a complex number $z = a + bi$, where $a, b \in \mathbb{R}$, the modulus of $z$, denoted by $|z|$ or sometimes by $r$, is calculated using the distance formula, which is an application of the Pythagorean theorem in the Argand plane.
$|z| = |a + bi| = \sqrt{a^2 + b^2}$
... (1)
Since $a$ and $b$ are real numbers, $a^2 \ge 0$ and $b^2 \ge 0$. Their sum $a^2 + b^2$ is also non-negative. Therefore, the modulus $|z|$ is always a real number greater than or equal to 0.
Examples of Calculating Modulus
Let's calculate the modulus for some complex numbers:
1. $z = 3 + 4i$. Here $a = 3, b = 4$.
$|z| = |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
2. $z = -2 - 5i$. Here $a = -2, b = -5$.
$|z| = |-2 - 5i| = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}$
3. $z = -7i$. This is a purely imaginary number, $z = 0 - 7i$. Here $a = 0, b = -7$.
$|z| = |-7i| = |0 - 7i| = \sqrt{0^2 + (-7)^2} = \sqrt{0 + 49} = \sqrt{49} = 7$
4. $z = 6$. This is a real number, $z = 6 + 0i$. Here $a = 6, b = 0$.
$|z| = |6| = |6 + 0i| = \sqrt{6^2 + 0^2} = \sqrt{36 + 0} = \sqrt{36} = 6$
This confirms that the modulus of a real number (considered as a complex number) is the same as its absolute value.
Properties of the Modulus
The modulus of a complex number has several important properties that simplify calculations and provide geometric insights. Let $z, z_1,$ and $z_2$ be any complex numbers.
1. Modulus of Zero
The modulus (length) of a complex number is 0 if and only if the complex number itself is the origin, $0+0i$.
$|z| = 0 \iff z = 0$
Explanation: The distance from the origin can only be zero if the point *is* the origin.
2. Equal Moduli
A complex number ($z$), its conjugate ($\overline{z}$), and its negative ($-z$) all have the same modulus.
$|z| = |\overline{z}| = |-z|$
Geometric Interpretation: The points for $z$, $\overline{z}$ (its reflection across the real axis), and $-z$ (its reflection through the origin) all lie on the same circle centered at the origin, meaning they are all the same distance from it.
3. Product with Conjugate
Multiplying a complex number by its own conjugate always results in a real number equal to the square of its modulus. This is a very useful identity.
$z \overline{z} = |z|^2$
... (2)
Explanation: For $z = a + bi$, we have $z \overline{z} = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - (-b^2) = a^2 + b^2$, which by definition is $|z|^2$. This property is the key to dividing complex numbers.
4. Modulus of a Product
The modulus of a product of complex numbers is the product of their individual moduli.
$|z_1 z_2| = |z_1| |z_2|$
Explanation: When you multiply two complex numbers, the length of the resulting vector is the product of the lengths of the original two vectors.
5. Modulus of a Quotient
The modulus of a quotient of complex numbers is the quotient of their individual moduli (as long as you are not dividing by zero).
$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$, provided $z_2 \neq 0$.
Explanation: When you divide two complex numbers, the length of the resulting vector is the quotient of the lengths of the original two vectors.
6. The Triangle Inequality
The modulus of a sum of two complex numbers is less than or equal to the sum of their moduli. This is called the triangle inequality.
$|z_1 + z_2| \le |z_1| + |z_2|$
... (3)
Geometric Interpretation: This property mirrors the rule for triangles: the length of any side of a triangle must be less than or equal to the sum of the lengths of the other two sides. A related property is $|z_1 - z_2| \ge ||z_1| - |z_2||$.
7. Modulus of a Power
The modulus of a complex number raised to an integer power is the modulus of that number raised to the same power.
$|z^n| = |z|^n$ for any integer $n$.
Explanation: This is just a repeated application of the product and quotient rules. For example, $|z^3| = |z \cdot z \cdot z| = |z| \cdot |z| \cdot |z| = |z|^3$.
Conjugate of a Complex Number and Its Properties
The conjugate of a complex number is its "mirror image" across the real axis. It is a fundamental concept used to simplify expressions, especially in division, and to connect complex numbers with real numbers in various proofs and formulas.
Definition of the Conjugate
The conjugate of a complex number $z = a + bi$ is found by simply flipping the sign of its imaginary part.
The conjugate of $z$ is denoted by $\overline{z}$ (read as "z-bar").
If $z = a + bi$, then its conjugate is $\overline{z} = a - bi$
Geometric Interpretation
In the Argand plane, the point $(a, -b)$ representing $\overline{z}$ is the perfect reflection of the point $(a, b)$ representing $z$ across the real (horizontal) axis.
Examples of Finding the Conjugate
1. If $z = 3 + 4i$, then $\overline{z} = 3 - 4i$.
2. If $z = -2 - 5i$, then $\overline{z} = -2 + 5i$.
3. If $z = -7i$ (which is $0 - 7i$), then $\overline{z} = 0 + 7i = 7i$.
4. If $z = 6$ (which is $6 + 0i$), then $\overline{z} = 6 - 0i = 6$. The conjugate of a real number is the number itself, as it lies on the axis of reflection.
Properties of the Conjugate
The complex conjugate has several useful algebraic properties. Let $z, z_1,$ and $z_2$ be complex numbers.
1. Double Conjugation
Taking the conjugate twice returns the original number.
$(\overline{\overline{z}}) = z$
Explanation: Reflecting a point across the real axis and then reflecting it back gives you the original point.
2. Sum with Conjugate
The sum of a complex number and its conjugate is twice its real part (always a real number).
$z + \overline{z} = 2 \text{Re}(z)$
Explanation: The imaginary parts ($bi$ and $-bi$) cancel each other out. $(a + bi) + (a - bi) = 2a$.
3. Difference with Conjugate
The difference between a complex number and its conjugate is $2i$ times its imaginary part (always a purely imaginary number, unless the number is real).
$z - \overline{z} = 2i \text{Im}(z)$
Explanation: The real parts ($a$ and $a$) cancel each other out. $(a + bi) - (a - bi) = 2bi$.
4. Condition for a Real Number
A complex number is purely real if and only if it is equal to its own conjugate.
$z$ is Real $\iff z = \overline{z}$
Explanation: If $z = \overline{z}$, then $a+bi = a-bi$, which means $2bi=0$, so $b=0$. This means $z$ has no imaginary part.
5. Condition for a Purely Imaginary Number
A non-zero complex number is purely imaginary if and only if it is equal to the negative of its conjugate.
$z$ is Purely Imaginary $\iff z = -\overline{z}$ (and $z \neq 0$)
Explanation: If $z = -\overline{z}$, then $a+bi = -(a-bi) = -a+bi$, which means $2a=0$, so $a=0$. This means $z$ has no real part.
6. Conjugate of a Sum
The conjugate of a sum is the sum of the conjugates. You can conjugate first and then add, or add first and then conjugate.
$\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
7. Conjugate of a Difference
The conjugate of a difference is the difference of the conjugates.
$\overline{z_1 - z_2} = \overline{z_1} - \overline{z_2}$
8. Conjugate of a Product
The conjugate of a product is the product of the conjugates.
$\overline{z_1 z_2} = \overline{z_1} \overline{z_2}$
9. Conjugate of a Quotient
The conjugate of a quotient is the quotient of the conjugates (provided the denominator is not zero).
$\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}$, provided $z_2 \neq 0$.
10. Conjugate of a Power
The conjugate of a complex number raised to a power is the conjugate raised to that power.
$\overline{(z^n)} = (\overline{z})^n$ for any integer $n$.
Square Root of a Complex Number
In the real number system, we cannot find the square root of a negative number. However, in the complex number system, every non-zero complex number has exactly two distinct square roots, which are themselves complex numbers.
To find the square root of a complex number $z = a + bi$, we are looking for a complex number $w = x + yi$ (where $x$ and $y$ are real numbers) such that $w^2 = z$.
Method to find the Square Root of $a + bi$
Let the square root of $a + bi$ be $x + yi$.
$\sqrt{a + bi} = x + yi$
Square both sides of the equation:
$(x + yi)^2 = a + bi$
Expand the left side using $(p+q)^2 = p^2 + 2pq + q^2$:
$x^2 + 2(x)(yi) + (yi)^2 = a + bi$
Since $i^2 = -1$, we have $(yi)^2 = y^2i^2 = -y^2$. Substitute this back:
$x^2 + 2xyi - y^2 = a + bi$
Group the real and imaginary parts on the left side:
$(x^2 - y^2) + i(2xy) = a + bi$
By the definition of equality for complex numbers, we can equate the real parts and the imaginary parts separately:
Real Part: $x^2 - y^2 = a$
... (A)
Imaginary Part: $2xy = b$
... (B)
To solve this system, we introduce a third equation using the modulus property $|w^2| = |w|^2$.
$|(x + yi)^2| = |a + bi|$
$|x + yi|^2 = |a + bi|$
Using the definition of modulus, $|z| = \sqrt{\text{Re}(z)^2 + \text{Im}(z)^2}$:
$(\sqrt{x^2 + y^2})^2 = \sqrt{a^2 + b^2}$
Modulus Part: $x^2 + y^2 = \sqrt{a^2 + b^2}$
... (C)
Now we have a simple system of linear equations for $x^2$ and $y^2$ using (A) and (C):
- Add (A) and (C) to find $x^2$: $2x^2 = a + \sqrt{a^2+b^2} \ $$ \implies x^2 = \frac{a + \sqrt{a^2+b^2}}{2}$.
- Subtract (A) from (C) to find $y^2$: $2y^2 = \sqrt{a^2+b^2} - a \ $$ \implies y^2 = \frac{\sqrt{a^2+b^2} - a}{2}$.
From these, we find the possible values for $x$ and $y$. Finally, we use equation (B), $2xy=b$, to determine the correct pairing of signs for $x$ and $y$.
- If $b$ is positive, $x$ and $y$ must have the same sign (both positive or both negative).
- If $b$ is negative, $x$ and $y$ must have opposite signs (one positive, one negative).
Example 1. Find the square root of $3 + 4i$.
Answer:
Given:
The complex number $z = 3 + 4i$. Here $a = 3$ and $b = 4$.
To Find:
The square roots of $3 + 4i$.
Solution:
Let $\sqrt{3 + 4i} = x + yi$. By equating parts, we get:
$x^2 - y^2 = 3$
... (A)
$2xy = 4$
... (B)
Now, find the modulus: $|3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
This gives us our third equation:
$x^2 + y^2 = 5$
... (C)
Solve for $x^2$ and $y^2$ using (A) and (C):
Add (A) and (C): $(x^2 - y^2) + (x^2 + y^2) = 3 + 5 \ $$ \implies 2x^2 = 8 \ $$ \implies x^2 = 4$.
Subtract (A) from (C): $(x^2 + y^2) - (x^2 - y^2) = 5 - 3 \ $$ \implies 2y^2 = 2 \ $$ \implies y^2 = 1$.
From these results, we get $x = \pm 2$ and $y = \pm 1$.
Now, we use equation (B), $2xy = 4$, which simplifies to $xy = 2$. Since the product $xy$ is positive, $x$ and $y$ must have the same sign.
- Possibility 1: $x$ and $y$ are both positive. This gives $x=2$ and $y=1$.
- Possibility 2: $x$ and $y$ are both negative. This gives $x=-2$ and $y=-1$.
So, the two square roots are $2 + 1i$ and $-2 - 1i$.
This can be written concisely as $\pm(2 + i)$.
The final answer is $\pm (2 + i)$.
Example 2. Find the square root of $-5 - 12i$.
Answer:
Given:
The complex number $z = -5 - 12i$. Here $a = -5$ and $b = -12$.
To Find:
The square roots of $-5 - 12i$.
Solution:
Let $\sqrt{-5 - 12i} = x + yi$. Equating parts gives:
$x^2 - y^2 = -5$
... (A')
$2xy = -12$
... (B')
Find the modulus: $|-5 - 12i| \ $$ = \sqrt{(-5)^2 + (-12)^2} \ $$ = \sqrt{25 + 144} \ $$ = \sqrt{169} = 13$.
This gives the third equation:
$x^2 + y^2 = 13$
... (C')
Solve for $x^2$ and $y^2$ using (A') and (C'):
Add (A') and (C'): $2x^2 = -5 + 13 \implies 2x^2 = 8 \implies x^2 = 4$.
Subtract (A') from (C'): $2y^2 = 13 - (-5) \implies 2y^2 = 18 \implies y^2 = 9$.
From these results, we get $x = \pm 2$ and $y = \pm 3$.
Now, use equation (B'), $2xy = -12$, which simplifies to $xy = -6$. Since the product $xy$ is negative, $x$ and $y$ must have opposite signs.
- Possibility 1: $x$ is positive and $y$ is negative. This gives $x=2$ and $y=-3$.
- Possibility 2: $x$ is negative and $y$ is positive. This gives $x=-2$ and $y=3$.
So, the two square roots are $2 - 3i$ and $-2 + 3i$.
This can be written concisely as $\pm(2 - 3i)$.
The final answer is $\pm (2 - 3i)$.
Quadratic Equations with Real Coefficients
We have already learned how to solve quadratic equations of the form $ax^2 + bx + c = 0$ in earlier classes, where $a, b,$ and $c$ are real numbers and $a \neq 0$. The solutions (roots) of such an equation are given by the well-known quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
... (1)
The nature of the roots of the quadratic equation is determined by the value of the expression under the square root, which is called the discriminant, denoted by $\Delta$ (Delta).
$\Delta = b^2 - 4ac$
Based on the sign of the discriminant, we categorized the roots as follows:
- If $\Delta = b^2 - 4ac > 0$, the equation has two distinct real roots.
- If $\Delta = b^2 - 4ac = 0$, the equation has two equal real roots.
- If $\Delta = b^2 - 4ac < 0$, the equation has no real roots. However, with the introduction of complex numbers, we can now find solutions in the complex number system.
Case when Discriminant is Negative ($\Delta < 0$)
When the discriminant is negative, $\Delta < 0$. We can express this negative value as a positive value multiplied by -1: $b^2 - 4ac = - (4ac - b^2)$, where $(4ac - b^2)$ is a positive real number.
In the quadratic formula, the term $\sqrt{b^2 - 4ac}$ becomes:
$\sqrt{b^2 - 4ac} = \sqrt{-(4ac - b^2)}$
Using the property $\sqrt{-k} = i\sqrt{k}$ for $k > 0$:
$\sqrt{-(4ac - b^2)} = \sqrt{-1} \cdot \sqrt{4ac - b^2} = i\sqrt{4ac - b^2}$
Substitute this expression back into the quadratic formula:
$x = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}$
... (2)
This formula gives two complex roots for the quadratic equation. The two roots are:
$x_1 = \frac{-b + i\sqrt{4ac - b^2}}{2a} = -\frac{b}{2a} + i\frac{\sqrt{4ac - b^2}}{2a}$
$x_2 = \frac{-b - i\sqrt{4ac - b^2}}{2a} = -\frac{b}{2a} - i\frac{\sqrt{4ac - b^2}}{2a}$
Notice that these two roots are complex numbers. The real part of both roots is $-\frac{b}{2a}$. The imaginary parts are opposites of each other. This means that when the coefficients $a, b, c$ of a quadratic equation are real and the discriminant is negative, the complex roots always occur in conjugate pairs.
Example 1. Solve the quadratic equation $x^2 + x + 1 = 0$.
Answer:
Given:
The quadratic equation $x^2 + x + 1 = 0$.
To Solve:
Find the roots of the equation.
Solution:
The equation is in the standard quadratic form $ax^2 + bx + c = 0$. By comparing the given equation with the standard form, we identify the coefficients:
$a = 1, b = 1, c = 1$
First, calculate the discriminant $\Delta = b^2 - 4ac$:
$\Delta = (1)^2 - 4(1)(1) = 1 - 4 = -3$
Since the discriminant $\Delta = -3 < 0$, the equation has complex roots.
Use the quadratic formula:
$x = \frac{-b \pm \sqrt{\Delta}}{2a}$
Substitute the values of $a, b,$ and $\Delta$:
$x = \frac{-1 \pm \sqrt{-3}}{2(1)}$
Now, express $\sqrt{-3}$ using the imaginary unit $i$: $\sqrt{-3} = \sqrt{-1 \times 3} = \sqrt{-1} \times \sqrt{3} = i\sqrt{3}$.
$x = \frac{-1 \pm i\sqrt{3}}{2}$
The two roots are obtained by taking the $+$ and $-$ signs:
$x_1 = \frac{-1 + i\sqrt{3}}{2} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$
$x_2 = \frac{-1 - i\sqrt{3}}{2} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$
These are the two complex roots of the quadratic equation. They are a pair of complex conjugates, as expected.
These specific roots are also known as the non-real cube roots of unity, often denoted by $\omega$ and $\omega^2$ respectively, which satisfy the property $1 + \omega + \omega^2 = 0$.
The final answer is $x = \frac{-1 \pm i\sqrt{3}}{2}$.