The Hook: When Reality Isn’t Enough
Remember Stranger Things? The Upside Down is a parallel dimension hidden beneath the real world. Complex numbers are just like that — a hidden mathematical dimension beyond the real numbers!
When you try to solve $x^2 = -1$ using real numbers, it’s impossible. But just like Eleven accessing the Upside Down, mathematicians discovered a whole new world: imaginary numbers. Engineers use these to design AC circuits, analyze vibrations, and even process the signals in your smartphone!
Why this matters for JEE: Complex numbers appear in 2-3 questions in JEE Main and are fundamental for JEE Advanced problems in algebra, coordinate geometry, and even calculus.
Prerequisites
Before diving into complex numbers, you should be comfortable with:
- Real number system — Understanding of $\mathbb{R}$
- Basic algebra — Solving quadratic equations
- Coordinate geometry basics — Plotting points on a plane
The Birth of $i$: The Imaginary Unit
The Problem
Consider the equation: $x^2 = -1$
In the real number system $\mathbb{R}$, this has no solution because:
- If $x > 0$, then $x^2 > 0$
- If $x < 0$, then $x^2 > 0$ (negative times negative is positive)
- If $x = 0$, then $x^2 = 0$
So $x^2$ can never be negative!
Interactive Demo: Visualize Complex Numbers
Explore complex numbers on the complex plane and see algebraic operations geometrically.
The Solution: Define a New Number
Mathematicians defined a new number called the imaginary unit:
$$\boxed{i = \sqrt{-1} \quad \text{or equivalently} \quad i^2 = -1}$$In simple terms: $i$ is a number that, when squared, gives $-1$. It doesn’t exist on the real number line — it lives in a new dimension!
Powers of $i$ — The Magic Cycle
Powers of $i$ follow a repeating pattern of 4:
$$i^1 = i$$ $$i^2 = -1$$ $$i^3 = i^2 \cdot i = -1 \cdot i = -i$$ $$i^4 = i^2 \cdot i^2 = (-1)(-1) = 1$$ $$i^5 = i^4 \cdot i = 1 \cdot i = i$$(cycle repeats!)
The Pattern: $1, i, -1, -i, 1, i, -1, -i, ...$
Shortcut: To find $i^n$, divide $n$ by 4 and use the remainder:
- Remainder 0: $i^n = 1$
- Remainder 1: $i^n = i$
- Remainder 2: $i^n = -1$
- Remainder 3: $i^n = -i$
Quick Formula
$$\boxed{i^{4k} = 1, \quad i^{4k+1} = i, \quad i^{4k+2} = -1, \quad i^{4k+3} = -i}$$where $k$ is any integer.
Example: Find $i^{2025}$
$2025 = 4(506) + 1$
So $i^{2025} = i^{4(506)+1} = i^1 = i$
What is a Complex Number?
Definition
A complex number is a number of the form:
$$\boxed{z = a + ib}$$where:
- $a$ is the real part, denoted $\text{Re}(z)$ or $a = \text{Re}(z)$
- $b$ is the imaginary part, denoted $\text{Im}(z)$ or $b = \text{Im}(z)$
- $i$ is the imaginary unit
- $a, b \in \mathbb{R}$ (both are real numbers!)
The Set of Complex Numbers
We denote the set of all complex numbers by $\mathbb{C}$:
$$\mathbb{C} = \{a + ib : a, b \in \mathbb{R}\}$$Examples
| Complex Number | Real Part | Imaginary Part |
|---|---|---|
| $3 + 4i$ | $3$ | $4$ |
| $-2 + 5i$ | $-2$ | $5$ |
| $7i$ | $0$ | $7$ |
| $5$ | $5$ | $0$ |
| $0$ | $0$ | $0$ |
- When $b = 0$: $z = a$ is a pure real number
- When $a = 0$ and $b \neq 0$: $z = ib$ is a pure imaginary number
- The imaginary part is $b$, NOT $ib$!
Complex Numbers as Ordered Pairs
Alternative Representation
A complex number $z = a + ib$ can be represented as an ordered pair:
$$z = (a, b)$$This is just like coordinates on a plane!
- $(3, 4)$ represents $3 + 4i$
- $(0, 1)$ represents $i$
- $(5, 0)$ represents $5$
Equality of Complex Numbers
Two complex numbers are equal if and only if their real parts are equal AND their imaginary parts are equal:
$$\boxed{a + ib = c + id \iff a = c \text{ and } b = d}$$Example: If $x + 2yi = 3 - 6i$, find $x$ and $y$.
Comparing real parts: $x = 3$
Comparing imaginary parts: $2y = -6 \Rightarrow y = -3$
Algebra of Complex Numbers
Addition and Subtraction
Add or subtract the real parts and imaginary parts separately:
$$(a + ib) + (c + id) = (a + c) + i(b + d)$$ $$(a + ib) - (c + id) = (a - c) + i(b - d)$$Example: $(3 + 5i) + (2 - 7i) = (3+2) + i(5-7) = 5 - 2i$
Multiplication
Use the distributive property and remember that $i^2 = -1$:
$$(a + ib)(c + id) = ac + iad + ibc + i^2bd$$ $$= ac + i(ad + bc) + (-1)bd$$ $$= (ac - bd) + i(ad + bc)$$ $$\boxed{(a + ib)(c + id) = (ac - bd) + i(ad + bc)}$$Example: $(2 + 3i)(1 + 4i)$
$= 2(1) + 2(4i) + 3i(1) + 3i(4i)$ $= 2 + 8i + 3i + 12i^2$ $= 2 + 11i + 12(-1)$ $= 2 + 11i - 12$ $= -10 + 11i$
This special product appears frequently:
$$(a + ib)(a - ib) = a^2 - (ib)^2 = a^2 - i^2b^2 = a^2 + b^2$$This is always a real number! We’ll use this for division.
Conjugate of a Complex Number
The conjugate of $z = a + ib$ is denoted $\bar{z}$ (read as “z bar”):
$$\boxed{\bar{z} = a - ib}$$Geometrically, the conjugate is the mirror image across the real axis.
Properties of Conjugates:
- $\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2}$
- $\overline{z_1 - z_2} = \bar{z_1} - \bar{z_2}$
- $\overline{z_1 \cdot z_2} = \bar{z_1} \cdot \bar{z_2}$
- $\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z_1}}{\bar{z_2}}$
- $z \cdot \bar{z} = a^2 + b^2$ (always real and non-negative!)
- $z + \bar{z} = 2a = 2\text{Re}(z)$
- $z - \bar{z} = 2ib = 2i\text{Im}(z)$
- $\overline{\bar{z}} = z$ (conjugate of conjugate is the original)
- $z = \bar{z}$ if and only if $z$ is real
Division
To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator:
$$\frac{a + ib}{c + id} = \frac{(a + ib)(c - id)}{(c + id)(c - id)} = \frac{(ac + bd) + i(bc - ad)}{c^2 + d^2}$$ $$\boxed{\frac{a + ib}{c + id} = \frac{ac + bd}{c^2 + d^2} + i\frac{bc - ad}{c^2 + d^2}}$$Example: Find $\frac{3 + 2i}{1 - i}$
Multiply by $\frac{1 + i}{1 + i}$ (conjugate of denominator):
$$\frac{3 + 2i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{(3 + 2i)(1 + i)}{(1 - i)(1 + i)}$$Numerator: $(3 + 2i)(1 + i) = 3 + 3i + 2i + 2i^2 = 3 + 5i - 2 = 1 + 5i$
Denominator: $(1 - i)(1 + i) = 1 - i^2 = 1 + 1 = 2$
Result: $\frac{1 + 5i}{2} = \frac{1}{2} + \frac{5}{2}i$
Important Algebraic Identities
All standard algebraic identities work with complex numbers:
| Identity | Formula |
|---|---|
| $(z_1 + z_2)^2$ | $z_1^2 + 2z_1z_2 + z_2^2$ |
| $(z_1 - z_2)^2$ | $z_1^2 - 2z_1z_2 + z_2^2$ |
| $z_1^2 - z_2^2$ | $(z_1 + z_2)(z_1 - z_2)$ |
| $(z_1 + z_2)^3$ | $z_1^3 + 3z_1^2z_2 + 3z_1z_2^2 + z_2^3$ |
Common Mistakes to Avoid
Wrong: The imaginary part of $3 + 4i$ is $4i$
Right: The imaginary part of $3 + 4i$ is $4$ (a real number!)
The imaginary part is the coefficient of $i$, not the term with $i$.
Wrong: $\sqrt{-4} = 2i$ or $-2i$ (two values)
Right: $\sqrt{-4} = \sqrt{4 \cdot (-1)} = \sqrt{4} \cdot \sqrt{-1} = 2i$ (principal value)
For negative numbers, we always take the positive imaginary value as principal square root.
JEE Note: $\sqrt{-a} \cdot \sqrt{-b} \neq \sqrt{ab}$ when $a, b > 0$!
Example: $\sqrt{-1} \cdot \sqrt{-1} = i \cdot i = i^2 = -1$, but $\sqrt{(-1)(-1)} = \sqrt{1} = 1$ ✗
Wrong: $\overline{z_1 / z_2} = \bar{z_1} / z_2$
Right: $\overline{z_1 / z_2} = \bar{z_1} / \bar{z_2}$
The conjugate distributes over division, but you must take conjugate of both numerator and denominator!
When to Use Complex Numbers
Use complex numbers when:
- Solving equations with no real solutions ($x^2 + 4 = 0$)
- Working with polynomial roots (Fundamental Theorem of Algebra)
- Simplifying trigonometric expressions (Euler’s formula)
- Analyzing periodic phenomena (AC circuits, waves)
- Finding roots of unity
Key insight: Every polynomial equation has solutions in $\mathbb{C}$!
Practice Problems
Level 1: Foundation (NCERT)
Find the value of $i^{538}$.
Solution:
$538 = 4(134) + 2$
So $i^{538} = i^{4(134)+2} = i^2 = -1$
Answer: $-1$
Express $(2 + 3i) + (4 - 5i)$ in the form $a + ib$.
Solution:
$(2 + 3i) + (4 - 5i) = (2 + 4) + i(3 - 5) = 6 - 2i$
Answer: $6 - 2i$
Find the conjugate of $z = 7 - 3i$.
Solution:
$\bar{z} = 7 - (-3)i = 7 + 3i$
Answer: $7 + 3i$
Level 2: JEE Main
If $\frac{(1 + i)(2 + i)}{3 + i}$ is equal to $a + ib$, find $a$ and $b$.
Solution:
First, simplify numerator: $(1 + i)(2 + i) = 2 + i + 2i + i^2 = 2 + 3i - 1 = 1 + 3i$
Now divide:
$$\frac{1 + 3i}{3 + i} \cdot \frac{3 - i}{3 - i} = \frac{(1 + 3i)(3 - i)}{(3 + i)(3 - i)}$$Numerator: $(1 + 3i)(3 - i) = 3 - i + 9i - 3i^2 = 3 + 8i + 3 = 6 + 8i$
Denominator: $(3 + i)(3 - i) = 9 - i^2 = 9 + 1 = 10$
Result: $\frac{6 + 8i}{10} = \frac{3}{5} + \frac{4}{5}i$
Answer: $a = \frac{3}{5}$, $b = \frac{4}{5}$
If $z = x + iy$ and $|z - 1| = |z + 1|$, find the locus of $z$.
Solution:
$|z - 1| = |x - 1 + iy|$ and $|z + 1| = |x + 1 + iy|$
$\sqrt{(x-1)^2 + y^2} = \sqrt{(x+1)^2 + y^2}$
Squaring both sides: $(x-1)^2 + y^2 = (x+1)^2 + y^2$
$x^2 - 2x + 1 = x^2 + 2x + 1$
$-2x = 2x$
$x = 0$
Answer: The locus is the imaginary axis (y-axis)
Find the value of $\frac{1 + i^2 + i^4 + i^6 + ... + i^{2020}}{1 + i}$.
Solution:
Notice that $i^{2k}$ alternates between $1$ (when $k$ is even) and $-1$ (when $k$ is odd).
Pattern: $i^0 = 1, i^2 = -1, i^4 = 1, i^6 = -1, ...$
From $i^0$ to $i^{2020}$, we have 1011 terms (even powers from 0 to 2020).
They pair up: $(1 - 1) + (1 - 1) + ... + 1$
There are 505 pairs that sum to 0, plus one extra term $i^{2020}$.
$i^{2020} = i^{4(505)} = 1$
So numerator $= 1$
$$\frac{1}{1 + i} = \frac{1}{1+i} \cdot \frac{1-i}{1-i} = \frac{1-i}{1-i^2} = \frac{1-i}{2}$$Answer: $\frac{1-i}{2} = \frac{1}{2} - \frac{1}{2}i$
Level 3: JEE Advanced
If $z_1$ and $z_2$ are two complex numbers such that $|z_1| = |z_2|$ and $\text{arg}(z_1) + \text{arg}(z_2) = \pi$, prove that $z_1 = -\bar{z_2}$.
Solution:
Let $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$
Given: $r_1 = r_2 = r$ and $\theta_1 + \theta_2 = \pi$
So $\theta_1 = \pi - \theta_2$
$$z_1 = r[\cos(\pi - \theta_2) + i\sin(\pi - \theta_2)]$$Using: $\cos(\pi - \theta) = -\cos\theta$ and $\sin(\pi - \theta) = \sin\theta$
$$z_1 = r[-\cos\theta_2 + i\sin\theta_2]$$Now, $\bar{z_2} = r(\cos\theta_2 - i\sin\theta_2)$
$$-\bar{z_2} = -r(\cos\theta_2 - i\sin\theta_2) = r[-\cos\theta_2 + i\sin\theta_2]$$Therefore, $z_1 = -\bar{z_2}$ Proved!
If $\left|\frac{z - 1}{z + 1}\right| = 1$, show that $z$ is purely imaginary or $z$ is on the imaginary axis.
Solution:
Let $z = x + iy$
$$\left|\frac{z - 1}{z + 1}\right| = \left|\frac{x - 1 + iy}{x + 1 + iy}\right| = 1$$ $$|z - 1| = |z + 1|$$ $$\sqrt{(x-1)^2 + y^2} = \sqrt{(x+1)^2 + y^2}$$Squaring:
$$(x-1)^2 + y^2 = (x+1)^2 + y^2$$ $$x^2 - 2x + 1 = x^2 + 2x + 1$$ $$-2x = 2x$$ $$x = 0$$When $x = 0$, $z = iy$ which is purely imaginary (assuming $y \neq 0$).
Proved!
Quick Revision Box
| Concept | Formula/Key Point |
|---|---|
| Imaginary unit | $i^2 = -1$ |
| Powers of $i$ | $i^{4k+r} = i^r$ where $r \in \{0,1,2,3\}$ |
| Complex number | $z = a + ib$, $a = \text{Re}(z)$, $b = \text{Im}(z)$ |
| Equality | $a + ib = c + id \iff a=c, b=d$ |
| Addition | $(a+ib) + (c+id) = (a+c) + i(b+d)$ |
| Multiplication | $(a+ib)(c+id) = (ac-bd) + i(ad+bc)$ |
| Conjugate | $\overline{a+ib} = a - ib$ |
| Key property | $(a+ib)(a-ib) = a^2 + b^2$ |
| Division | Multiply by $\frac{\bar{\text{denominator}}}{\bar{\text{denominator}}}$ |
Related Topics
Within Complex Numbers Chapter
- Argand Diagram - Geometric representation of complex numbers
- Modulus and Argument - Polar form and properties
- De Moivre’s Theorem - Powers and roots
- Cube Roots of Unity - Special roots
- Quadratic Equations - Solving equations with complex roots
Math Connections
- Trigonometric Identities - Euler’s formula connects complex numbers and trig
- Circles - Argand plane is like coordinate plane
- Binomial Expansion - $(a+ib)^n$ expansions
- Sets Basics - Understanding number sets
Physics Applications
- Electromagnetic Waves - Complex exponentials for wave solutions
- AC Circuits - Impedance represented as complex numbers
Teacher’s Summary
- The imaginary unit $i$ is defined by $i^2 = -1$, and powers of $i$ cycle every 4: $(1, i, -1, -i)$
- Complex numbers $z = a + ib$ extend real numbers, with both parts being real numbers
- Algebraic operations follow the same rules as real numbers, with $i^2 = -1$
- Conjugate $\bar{z} = a - ib$ is crucial for division and many properties
- Key property: $(a+ib)(a-ib) = a^2 + b^2$ (always real and non-negative)
“Complex numbers aren’t complex — they’re just numbers with an extra dimension. Master the basics, and the rest becomes simple!”
Exam Strategy: For JEE, be lightning fast with powers of $i$ and division using conjugates. These appear in 90% of complex number problems!