Complex Numbers

Concept Core

From imaginary unit to Argand plane geometry — everything you need.

The Imaginary Unit & Powers of i

Define i = √(−1). Then i² = −1, i³ = −i, i⁴ = 1. Powers of i cycle with period 4.

i⁰=1 | i¹=i | i²=−1 | i³=−i | i⁴=1 | i⁴ᵏ=1 always

To find iⁿ: divide n by 4, use remainder: r=0→1, r=1→i, r=2→−1, r=3→−i.

Standard Form & Operations

A complex number: z = a + ib where a = real part, b = imaginary part.

Addition: (a+ib)+(c+id) = (a+c) + i(b+d)

Multiplication: (a+ib)(c+id) = (ac−bd) + i(ad+bc)

Division: Multiply numerator and denominator by conjugate of denominator.

Modulus & Argument

Modulus: |z| = √(a² + b²) — distance from origin in Argand plane

Argument: arg(z) = θ = tan⁻¹(b/a) — angle from positive real axis

Polar form: z = r(cosθ + i sinθ) = re^(iθ) where r = |z|

|z₁z₂| = |z₁||z₂| arg(z₁z₂) = arg(z₁) + arg(z₂)

Conjugate of a Complex Number

If z = a + ib, then conjugate z̄ = a − ib (flip sign of imaginary part).

Key properties:

z · z̄ = a² + b² = |z|² | z + z̄ = 2a (real) | z − z̄ = 2ib (imaginary)

Use z̄ to rationalise: divide by (a+ib) → multiply by (a−ib)/(a²+b²)

Argand Plane (Complex Plane)

Plot z = a + ib as the point (a, b). Real axis = x-axis, imaginary axis = y-axis.

Distance between z₁ and z₂: |z₁ − z₂|

Midpoint of z₁, z₂: (z₁ + z₂)/2

|z| = r represents a circle of radius r centred at origin

|z − z₀| = r represents a circle of radius r centred at z₀

De Moivre's Theorem

For any integer n: (cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ)

(re^iθ)ⁿ = rⁿ e^(inθ) = rⁿ(cos nθ + i sin nθ)

Use to find nth roots of unity and powers of complex numbers in polar form.

Cube Roots of Unity (ω) — EAPCET Favourite

The cube roots of 1 are: 1, ω, ω² where ω = (−1 + i√3)/2

1 + ω + ω² = 0 ω³ = 1 ω̄ = ω²

These identities unlock factorisation problems. If any expression = 1+ω+ω² appearing in a sum, it equals zero.

Also: |ω| = 1, arg(ω) = 120°, arg(ω²) = 240°. They lie on unit circle.

Formula Vault

Every complex number formula — from basic to De Moivre's.

Imaginary Unit

i = √(−1); i² = −1

Powers cycle with period 4

Standard Form

z = a + ib

Re(z) = a; Im(z) = b

Modulus

|z| = √(a² + b²)

Distance from origin

Conjugate

z̄ = a − ib

z · z̄ = |z|²

Argument

arg(z) = tan⁻¹(b/a)

Adjust for correct quadrant

Polar Form

z = r(cosθ + i sinθ)

r = |z|; θ = arg(z)

De Moivre's Theorem

(cosθ + i sinθ)ⁿ = cos nθ + i sin nθ

For any integer n

Cube Roots of Unity

1 + ω + ω² = 0; ω³ = 1

ω = (−1+i√3)/2

Modulus of Product

|z₁z₂| = |z₁| · |z₂|

arg(z₁z₂) = arg z₁ + arg z₂

Division

z₁/z₂ = z₁·z̄₂ / |z₂|²

Multiply by conjugate

Triangle Inequality

|z₁+z₂| ≤ |z₁| + |z₂|

||z₁|−|z₂|| ≤ |z₁−z₂|

nth Root of Unity

e^(2πik/n), k=0..n−1

Sum of all nth roots = 0

Worked Examples

5 problems from powers of i to cube roots — all EAPCET patterns.

EasyFind the value of i⁵⁷ + i⁵⁸ + i⁵⁹ + i⁶⁰▾

Evaluate: i⁵⁷ + i⁵⁸ + i⁵⁹ + i⁶⁰

60 ÷ 4 = 15 remainder 0, so i⁶⁰ = i⁰ = 1

57 ÷ 4 = 14 r 1 → i⁵⁷ = i; 58 ÷ 4 = 14 r 2 → i⁵⁸ = −1; 59 ÷ 4 = 14 r 3 → i⁵⁹ = −i

Sum = i + (−1) + (−i) + 1 = 0

✓ i⁵⁷ + i⁵⁸ + i⁵⁹ + i⁶⁰ = 0 (any 4 consecutive powers of i sum to zero)

EasyFind modulus and argument of z = 1 − i▾

For z = 1 − i, find |z| and arg(z).

|z| = √(1² + (−1)²) = √2

Basic angle: tan⁻¹(|−1|/1) = tan⁻¹(1) = 45° = π/4

z = 1 − i is in 4th quadrant (positive real, negative imaginary). So arg(z) = −π/4 (or 315°)

✓ |z| = √2, arg(z) = −π/4

MediumSimplify (1+i)/(1−i) and find its modulus▾

Simplify (1+i)/(1−i) to a+ib form and find its modulus.

Multiply numerator and denominator by conjugate (1+i): (1+i)²/((1−i)(1+i))

Denominator: 1 − i² = 1+1 = 2

Numerator: (1+i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i

Result: 2i/2 = i = 0 + 1·i

✓ (1+i)/(1−i) = i, |i| = 1

EAPCET LevelIf 1, ω, ω² are cube roots of unity, find (1−ω+ω²)⁴ + (1+ω−ω²)⁴▾

Find the value of (1−ω+ω²)⁴ + (1+ω−ω²)⁴ where ω is a cube root of unity.

Use: 1 + ω + ω² = 0, so 1 + ω² = −ω and 1 + ω = −ω²

1 − ω + ω² = (1+ω²) − ω = −ω − ω = −2ω

1 + ω − ω² = (1+ω) − ω² = −ω² − ω² = −2ω²

(−2ω)⁴ + (−2ω²)⁴ = 16ω⁴ + 16ω⁸ = 16ω + 16ω² (since ω³=1)

= 16(ω + ω²) = 16(−1) = −16

✓ Result = −16

Trap QuestionFind the locus of z if |z−2| = |z+2|. Students get the geometry wrong.▾

Find the locus of z = x+iy such that |z−2| = |z+2|.

The trap: students try to expand algebraically and get messy. Use the geometric meaning.

|z−2| = distance of z from point (2,0). |z+2| = distance of z from (−2,0).

Equal distances from two points → locus is the perpendicular bisector of the segment joining (2,0) and (−2,0).

That perpendicular bisector is the y-axis, i.e., x = 0, i.e., Re(z) = 0.

✓ Locus is the imaginary axis (y-axis): Re(z) = 0

Mistake DNA

4 errors from distractor analysis — where EAPCET candidates lose marks.

🔢

Wrong Powers of i — Not Using the Cycle

Students compute i²³ as if i repeats with period 2 or forget to use modular arithmetic.

❌ Wrong

i²³ = i²² × i
= (i²)¹¹ × i = (−1)¹¹ × i
= −i (correct here but
method is fragile) ✗

✓ Correct

23 ÷ 4 = 5 r 3
i²³ = i³ = −i ✓
Use remainder directly.

Always use remainder when 4 divides n. This is faster and less error-prone than repeated squaring.

📐

Argument Quadrant Error

Using tan⁻¹(b/a) without checking which quadrant z is in gives wrong argument for 2nd/3rd quadrant numbers.

❌ Wrong

z = −1 + i√3:
arg = tan⁻¹(√3/−1)
= tan⁻¹(−√3) = −60° ✗
(4th quadrant angle)

✓ Correct

z is in 2nd quadrant
arg = 180° − 60°= 120°
= 2π/3 ✓

tan⁻¹(b/a) gives a reference angle. Adjust: 2nd quadrant → π−θ, 3rd quadrant → π+θ (or −π+θ), 4th quadrant → −θ.

🔄

Forgetting ω³ = 1 and 1+ω+ω² = 0

Not substituting these key identities leads to very messy algebra.

❌ Wrong

Trying to expand
(1+ω)⁸ by expanding
binomially without using
1+ω = −ω² ✗

✓ Correct

1+ω = −ω²
(1+ω)⁸ = (−ω²)⁸
= ω¹⁶ = ω (since ω³=1) ✓

Commit to memory: 1+ω+ω²=0, ω³=1, conjugate of ω is ω². These three identities solve 90% of cube root problems in under 30 seconds.

🌀

Confusing |z₁+z₂| with |z₁|+|z₂|

The triangle inequality is an inequality, not equality — equality holds only when z₁ and z₂ have the same argument.

❌ Wrong

|z₁+z₂| = |z₁|+|z₂|
always ✗
(only in special case)

✓ Correct

|z₁+z₂| ≤ |z₁|+|z₂| ✓
Equality when arg(z₁)=arg(z₂)
(same direction)

The triangle inequality gives bounds. For exact computation, use |z|² = z·z̄ to avoid angle complications.

Chapter Intelligence

Weightage, PYQ patterns, and exam strategy for Complex Numbers.

EAPCET Topic Weightage (2019–2024)

Cube roots of unity (ω)

Modulus & argument

Argand plane/locus

Powers of i

De Moivre's theorem

High-Yield PYQ Patterns

Find (1+ω)ⁿ or (1−ω+ω²)ⁿ arg of complex number Locus: |z−a| = |z−b| Powers of i (large exponent) Simplify using conjugate |z₁/z₂| and arg(z₁/z₂)

Exam Strategy

For iⁿ questions: find n mod 4 first. This takes 5 seconds and is never wrong.
Cube root of unity questions: write 1+ω = −ω² and 1+ω² = −ω immediately. These substitutions collapse complex expressions in one step.
Locus questions: think geometrically before expanding algebraically. |z−a| = |z−b| is always a perpendicular bisector.
Modulus questions: use |z|² = z·z̄ to avoid square roots. Compute modulus squared, then take square root at the end.
Complex numbers connect to Trigonometry (De Moivre's, nth roots) and Coordinate Geometry (Argand plane loci).

Concept Core

Formula Vault

Worked Examples

Mistake DNA

Chapter Intelligence

💡 Suggestions & Feedback

Thank you!