Complex Numbers — vidhyapath 2027

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).