Complex Numbers

Complex numbers deliver 2–4 direct questions in NDA Paper I, almost all solvable in under a minute once you know modulus, argument and the powers of i. It is one of the highest reward-to-effort chapters in Algebra.

Core ideas

A complex number is z = a + ib where i = √(−1), so i² = −1. Powers of i cycle every four: i¹=i, i²=−1, i³=−i, i⁴=1. To evaluate iⁿ, divide n by 4 and use the remainder.

Modulus and conjugate: |z| = √(a²+b²), conjugate z̄ = a − ib, and z·z̄ = |z|² = a²+b².

Argument: arg(z) = tan⁻¹(b/a), adjusted for the quadrant of (a, b).

Polar / Euler form: z = r(cos θ + i sin θ) = r·e^(iθ), where r = |z|.

De Moivre's theorem: (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ — the workhorse for powers and roots.

Cube roots of unity: 1, ω, ω² where ω = (−1 + i√3)/2. Key facts: 1 + ω + ω² = 0 and ω³ = 1.

Exam Tricks & Tips

• 🎯 |z₁z₂| = |z₁||z₂| and |z₁/z₂| = |z₁|/|z₂|; use these instead of expanding messy products.
• 🎯 arg(z₁z₂) = arg z₁ + arg z₂ — arguments ADD on multiplication, subtract on division.
• 🎯 For any question with ω, replace ω³ by 1 and ω² by (−1−ω) to collapse it instantly.
• 🎯 To rationalise 1/(a+ib), multiply top and bottom by the conjugate a−ib; denominator becomes a²+b².
• 🎯 |z₁ + z₂| ≤ |z₁| + |z₂| (triangle inequality) helps in maximum/minimum modulus questions.
• ❌ Common mistake: writing √(−4)·√(−9) = √36 = 6. The rule √a·√b = √(ab) fails for negatives; correct is (2i)(3i) = −6.

Expected exam pattern

Typical items: find iⁿ or a power like (1+i)⁸, compute modulus/argument, simplify an ω-expression, or find the conjugate/multiplicative inverse. Occasionally a De Moivre root-of-unity problem.

Quick recap

Remember i cycles mod 4, |z|²=a²+b²=z·z̄, moduli multiply and arguments add, De Moivre for powers, and 1+ω+ω²=0. Never split square roots of negatives.

Cover, recall, check. 11 cards on complex-number results NDA loves.

Q1. Value of i² , i³ , i⁴?
A1. i² = −1, i³ = −i, i⁴ = 1 (cycle repeats every 4).

Q2. How to find iⁿ for large n?
A2. Take n mod 4; the remainder 0,1,2,3 gives 1, i, −1, −i.

Q3. Modulus of z = a + ib?
A3. |z| = √(a² + b²).

Q4. z·z̄ equals?
A4. |z|² = a² + b².

Q5. Multiplicative inverse of z?
A5. z⁻¹ = z̄ / |z|² = (a − ib)/(a² + b²).

Q6. De Moivre's theorem?
A6. (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ.

Q7. The three cube roots of unity?
A7. 1, ω, ω² where ω = (−1 + i√3)/2.

Q8. Two key identities for ω?
A8. 1 + ω + ω² = 0 and ω³ = 1.

Q9. arg(z₁z₂) in terms of individual arguments?
A9. arg z₁ + arg z₂ (arguments add on multiplication).

Q10. Polar form of z?
A10. z = r(cos θ + i sin θ), r = |z|, θ = arg z.

Q11. Why is √(−4)·√(−9) ≠ 6?
A11. √a·√b = √(ab) fails for negatives; (2i)(3i) = 6i² = −6.

When x² + 1 = 0 has no real solution, mathematics invents i = √(−1) and a whole new plane opens up. Complex numbers are a guaranteed 2–3 mark block in NDA.

Core idea / what this tests: Algebra of a + ib, modulus–argument (polar) form, and the elegant results — De Moivre's theorem and roots of unity — that turn ugly powers into one-line answers.

Deep explanation

Beginner — The number a + ib

A complex number z = a + ib has real part a and imaginary part b, with i² = −1. Powers of i cycle with period 4: i, −1, −i, 1. So i^n depends only on n mod 4.

• Addition: add real and imaginary parts.
• Multiplication: use i² = −1.
• Conjugate z̄ = a − ib; z·z̄ = a² + b² is real.
• To divide, multiply numerator and denominator by the conjugate of the denominator.

Intermediate — Modulus and argument

• Modulus |z| = √(a² + b²) = distance from origin.
• Argument θ = arg(z) with tan θ = b/a (fix the quadrant from the signs of a, b).
• Polar form: z = r(cos θ + i sin θ), often written r·cis θ.

Useful identities: |z₁z₂| = |z₁||z₂|, arg(z₁z₂) = arg z₁ + arg z₂, and |z|² = z·z̄.

Advanced — De Moivre and roots of unity

De Moivre's Theorem: (cos θ + i sin θ)^n = cos nθ + i sin nθ.
The n n-th roots of unity are cos(2kπ/n) + i sin(2kπ/n), k = 0…n−1. They sum to zero and are spaced equally on the unit circle. The cube roots are 1, ω, ω² with 1 + ω + ω² = 0 and ω³ = 1.

Worked example

Evaluate (1 + i)^8.
Write 1 + i in polar form: modulus = √2, argument = 45°. So 1 + i = √2·cis 45°.
By De Moivre, (1+i)^8 = (√2)^8 · cis(8·45°) = 16 · cis 360° = 16·(1) = 16.

NDA relevance

Expect one direct simplification (powers of i or (1+i)^n), one modulus/argument question, and frequently a cube-root-of-unity identity such as evaluating (1 + ω − ω²)^n. These are fast marks if the polar shortcut is reflex.

Exam tricks & shortcuts

• For high powers, always convert to polar — De Moivre beats binomial expansion every time.
• Memorise 1 + ω + ω² = 0 and ω³ = 1; replace ω² by (−1 − ω) instantly.
• i^n: reduce n mod 4 mentally.

Key formulas

• z = a + ib; modulus |z| = √(a²+b²); conjugate z̄ = a − ib; z·z̄ = |z|².
• Polar: z = r(cosθ + i sinθ) = re^{iθ}, θ = arg z = tan⁻¹(b/a).
• Multiply: moduli multiply, arguments add; divide: moduli divide, arguments subtract.
• De Moivre: (cosθ + i sinθ)ⁿ = cos nθ + i sin nθ.
• |z₁z₂| = |z₁||z₂|; |z₁+z₂| ≤ |z₁|+|z₂| (triangle inequality).
• Cube roots of unity 1, ω, ω²: 1+ω+ω²=0, ω³=1.

Usage: switch to polar form for powers and roots of complex numbers.

Worked Example

Problem: Express z = −1 + i√3 in polar (modulus–argument) form, and hence find z³.

Solution:
Write z = x + iy with x = −1, y = √3.

Modulus: |z| = √(x² + y²) = √((−1)² + (√3)²) = √(1 + 3) = √4 = 2.

Argument: the reference angle α satisfies tan α = |y| / |x| = √3 / 1, so α = 60°. Since x < 0 and y > 0, z lies in the second quadrant, so
arg(z) = 180° − 60° = 120° = 2π/3.

Polar form: z = 2(cos 120° + i sin 120°).

Now apply De Moivre's theorem, (r(cos θ + i sin θ))ⁿ = rⁿ(cos nθ + i sin nθ):
z³ = 2³ (cos(3·120°) + i sin(3·120°)) = 8(cos 360° + i sin 360°).
Since cos 360° = 1 and sin 360° = 0,
z³ = 8(1 + i·0) = 8.

Answer: z = 2(cos 120° + i sin 120°) and z³ = 8.