Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Overview
A complex number z = a + ib has a real part a and an imaginary part b, where i = √(−1) and i² = −1. This chapter covers algebra, conjugate, modulus, argument, polar/Euler form, and De Moivre's theorem.
1. Algebra and Conjugate
i² = −1, i³ = −i, i⁴ = 1 (powers of i repeat every 4)
Conjugate: z̄ = a − ib
z · z̄ = a² + b² = |z|²
2. Modulus and Argument
Modulus: |z| = √(a² + b²)
Argument: θ = arg(z) = tan⁻¹(b/a) (fix quadrant using signs of a, b)
Properties: |z₁z₂| = |z₁||z₂| , arg(z₁z₂) = arg z₁ + arg z₂
3. Polar and Euler Form
Polar: z = r(cos θ + i sin θ)
Euler: z = r e^{iθ}
4. De Moivre's Theorem
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
Used for powers and the n nth-roots of a complex number.
Worked Examples
Example 1 — Conjugate and division
Express (3 + 2i)/(1 − i) in a + ib form.
Multiply by conjugate (1 + i):
= (3 + 2i)(1 + i) / (1 − i)(1 + i) = (3 + 3i + 2i + 2i²)/(1 + 1)
= (3 + 5i − 2)/2 = (1 + 5i)/2 = 1/2 + (5/2)i
Example 2 — Modulus
| **Find | z | for z = 3 + 4i.** |
|---|
|z| = √(3² + 4²) = √25 = 5
Example 3 — Argument
Find arg(1 + i).
θ = tan⁻¹(1/1) = π/4 (first quadrant)
Example 4 — Polar form
Write z = 1 + i√3 in polar form.
r = √(1 + 3) = 2 ; θ = tan⁻¹(√3/1) = π/3
z = 2(cos π/3 + i sin π/3)
Example 5 — De Moivre's theorem
Find (cos 30° + i sin 30°)⁶.
= cos(6×30°) + i sin(6×30°) = cos 180° + i sin 180° = −1
Common Mistakes
- Writing i² = 1 instead of −1.
- Forgetting to adjust the argument for the correct quadrant.
- Not rationalising (multiplying by conjugate) when dividing.
- Confusing modulus (√(a²+b²)) with the argument.
Exam Tips & Revision Checklist
- z·z̄ = |z|² is the key trick for division.
- Memorise the powers of i cycle and De Moivre's theorem.
- Practised: algebra of i, conjugate division, modulus, argument, polar form, De Moivre.
Frequently Asked Questions
Modulus |z| = √(a² + b²); conjugate z̄ = a − ib. Also z·z̄ = |z|² = a² + b².