FM1: Complex Numbers and De Moivre's Theorem

Year 1 / ASYear 2 / A-Level All Boards (AQA, Edexcel, OCR, WJEC, CCEA) AQA

Modulus-argument form, De Moivre's Theorem, roots of unity, loci in the complex plane.

Exam Board:

📋 Key Definitions and Core Concepts

De Moivre's Theorem: [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) for all n ∈ ℤ.

Roots of Unity: Solutions to zⁿ = 1: z_k = e^{i 2kπ/n} for k = 0, 1, ..., n-1.

Euler's formula: e^{iθ} = cos θ + i sin θ; e^{iπ} + 1 = 0.
Loci: |z - z₁| = r is a circle centre z₁; |z - z₁| = |z - z₂| is perpendicular bisector.
Non-real roots of real polynomials occur in complex conjugate pairs z and z̄.

Exam-Style Question

Question:

Express cos 3θ in terms of cos θ using De Moivre's Theorem.

Model Solution & Mark Scheme:

(cos θ + i sin θ)³ = cos³θ + 3i cos²θ sin θ - 3 cos θ sin²θ - i sin³θ.
Real part: cos 3θ = cos³θ - 3 cos θ (1 - cos²θ) = 4 cos³θ - 3 cos θ.

Q1: Find all solutions to z⁴ = -16.

Show Model Answer

Answer: z = 2 e^{i(π/4 + kπ/2)} => √2(±1 ± i).

Q2: Describe the locus |z - 3i| = 2.

Show Model Answer

Answer: A circle with centre (0, 3) and radius 2.

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