Complex numbers

Further Maths · Further Pure Mathematics 2. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

When two complex numbers are multiplied, their moduli are multiplied and their arguments are added. So raising a number to the power n raises the modulus to the power n and multiplies the argument by n. This is de Moivre's theorem, and it holds for positive and negative integers and for rational powers.

The theorem has four main uses: writing cos nθ, sin nθ or tan nθ in powers of cos θ, sin θ or tan θ; writing powers of sin θ and cos θ in multiple angles; summing series of sines and cosines by the C + iS method; and finding roots.

The nth roots of unity are the n solutions of zⁿ = 1. They lie equally spaced on the unit circle.

Rules to remember

de Moivre: [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ)
Multiply: multiply the moduli and add the arguments. Divide: divide the moduli and subtract the arguments
Multiple angles: expand (cos θ + i sin θ)ⁿ by the binomial theorem; the real part is cos nθ and the imaginary part is sin nθ
If z = cos θ + i sin θ: zⁿ + z⁻ⁿ = 2 cos nθ and zⁿ − z⁻ⁿ = 2i sin nθ; so (2 cos θ)ⁿ = (z + 1/z)ⁿ and (2i sin θ)ⁿ = (z − 1/z)ⁿ
nth roots of unity: z = e^2πki/n for k = 0, 1, …, n − 1; their sum is 0 (for n ≥ 2)
nth roots of re^iα: z = r^1/n e^{i(α + 2πk)/n} for k = 0, 1, …, n − 1
C + iS method: C + iS is a geometric or binomial series in e^iθ; sum it, then take the real part for C and the imaginary part for S

Common mistake

When finding nth roots students give only one root. Add 2πk to the argument before dividing by n, with k = 0, 1, …, n − 1, to get all n roots.

Worked example

Use de Moivre's theorem to find (√3 + i)⁶.

Modulus: √((√3)² + 1²) = √4 = 2. Argument: tan⁻¹(1/√3) = π/6.
(√3 + i)⁶ = 2⁶[cos(6 × π/6) + i sin(6 × π/6)] = 64(cos π + i sin π).
cos π = −1 and sin π = 0, so the value is 64 × (−1) = −64.

Answer: −64

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Complex Numbers II (de Moivre, roots)

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