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Further Complex Numbers

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Edexcel IAL Further Pure Mathematics 2 (WFM02)

Further Complex Numbers

The big idea: Writing complex numbers as re^iθ turns powers and roots into simple arithmetic on r and θ — and that single idea produces de Moivre's theorem, trig identities and nth roots.

Summary — What This Chapter Covers

Euler's relation: e^iθ = cos θ + i sin θ, so z = r(cos θ + i sin θ) = re^iθ.
Multiplying multiplies moduli and adds arguments; dividing divides moduli and subtracts arguments.
De Moivre: (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ for any integer n (proved by induction for positive n).
With z = e^iθ: zⁿ + z⁻ⁿ = 2 cos nθ and zⁿ − z⁻ⁿ = 2i sin nθ.
The nth roots of a complex number are equally spaced on a circle, 2π/n apart.

1. Exponential Form and Euler's Relation

Exponential form

z = re^iθ

Every non-zero complex number can be written z = re^iθ, where r = |z| and θ = arg z. Products and quotients become easy:

Rules z₁z₂ = r₁r₂e^{i(θ₁ + θ₂)} z₁/z₂ = (r₁/r₂)e^{i(θ₁ − θ₂)}

Useful forms cos θ = ½(e^iθ + e^−iθ) sin θ = (1/(2i))(e^iθ − e^−iθ)

2. De Moivre's Theorem

De Moivre

Powers of cos θ + i sin θ

(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. For positive integers it is proved by induction: assume it for n = k, multiply by (cos θ + i sin θ), and use the compound-angle formulae to get cos(k + 1)θ + i sin(k + 1)θ. For negative integers, write n = −m and use 1/(cos mθ + i sin mθ) = cos mθ − i sin mθ.

Worked Example

Express cos 3θ in terms of cos θ.

3. Trigonometric Identities and Series

Identities

Powers of cos θ in terms of multiple angles

To write cosⁿθ or sinⁿθ in terms of multiple angles, let z = e^iθ, use 2 cos θ = z + 1/z (or 2i sin θ = z − 1/z), expand with the binomial theorem, and pair terms as z^k + z^−k = 2 cos kθ.

Worked Example

Show that cos³θ = ¼ cos 3θ + ¾ cos θ.

Sums of series

A sum like C = Σ cos kθ is found by forming C + iS = Σ e^ikθ, which is a geometric series. Sum it, then take the real part.

4. nth Roots of a Complex Number

Roots

Equally spaced on a circle

To solve zⁿ = re^iθ, write the right side with a general argument θ + 2kπ:

Roots z = r^1/n e^{i(θ + 2kπ)/n}, k = 0, 1, …, n − 1

Worked Example

Solve z³ = 8i, giving answers in the form a + bi.

Roots of unity

ⁿ ² ⁿ⁻¹

Memorise

Euler

e^iθ = cos θ + i sin θ

De Moivre

ⁿ

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Further Complex Numbers

Further Complex Numbers

z = reiθ

Powers of cos θ + i sin θ

Powers of cos θ in terms of multiple angles

Equally spaced on a circle

Euler

De Moivre

More Further Pure Mathematics 2 topics

z = re^iθ