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How does de Moivre's theorem let you take powers and roots of complex numbers?

De Moivre's theorem, the exponential (Euler) form of a complex number, using de Moivre to derive trigonometric identities, and finding the nth roots of a complex number.

A CCEA A2 Further Maths answer on de Moivre's theorem, the exponential form of a complex number, deriving trigonometric identities such as multiple-angle formulae, and finding the nth roots of a complex number on the Argand diagram.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

CCEA wants you to use de Moivre's theorem to raise complex numbers to powers, work in the exponential (Euler) form, derive trigonometric identities such as multiple-angle formulae, and find the nnth roots of a complex number, knowing they are equally spaced on a circle in the Argand diagram.

The answer

De Moivre's theorem

For a general complex number z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta), zn=rn(cos⁡nθ+isin⁡nθ)z^n = r^n(\cos n\theta + i\sin n\theta): the modulus is raised to the power and the argument is multiplied.

The exponential form

Deriving trigonometric identities

The nth roots of a complex number

Examples in context

Example 1. Roots of unity in signal processing. The nnth roots of 11, equally spaced around the unit circle, are the backbone of the discrete Fourier transform that decomposes audio and images into frequencies. De Moivre's geometry of equally spaced roots is exactly why the transform works.

Example 2. Deriving identities without memorising them. Rather than memorising cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta, a student rederives it in seconds from (cos⁡θ+isin⁡θ)3(\cos\theta + i\sin\theta)^3. De Moivre turns a long list of identities into one method.

Try this

Q1. Use de Moivre to find (cos⁡π6+isin⁡π6)3\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)^3. [2 marks]

  • Cue. cos⁡π2+isin⁡π2=i\cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i.

Q2. How many distinct nnth roots does a non-zero complex number have? [1 mark]

  • Cue. Exactly nn, equally spaced on a circle.

Q3. Write z=2eiπ/3z = 2e^{i\pi/3} in the form r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta). [1 mark]

  • Cue. 2(cos⁡π3+isin⁡π3)2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right).

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA A2 20216 marksUse de Moivre's theorem to express cos⁡3θ\cos 3\theta in terms of cos⁡θ\cos\theta.
Show worked answer →

By de Moivre's theorem, (cos⁡θ+isin⁡θ)3=cos⁡3θ+isin⁡3θ(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta.

Expand the left side with the binomial theorem:

(cos⁡θ+isin⁡θ)3=cos⁡3θ+3cos⁡2θ(isin⁡θ)+3cos⁡θ(isin⁡θ)2+(isin⁡θ)3.(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3.

Using i2=−1i^2 = -1 and i3=−ii^3 = -i, the real part is cos⁡3θ−3cos⁡θsin⁡2θ\cos^3\theta - 3\cos\theta\sin^2\theta.

Equating real parts: cos⁡3θ=cos⁡3θ−3cos⁡θsin⁡2θ\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta. Replace sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta:

cos⁡3θ=cos⁡3θ−3cos⁡θ(1−cos⁡2θ)=4cos⁡3θ−3cos⁡θ.\cos 3\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta.

Markers reward applying de Moivre, the binomial expansion, taking the real part, and substituting for sin⁡2θ\sin^2\theta.

CCEA A2 20197 marksFind the three cube roots of 88, giving each in the form r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta), and show them on an Argand diagram.
Show worked answer →

Write 88 in modulus-argument form: 8=8(cos⁡0+isin⁡0)8 = 8(\cos 0 + i\sin 0), and add multiples of 2π2\pi to the argument: 8=8(cos⁡(2kπ)+isin⁡(2kπ))8 = 8\big(\cos(2k\pi) + i\sin(2k\pi)\big).

The cube roots have modulus 81/3=28^{1/3} = 2 and arguments 0+2kπ3\dfrac{0 + 2k\pi}{3} for k=0,1,2k = 0, 1, 2:

k=0:  2(cos⁡0+isin⁡0);k=1:  2(cos⁡2π3+isin⁡2π3);k=2:  2(cos⁡4π3+isin⁡4π3).k = 0:\; 2(\cos 0 + i\sin 0); \quad k = 1:\; 2\left(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\right); \quad k = 2:\; 2\left(\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}\right).

These are 22, −1+i3-1 + i\sqrt{3} and −1−i3-1 - i\sqrt{3}, equally spaced at 120∘120^\circ around a circle of radius 22 centred at the origin.

Markers reward the modulus-argument form with the added 2kπ2k\pi, the three roots, and the equal spacing on the Argand diagram.

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