De Moivre's theorem for integer and rational powers, using it to find powers of complex numbers, and applying it with the binomial theorem to derive multiple-angle identities and to express powers of sine and cosine.

What this dot point is asking

OCR wants you to state and use de Moivre's theorem, $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$, to compute powers of complex numbers, to derive multiple-angle identities (such as $\cos 3\theta$ and $\sin 3\theta$) by expanding with the binomial theorem and equating real or imaginary parts, and to express powers of $\cos\theta$ or $\sin\theta$ in terms of multiple angles using $z = \cos\theta + i\sin\theta$ together with $z^n + z^{-n} = 2\cos n\theta$ and $z^n - z^{-n} = 2i\sin n\theta$.

The theorem

De Moivre's theorem extends "moduli multiply, arguments add" to powers. Raising $\cos\theta + i\sin\theta$ to the power $n$ multiplies the argument by $n$, which is why it produces $\cos n\theta + i\sin n\theta$. It is proved for positive integers by induction (a standard exam proof) and extends to negative and rational $n$.

Powers of complex numbers

To compute a power such as $(1 + i)^8$, convert to modulus-argument form, apply de Moivre, then convert back if needed. This is far quicker than repeated multiplication.

Deriving multiple-angle identities

To get $\cos n\theta$ or $\sin n\theta$ in terms of $\cos\theta$ and $\sin\theta$, expand $(\cos\theta + i\sin\theta)^n$ with the binomial theorem and take the real or imaginary part. Real parts give the cosine identity; imaginary parts give the sine identity. A final substitution of $\sin^2\theta = 1 - \cos^2\theta$ (or vice versa) tidies the result.

This is the method behind the standard results $\cos 2\theta = 2\cos^2\theta - 1$, $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ and $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$.

Powers of sine and cosine (linearising)

The reverse problem, writing $\cos^n\theta$ or $\sin^n\theta$ as a sum of cosines or sines of multiple angles, is solved by setting $z = \cos\theta + i\sin\theta$. Then de Moivre gives the crucial identities below, you expand the relevant power of $(z \pm z^{-1})$ with the binomial theorem, and you regroup the terms into multiple-angle pairs. This is exactly what you need before integrating a power of sine or cosine.

De Moivre's theorem ties the complex strand to trigonometry and is the engine for the roots of unity, where the same argument-multiplying idea finds all the $n$th roots of a complex number.

Try this

Q1. Use de Moivre to evaluate $\left(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\right)^6$. [2 marks]

• Cue. The argument is multiplied by $6$: $\cos\pi + i\sin\pi = -1$.

Q2. State the real and imaginary parts you would equate to derive $\cos 4\theta$ from $(\cos\theta + i\sin\theta)^4$. [2 marks]

• Cue. Take the real part; $\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta$.