De Moivre's theorem (WACE Year 12 Mathematics Specialist Unit 3)

What this dot point is asking

SCSA wants you to apply de Moivre's theorem to integer powers and to use it to derive trigonometric identities by equating real and imaginary parts.

The theorem

For $n \ge 1$ this follows by induction from the polar multiplication rule: each extra factor multiplies the modulus by $r$ and adds $\theta$ to the argument. The base case $n = 1$ is trivial, and if the result holds for $n = k$ then

It extends to negative integers using $\tfrac{1}{z} = \tfrac{1}{r}\,\text{cis}(-\theta)$ and to $n = 0$ trivially.

Evaluating powers

To compute $z^n$, convert $z$ to polar form, apply the theorem, then convert back if a Cartesian answer is required. This is far faster than expanding $(x + iy)^n$ for large $n$.

Deriving trigonometric identities

Take $r = 1$ and write $(\cos\theta + i\sin\theta)^n$ two ways: once by de Moivre as $\cos n\theta + i\sin n\theta$, and once by the binomial theorem. Equating real parts gives a formula for $\cos n\theta$ and equating imaginary parts gives $\sin n\theta$, both in terms of powers of $\cos\theta$ and $\sin\theta$.

Powers of i and roots of unity

A special case worth knowing is $z = \text{cis}\,\theta$ with modulus $1$: every power then stays on the unit circle, since $z^n = \text{cis}(n\theta)$. Setting $z = \text{cis}\,\dfrac{2\pi}{n}$ gives the primitive $n$th root of unity, whose powers $1, z, z^2, \dots, z^{n-1}$ are the $n$ solutions of $w^n = 1$, equally spaced around the unit circle. De Moivre's theorem is the engine behind this: it converts the root problem into adding equal angles.

Negative and fractional powers

The theorem holds for every integer $n$, including negatives. For $n = -1$, $[r\,\text{cis}\,\theta]^{-1} = r^{-1}\,\text{cis}(-\theta)$, which agrees with the division rule. For example $\left(2\,\text{cis}\,\dfrac{\pi}{3}\right)^{-2} = 2^{-2}\,\text{cis}\!\left(-\dfrac{2\pi}{3}\right) = \dfrac{1}{4}\,\text{cis}\!\left(-\dfrac{2\pi}{3}\right)$. Strictly, de Moivre's theorem as stated is for integer exponents; for fractional powers it gives one of several roots, which is why finding all the $n$th roots of a complex number is handled separately by adding $\dfrac{2\pi k}{n}$ to the argument.

A second worked identity

The same method gives $\cos n\theta$ for any $n$. For $n = 4$, expanding $(\cos\theta + i\sin\theta)^4$ and taking the real part gives $\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta$, which simplifies using $\sin^2\theta = 1 - \cos^2\theta$ to $\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$. These identities are the bridge from de Moivre's theorem to the integration of powers of $\cos$ and $\sin$ in Unit 4.