Core Pure · Complex Numbers · Year 2

De Moivre's Theorem

One compact identity that turns raising a complex number to a power into simple arithmetic on its modulus and argument — proved by induction and put straight to work.

13 min read 3 worked examples

Statement of the theorem

Theorem 5.1 — De Moivre's theorem

For any integer $n$, $$(\cos\theta+i\sin\theta)^n = \cos(n\theta)+i\sin(n\theta).$$

In modulus–argument form this says that raising $z=r(\cos\theta+i\sin\theta)$ to the power $n$ gives $z^n = r^n(\cos n\theta+i\sin n\theta)$: the modulus is raised to the power $n$, and the argument is simply multiplied by $n$. This is an immediate consequence of repeatedly applying the multiplication rule from the Modulus–Argument Form lesson ($n-1$ times), but it is instructive — and required by the specification — to prove it properly by induction.

Proof by induction (positive integers)

ProofBy induction on n, for $n\in\mathbb{Z}^{+}$

Let $P(n)$ be the statement $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$.

Define the statement to be proved for each n.

Base case: for $n=1$, LHS $=\cos\theta+i\sin\theta=$ RHS. So $P(1)$ is true.

Both sides are identical when n = 1.

Inductive step: assume $P(k)$ is true for some $k\in\mathbb{Z}^{+}$, i.e. $(\cos\theta+i\sin\theta)^k=\cos k\theta+i\sin k\theta$.

Standard induction hypothesis.

Consider $(\cos\theta+i\sin\theta)^{k+1} = (\cos\theta+i\sin\theta)^k(\cos\theta+i\sin\theta)$.

Split off one extra factor.

$= (\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta)$

Substitute the inductive hypothesis P(k).

$= (\cos k\theta\cos\theta-\sin k\theta\sin\theta) + i(\sin k\theta\cos\theta+\cos k\theta\sin\theta)$

Expand and group real/imaginary parts, using i² = −1.

$= \cos(k\theta+\theta)+i\sin(k\theta+\theta) = \cos\big((k{+}1)\theta\big)+i\sin\big((k{+}1)\theta\big)$

These are exactly the compound-angle formulae for cos and sin of a sum.

So $P(k+1)$ is true whenever $P(k)$ is true.

Completes the inductive step.

Since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all $k\in\mathbb{Z}^{+}$, by the principle of mathematical induction $P(n)$ is true for all $n\in\mathbb{Z}^{+}$.

Standard induction conclusion, referencing base case and inductive step.

▪ QED

Extending to n = 0 and negative integers

For $n=0$: $(\cos\theta+i\sin\theta)^0=1=\cos0+i\sin0$, so the theorem holds trivially. For negative integers, write $n=-m$ with $m\in\mathbb{Z}^{+}$: since $(\cos\theta+i\sin\theta)^{-1}=\cos\theta-i\sin\theta=\cos(-\theta)+i\sin(-\theta)$ (the reciprocal of a modulus-1 number is its conjugate), applying the positive-integer case (already proved) to angle $-\theta$ gives $(\cos\theta+i\sin\theta)^{-m}=\big(\cos(-\theta)+i\sin(-\theta)\big)^m=\cos(-m\theta)+i\sin(-m\theta)=\cos(n\theta)+i\sin(n\theta)$. So the theorem holds for every integer $n$.

Using De Moivre's theorem to simplify expressions

Example 1 · Simplifying a quotient of powers

Simplify $\dfrac{(\cos3\theta+i\sin3\theta)^2}{(\cos\theta+i\sin\theta)^5}$.

By De Moivre, the numerator is $(\cos3\theta+i\sin3\theta)^2=\cos6\theta+i\sin6\theta$, and the denominator is $(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta$.

Using the division rule (moduli both $1$, arguments subtract): $\dfrac{\cos6\theta+i\sin6\theta}{\cos5\theta+i\sin5\theta} = \cos(6\theta-5\theta)+i\sin(6\theta-5\theta) = \cos\theta+i\sin\theta$.

Using De Moivre's theorem to compute powers

Converting a complex number to modulus–argument form first, then applying De Moivre, is dramatically faster than repeated binomial expansion for large powers.

Argand diagram showing a point z with modulus r and argument theta, and the point z to the power n with modulus r to the n and argument n theta, illustrating de Moivre's theorem as scaling and rotation

Fig. 1 — raising z to the power n scales the modulus to rⁿ and multiplies the argument by n: repeated rotation and scaling about the origin.

Example 2 · Computing a high power

Find $(\sqrt3+i)^6$, giving your answer in the form $a+bi$.

$|\sqrt3+i| = \sqrt{3+1}=2$; first quadrant, $\arg = \arctan\left(\dfrac{1}{\sqrt3}\right)=\dfrac{\pi}{6}$.

So $\sqrt3+i = 2\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)$, and by De Moivre:

$$(\sqrt3+i)^6 = 2^6\left(\cos\frac{6\pi}{6}+i\sin\frac{6\pi}{6}\right) = 64(\cos\pi+i\sin\pi) = 64(-1+0i) = -64$$

Example 3 · A negative integer power

Find $(1-i)^{-4}$, giving your answer in the form $a+bi$.

$|1-i|=\sqrt2$; fourth quadrant, $\arg = -\dfrac{\pi}{4}$. So $1-i = \sqrt2\left(\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right)\right)$.

$$(1-i)^{-4} = (\sqrt2)^{-4}\left(\cos\left(-4\times-\frac{\pi}{4}\right)+i\sin\left(-4\times-\frac{\pi}{4}\right)\right) = \frac14(\cos\pi+i\sin\pi) = -\frac14$$

Common mistake

De Moivre's theorem applies to a number already in the form $\cos\theta+i\sin\theta$ or $r(\cos\theta+i\sin\theta)$ — it is easy to forget the modulus and write $r^n$ as just $r$, especially when $r=1$ is the more familiar textbook case. Always raise the modulus to the power $n$ as a separate, explicit step.

Looking ahead

De Moivre's theorem is also the engine behind finding roots of complex numbers (fractional powers) and deriving multiple-angle trig identities by comparing real and imaginary parts of $(\cos\theta+i\sin\theta)^n$ — both are developed fully in the next two lessons.

Exam-style questions

Question 1 6 marks

Prove by induction that, for all positive integers $n$, $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$.

This is the full induction proof given in the lesson above: state $P(n)$, verify the base case $P(1)$, assume $P(k)$, use $(\cos\theta+i\sin\theta)^{k+1}=(\cos\theta+i\sin\theta)^k(\cos\theta+i\sin\theta)$ and the compound-angle formulae to reach $P(k+1)$, then conclude by the principle of mathematical induction. (B1 define P(n), B1 base case, M1 split off factor, M1 substitute hypothesis, M1 expand using compound-angle formulae, A1 correct conclusion with reference to induction)

Question 2 5 marks

Find the exact value of $(1+i\sqrt3)^5$, giving your answer in the form $a+bi$.
(Hint: convert to modulus–argument form first, and remember to bring the argument back into (−π, π] before evaluating cos and sin.)

$|1+i\sqrt3|=\sqrt{1+3}=2$; first quadrant, $\arg=\arctan(\sqrt3)=\dfrac{\pi}{3}$ (M1, A1)

$(1+i\sqrt3)^5 = 2^5\left(\cos\dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}\right) = 32\left(\cos\dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}\right)$ (M1)

$\dfrac{5\pi}{3}-2\pi = -\dfrac{\pi}{3}$; $\cos\left(-\dfrac{\pi}{3}\right)=\dfrac12$, $\sin\left(-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt3}{2}$ (M1)

$(1+i\sqrt3)^5 = 32\left(\dfrac12-\dfrac{\sqrt3}{2}i\right) = 16-16\sqrt3\,i$ (A1)

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