Further Maths/Lessons/Complex Numbers (Year 2)/De Moivre's Theorem
Core Pure · Complex Numbers · Year 2

De Moivre's Theorem

One theorem turns raising a complex number to a power into simple multiplication of angles.

9 min read 1 worked example

De Moivre's theorem

Warm-up — squaring in modulus-argument form

Using $z=re^{i\theta}$, note that $z^2=(re^{i\theta})^2=r^2e^{i2\theta}=r^2(\cos2\theta+i\sin2\theta)$. Squaring a complex number doubles its argument and squares its modulus. The same pattern continues for any power.

Theorem 1.1 — De Moivre's theorem

For any integer $n$,

$$\big(r(\cos\theta+i\sin\theta)\big)^n=r^n(\cos n\theta+i\sin n\theta)$$

Equivalently, in exponential form: $\left(re^{i\theta}\right)^n=r^ne^{in\theta}$ — raising a complex number to an integer power multiplies its argument by that power and raises its modulus to that power.

ProofDe Moivre's theorem for positive integers, by induction
1
Base case: when $n=1$, $\big(r(\cos\theta+i\sin\theta)\big)^1=r(\cos\theta+i\sin\theta)=r^1(\cos1\theta+i\sin1\theta)$.
The statement holds trivially for $n=1$.
2
Assumption: suppose the result holds for $n=k$, i.e. $\big(r(\cos\theta+i\sin\theta)\big)^k=r^k(\cos k\theta+i\sin k\theta)$.
This is the inductive hypothesis.
3
$\big(r(\cos\theta+i\sin\theta)\big)^{k+1}=r^k(\cos k\theta+i\sin k\theta)\times r(\cos\theta+i\sin\theta)$
Write $z^{k+1}=z^k\cdot z$ and substitute the assumption.
4
$=r^{k+1}\big[(\cos k\theta\cos\theta-\sin k\theta\sin\theta)+i(\sin k\theta\cos\theta+\cos k\theta\sin\theta)\big]$
Multiply out and collect real and imaginary parts, using $i^2=-1$.
5
$=r^{k+1}\big[\cos((k+1)\theta)+i\sin((k+1)\theta)\big]$
Apply the cosine and sine addition formulae, so the statement holds for $n=k+1$.
6
Since the result holds for $n=1$, and $n=k$ true implies $n=k+1$ true, it holds for all $n\in\mathbb{Z}^+$ by mathematical induction.
The case $n=0$ is trivial ($z^0=1$), and negative integers follow by applying the positive-integer result to $\frac1z=z^{-1}$ and using $(z^{-1})^m=z^{-m}$.
∎
The quick proof, using exponential form

Once you're comfortable with $z=re^{i\theta}$, De Moivre's theorem is almost immediate: $z^n=(re^{i\theta})^n=r^ne^{in\theta}$, which is exactly the ordinary law of indices $(a^b)^c=a^{bc}$ applied to $e^{i\theta}$. Converting back with Euler's relation gives $r^n(\cos n\theta+i\sin n\theta)$. This shortcut is valid for any integer $n$ without needing a separate induction argument — the induction proof above is included because it's the classical, syllabus-expected derivation.

Example 1 · Applying De Moivre's theorem directly
Simplify $\left(\cos\dfrac{5\pi}{18}+i\sin\dfrac{5\pi}{18}\right)^4\left(\cos\dfrac{\pi}{9}-i\sin\dfrac{\pi}{9}\right)^{3}$, giving your answer in the form $\cos\theta+i\sin\theta$.

Rewrite the second bracket using $\cos\theta-i\sin\theta=\cos(-\theta)+i\sin(-\theta)$, so it becomes $\left(\cos\left(-\dfrac\pi9\right)+i\sin\left(-\dfrac\pi9\right)\right)^3$.

By De Moivre, the first bracket raised to the 4th power gives argument $4\times\dfrac{5\pi}{18}=\dfrac{10\pi}9$, and the second raised to the 3rd power gives argument $3\times\left(-\dfrac\pi9\right)=-\dfrac\pi3$.

Multiplying (moduli both $1$, arguments add): total argument $\dfrac{10\pi}9-\dfrac\pi3=\dfrac{10\pi}9-\dfrac{3\pi}9=\dfrac{7\pi}9$. Result: $\cos\dfrac{7\pi}9+i\sin\dfrac{7\pi}9$.

Exam-style questions

Question 1 5 marks

By expressing $z=1+i$ in modulus-argument form, find the smallest positive integer $n$ for which $z^n$ is real and positive.

$|z|=\sqrt2$, $\arg z=\dfrac\pi4$, so $z=\sqrt2\,e^{\frac\pi4 i}$. (M1, A1)

By De Moivre, $z^n=(\sqrt2)^ne^{\frac{n\pi}4 i}=(\sqrt2)^n\left(\cos\dfrac{n\pi}4+i\sin\dfrac{n\pi}4\right)$. (M1)

$z^n$ is real when $\sin\dfrac{n\pi}4=0$, i.e. $n$ is a multiple of $4$; it's positive when additionally $\cos\dfrac{n\pi}4>0$, which requires $n$ to be a multiple of $8$. The smallest such positive integer is $n=8$. (M1, A1)

Question 2 5 marks

Simplify $\left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)^5\left(\cos\dfrac{2\pi}3-i\sin\dfrac{2\pi}3\right)^2$, giving your answer in the form $\cos\theta+i\sin\theta$ where $-\pi<\theta\leq\pi$.

By De Moivre, the first bracket raised to the 5th power gives argument $5\times\dfrac\pi6=\dfrac{5\pi}6$. (M1, A1)

Rewriting $\cos\dfrac{2\pi}3-i\sin\dfrac{2\pi}3=\cos\left(-\dfrac{2\pi}3\right)+i\sin\left(-\dfrac{2\pi}3\right)$, raised to the 2nd power gives argument $2\times\left(-\dfrac{2\pi}3\right)=-\dfrac{4\pi}3$. (M1)

Multiplying (moduli both $1$, arguments add): total argument $\dfrac{5\pi}6-\dfrac{4\pi}3=\dfrac{5\pi}6-\dfrac{8\pi}6=-\dfrac\pi2$, already in range. Result: $\cos\left(-\dfrac\pi2\right)+i\sin\left(-\dfrac\pi2\right)$. (A1)

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