De Moivre’s theorem

De Moivre’s theorem. For any real number \(\theta\) and integer \(n\), it holds that $$(\cos\theta + i \sin\theta)^n = \cos{n\theta} + i \sin{n\theta}.$$

We prove this theorem for \(n \ge 0\) by induction.

Proof for positive integers. The statement clearly holds \(n=0\). Assume that $$(\cos(\theta) + i \sin(\theta))^k = \cos{(k\theta)} + i \sin{(k\theta)}$$ for a particular integer \(k \ge 1\) (the induction hypothesis). We must show that $$(\cos(\theta) + i \sin(\theta))^{k+1} = \cos{((k+1)\theta)} + i \sin{((k+1)\theta)}.$$ We have $$\begin{align*}
(\cos(\theta) + i \sin(\theta))^{k+1} &= (\cos(\theta) + i \sin(\theta))^{k} (\cos(\theta) + i \sin(\theta)) \\
&= (\cos(k\theta) + i \sin(k\theta)) (\cos(\theta) + i \sin(\theta)) \\
&=\cos(k\theta)\cos\theta-\sin(k\theta)\sin(\theta) \\
& \phantom{\cos(k\theta)\cos\theta}+ i(\cos{(k\theta)}\sin{(\theta)} + \sin{(k\theta)}\cos{(\theta)}) \\
&= \cos{((k+1)\theta)} + i \sin{((k+1)\theta)}.

Once we know it is true for positive integers, it is easier to prove it holds for negative integers.

Proof for negative integers. Let \(n\) be a positive integer. Then $$\begin{align*}
(\cos(\theta)+i\sin(\theta))^{-n} &= \frac{1}{(\cos(\theta)+i\sin(\theta))^{n}} \\
&= \frac{1}{(\cos(n\theta)+i\sin(n\theta))},
\end{align*}$$ where we apply De Moivre’s theorem for positive integers. We then realise the denominator to see that $$\begin{align*}
(\cos(\theta)+i\sin(\theta))^{-n} &= \frac{ \cos(n\theta)-i\sin(n\theta)}{(\cos(n\theta)+i\sin(n\theta))(\cos(n\theta)-i\sin(n\theta))} \\
&= \frac{ \cos(n\theta)-i\sin(n\theta)}{\cos^2{(n\theta)}+\sin^2{(n\theta)}} \\
&= \cos(n\theta)-i\sin(n\theta) \\
&= \cos(-n\theta)+i\sin(-n\theta).

Let us apply De Moivre’s theorem.

Example. To evaluate \(( \sqrt{2} + \sqrt{2} i)^6 \), we first convert \(\sqrt{2} + \sqrt{2} i \) to polar form. We find that it equals to $$2 \left(\cos\left(\frac{\pi}{4}\right)+ i \sin\left(\frac{\pi}{4}\right) \right) = 2e^{\frac{\pi i}{4}}.$$ Thus we have $$\begin{align*}
(\sqrt{2} + \sqrt{2} i)^6 &= \left(2 \left(\cos\left(\frac{\pi}{4}\right)+ i \sin\left(\frac{\pi}{4}\right) \right) \right)^6 \\
&= 2^6 \left(\cos\left(\frac{6\pi}{4}\right)+ i \sin\left(\frac{6\pi}{4}\right) \right) \\
&= -64 i. \end{align*} $$

GL tip: Does De Moivre’s theorem work for fractions? The answer is yes and no. The issue with taking a fractional power is that the result is multi-valued. For example, if try to find the square root of \(1\) by taking it power of \(\frac{1}{2}\) and applying De Moivre’s theorem, then we get \(1^{1/2} = 1\) when we use \(1 = \cos(0)+i \sin(0)\) and \(1^{1/2} = -1\) when we use \(1 = \cos(2\pi)+i\sin(2\pi)\). However, we do know applying De Moivre’s theorem gives you a root. We will use this when we learn how to find \(n\)th roots of complex numbers.


  1. Simplify the following using De Moivre’s theorem.
    1. \( \left( \cos{\frac{\pi}{5}} + i \sin{\frac{\pi}{5}}\right)^{10} \)
    2. \(( \sqrt{2} + \sqrt{2} i)^6 \)
    3. \((-1-i)^{20}\)
    4. \((\sqrt{3}-i)^{-3}\)
    5. \((2\sqrt{3}-2i)^{-4}\)
  2. Prove De Moivre’s theorem for positive integer \(n\) using induction.