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)}. \end{align*}

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). \end{align*}

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.

Problems

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.