de Moivre identity, proof of

To prove the de Moivre identity, we will first prove by induction on $n$ that the identity holds for all natural numbers.

For the case $n=0$ , observe that

$\cos(0\theta)+i\sin(0\theta)=1+i0=\left(\cos(\theta)+i\sin(\theta)\right)^{0}.$

Assume that the identity holds for a certain value of $n$ :

$\cos(n\theta)+i\sin(n\theta)=\left(\cos(\theta)+i\sin(\theta)\right)^{n}.$

Multiply both sides of this identity by $\cos(\theta)+i\sin(\theta)$ and expand the left side to obtain

	$\displaystyle\cos(\theta)\cos(n\theta)-\sin(\theta)\sin(n\theta)+i\cos(\theta)% \sin(n\theta)+i\sin(\theta)\cos(n\theta)$	$\displaystyle=\left(\cos(\theta)+i\sin(\theta)\right)\left(\cos(n\theta)+i\sin% (n\theta)\right)$
		$\displaystyle=\left(\cos(\theta)+i\sin(\theta)\right)^{n+1}.$

	$\displaystyle\cos(\theta)\cos(n\theta)-\sin(\theta)\sin(n\theta)$	$\displaystyle=\cos(n\theta+\theta)$
	$\displaystyle\cos(\theta)\sin(n\theta)+\sin(\theta)\cos(n\theta)$	$\displaystyle=\sin(n\theta+\theta)$

Therefore,

$\cos((n+1)\theta)+i\sin((n+1)\theta)=\left(\cos(\theta)+i\sin(\theta)\right)^{% n+1}.$

Hence by induction de Moivre’s identity holds for all natural $n$ .

Now let $-n$ be any negative integer. Then using the fact that $\cos$ is an even and $\sin$ an odd function, we obtain that

	$\displaystyle\cos(-n\theta)+i\sin(-n\theta)$	$\displaystyle=\cos(n\theta)-i\sin(n\theta)$
		$\displaystyle=\frac{\cos(n\theta)-i\sin(n\theta)}{\cos^{2}(n\theta)+\sin^{2}(n% \theta)}$
		$\displaystyle=\frac{1}{\cos(n\theta)+i\sin(n\theta)}\cdot\frac{\cos(n\theta)-i% \sin(n\theta)}{\cos(n\theta)-i\sin(n\theta)}$
		$\displaystyle=\frac{1}{\cos(n\theta)+i\sin(n\theta)},$

the denominator of which is $\left(\cos(n\theta)+i\sin(n\theta)\right)^{n}$ . Hence

$\cos(-n\theta)+i\sin(-n\theta)=\left(\cos(\theta)+i\sin(\theta)\right)^{-n}.$