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

$\cos (\theta )\cos (n\theta )-\sin (\theta )\sin (n\theta )+i\cos (\theta )\sin (n\theta )+i\sin (\theta )\cos (n\theta )$$=\left(\cos (\theta )+i\sin (\theta )\right)\left(\cos (n\theta )+i\sin (n\theta )\right)$

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

By the angle sum identities,

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

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

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

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

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

$={\displaystyle \frac{1}{\cos (n\theta )+i\sin (n\theta )}},$

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

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