## de Moivres theorem

Prove de Moivre's theorem by induction: $$(\cos \alpha + i \sin\alpha)^n = \cos(n\alpha)+ i \sin(n\alpha)$$

• #### Hint

$$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$$,
$$\sin(\alpha+\beta)=\cos(\alpha)\sin(\beta)+\sin(\alpha)\cos(\beta)$$.
For $$n=1$$ both sides are identical.
Assuming that the statement is true for $$n$$, we show that it holds for $$n+1$$.
$$(\cos \alpha + i \sin\alpha)^{n+1}= (\cos \alpha + i \sin\alpha)^n(\cos \alpha + i \sin\alpha)= (\cos(n\alpha)+ i \sin(n\alpha))(\cos \alpha + i \sin\alpha)=\\ \cos(n\alpha)\cos \alpha - \sin(n\alpha)\sin\alpha + i(\cos(n\alpha)\sin \alpha - \sin(n\alpha)\cos\alpha)= \cos((n+1)\alpha)+ i \sin((n+1)\alpha)$$