## De Moivre’s formula

By induction prove the de Moivre’s formula: $$(\cos \alpha + i \sin\alpha)^n = \cos(n\alpha)+ i \sin(n\alpha)$$

• #### Hint

Use these identities:

$$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$$,

$$\sin(\alpha+\beta)=\cos(\alpha)\sin(\beta)+\sin(\alpha)\cos(\beta)$$.

• #### Solution

There is nothing to prove for $$n = 1$$, because $$(\cos \alpha + i \sin\alpha)^1 = \cos(1\alpha)+ i \sin(1\alpha)$$.

Induction step from $$n$$ to $$n + 1$$:

$$$$(\cos \alpha + i \sin\alpha)^{n+1}=\\ = (\cos \alpha + i \sin\alpha)^n(\cos \alpha + i \sin\alpha)\\ = (\cos(n\alpha)+ i \sin(n\alpha))(\cos \alpha + i \sin\alpha)\\ = \cos(n\alpha)\cos \alpha + \cos(n\alpha) i \sin\alpha +i \sin(n\alpha)\cos \alpha +i \sin(n\alpha) i \sin\alpha \\ = \cos(n\alpha)\cos \alpha - \sin(n\alpha) \sin\alpha + i(\cos(n\alpha) \sin\alpha + \sin(n\alpha)\cos \alpha) \\ = \cos((n+1)\alpha)+ i \sin((n+1)\alpha)\\$$$$

In the first step, we used the inductive hypothesis, in the last summation formulas.