De Moivre’s formula
Task number: 3321
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 \( n + 1 \):
\( \begin{equation} (\cos \alpha + i \sin\alpha)^{n+1}=\\ = (\cos \alpha + i \sin\alpha)^n(\cos \alpha + i \sin\alpha)\\ = (\cos(n\alpha)+ i \sin(n\alpha))(\cos \alpha + i \sin\alpha)\\ = \cos(n\alpha)\cos \alpha + \cos(n\alpha) i \sin\alpha +i \sin(n\alpha)\cos \alpha +i \sin(n\alpha) i \sin\alpha \\ = \cos(n\alpha)\cos \alpha - \sin(n\alpha) \sin\alpha + i(\cos(n\alpha) \sin\alpha + \sin(n\alpha)\cos \alpha) \\ = \cos((n+1)\alpha)+ i \sin((n+1)\alpha)\\ \end{equation} \)
In the first step, we used the inductive hypothesis, in the last summation formulas.
