de Moivres theorem

Task number: 2797

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

  • Hint

    Use the additive identities:

  • Resolution

    For \(n=1\) both sides are identical.

    Assuming that the statement is true for \(n\), we show that it holds for \(n+1\).

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

Difficulty level: Easy task (using definitions and simple reasoning)
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