TheoremProved

De Moivre's formula

Statement

For any real $\theta$ and integer $n$ , $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ .

Why is it true?

$\cos\theta+i\sin\theta$ is just the point on the unit circle at angle $\theta$ . Multiplying two such points together adds their angles — that is what complex multiplication does geometrically — so raising one point to the $n$ -th power spins it around $n$ times as fast, landing exactly at angle $n\theta$ .

Proof sketch

Induction on $n\ge 0$ using the angle-addition identities $\cos(a+b)=\cos a\cos b - \sin a\sin b$ and $\sin(a+b)=\sin a\cos b+\cos a\sin b$ to show $(\cos\theta+i\sin\theta)^{k+1}=(\cos\theta+i\sin\theta)^k(\cos\theta+i\sin\theta)$ simplifies to $\cos((k+1)\theta)+i\sin((k+1)\theta)$ ; negative $n$ follows from $(\cos\theta+i\sin\theta)^{-1}=\cos\theta - i\sin\theta = \cos(-\theta)+i\sin(-\theta)$ .

Topics that use this theorem

Complex numbers

Related theorems

Euler's formula

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

John Stillwell (2010). Mathematics and Its History