De Moivre's Formula

Theorem

 * $\left({\cos x + i \sin x}\right)^n = \cos \left({n x}\right)+ i \sin \left({n x}\right)$, for $\begin{cases} n = 1, 2, 3, \ldots \\ x \in \R \end{cases}$

Also known as De Moivre's Theorem.

Proof from Euler's Formula
The formula can be derived from Euler's Formula, as follows:

{{eqn|r = \cos \left({n x}\right) + i \sin \left({n x}\right) |c = Euler's Formula {{end-eqn}} {{qed}}

Proof by Induction
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\left({\cos x + i \sin x}\right)^n = \cos \left({n x}\right)+ i \sin \left({n x}\right)$

Basis for the Induction
$P \left({1}\right)$ is the case:


 * $\left({\cos x + i \sin x}\right)^1 = \cos \left({1 x}\right) + i \sin \left({1 x}\right)$

which is trivially true.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\left({\cos x + i \sin x}\right)^k = \cos \left({k x}\right) + i \sin \left({k x}\right)$

Then we need to show:
 * $\left({\cos x + i \sin x}\right)^{k+1} = \cos \left({\left({k+1}\right) x}\right) + i \sin \left({\left({k+1}\right) x}\right)$

Induction Step
This is our induction step:

Hence, by induction, for all $n \in \N^*$:


 * $\left({\cos x + i \sin x}\right)^n = \cos \left({n x}\right)+ i \sin \left({n x}\right)$