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
The formula can be derived from Euler's formula:
 * $e^{ix} = \cos x + i \sin x$

and laws of exponentials:
 * $\left({e^{ix}}\right)^n = e^{i n x}$

Then, by Euler's formula:
 * $e^{i \left({n x}\right)} = \cos \left({n x}\right) + i \sin \left({n x}\right)$

Proof by Induction
Proof by induction:

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

$P \left({1}\right)$ 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.

Basis for the Induction
$P \left({2}\right)$ is the case:
 * $\left({\cos x + i \sin x}\right)^2 = \cos \left({2 x}\right) + i \sin \left({2 x}\right)$

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, 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:

So true for $n=k+1$, hence:
 * $\left({\cos x + i \sin x}\right)^n = \cos \left({n x}\right)+ i \sin \left({n x}\right)$

is true for all $n = 1, 2, 3, \ldots$ by mathematical induction.