De Moivre's Formula/Positive Integer Index/Proof 1

Proof
Proof by induction:

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
 * $\paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \, \map \sin {n x} }$

$\map P 1$ is the case:


 * $\paren {r \paren {\cos x + i \sin x} }^1 = r^1 \paren {\map \cos {1 x} + i \, \map \sin {1 x} }$

which is trivially true.

Basis for the Induction
$\map P 2$ is the case:


 * $\paren {r \paren {\cos x + i \sin x} }^2 = r^2 \paren {\map \cos {n x} + i \, \map \sin {2 x} }$

From Product of Complex Numbers in Polar Form, we have:
 * $r_1 \paren {\cos x_1 + i \sin x_1 } r_2 \paren {\cos x_2 + i \sin x_2} = r_1 r_2 \paren {\map \cos {x_1 + x_2} + i \, \map \sin {x_1 + x_2} }$

Setting $r_1 = r_2 = r$ and $x_1 = x_2 = x$ gives the result.

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\paren {r \paren {\cos x + i \sin x} }^k = r^k \paren {\map \cos {k x} + i \, \map \sin {k x} }$

Then we need to show:
 * $\paren {r \paren {\cos x + i \sin x} }^{k + 1} = r^{k + 1} \paren {\map \cos {\paren {k + 1} x} + i \, \map \sin {\paren {k + 1} x} }$

Induction Step
This is our induction step:

Hence, by induction, for all $n \in \Z_{> 0}$:


 * $\paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \, \map \sin {n x} }$