# De Moivre's Formula/Positive Integer Index

## Theorem

Let $z \in \C$ be a complex number expressed in polar form:

$z = r \paren {\cos x + i \sin x}$

Then:

$\forall n \in \Z_{>0}: \paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \map \sin {n x} }$

### Corollary

$\forall n \in \Z_{>0}: \paren {\cos x + i \sin x}^n = \map \cos {n x} + i \map \sin {n x}$

## Proof 1

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:

 $\ds \paren {r \paren {\cos x + i \sin x} }^{k + 1}$ $=$ $\ds \paren {r \paren {\cos x + i \sin x} }^k \paren {r \paren {\cos x + i \sin x} }$ $\ds$ $=$ $\ds r^k \paren {\map \cos {k x} + i \, \map \sin {k x} } \paren {r \paren {\cos x + i \sin x} }$ Induction Hypothesis $\ds$ $=$ $\ds r^{k + 1} \paren {\map \cos {\paren {k + 1} x} + i \, \map \sin {\paren {k + 1} x} }$ Product of Complex Numbers in Polar Form

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

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

$\blacksquare$

## Proof 2

$z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \, \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$

Setting $z_1 = z_2 = \cdots = z_n = r \paren {\cos x + i \sin x}$ gives the result.

## Also known as

De Moivre's Theorem.

## Source of Name

This entry was named for Abraham de Moivre.