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:

	\(\displaystyle \)	\(\displaystyle \left({\cos x + i \sin x}\right)^n\)	\(=\)	\(\displaystyle \left({e^{ix} }\right)^n\)	\(\displaystyle \)	Euler's Formula
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^{i n x}\)	\(\displaystyle \)	Power of Power
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos \left({n x}\right) + i \sin \left({n x}\right)\)	\(\displaystyle \)	Euler's Formula

$\blacksquare$

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:

	\(\displaystyle \)	\(\displaystyle \left({\cos x + i \sin x}\right)^{k+1}\)	\(=\)	\(\displaystyle \left({\cos x + i \sin x}\right) \left({\cos x + i \sin x}\right)^k\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\cos x + i \sin x}\right) \left({\cos \left({k x}\right) + i \sin \left({k x}\right)}\right)\)	\(\displaystyle \)	by the induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos \left({k x}\right) \cos x - \sin \left({k x}\right) \sin x + i \left({\cos \left({k x}\right) \sin x + \sin \left({k x}\right) \cos x}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos \left({k x + x}\right) + i \sin \left({k x + x}\right)\)	\(\displaystyle \)	Sine and Cosine of Sum
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos \left({\left({k + 1}\right) x}\right) + i \sin \left({\left({k + 1}\right) x}\right)\)	\(\displaystyle \)

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

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

$\blacksquare$

Source of Name

This entry was named for Abraham de Moivre.

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $6.9$