(related to Theorem: De Moivre's Identity, Complex Powers)

- By hypothesis, $\phi$ is a real number.
- We prove the formula by induction.
- Base case $n=1$
- By the multiplication of complex numbers using polar coordinates (with $r=1$) $$1=[\cos(\phi)+i\sin(\phi)].$$

- Induction step
- Assume, the formula $[\cos(\phi)+i\sin(\phi)]^n=\cos(n\phi)+i\sin(n\phi)$ holds for all natural numbers $n\in\mathbb N$ not exceeding some number $N.$
- Then, by additivity theorems for cosine and sine: $$\begin{array}{rcl}[\cos(\phi)+i\sin(\phi)]^{n+1}&=&[\cos(\phi)+i\sin(\phi)]^n\cdot [\cos(\phi)+i\sin(\phi)]\\ &=&[\cos(n\phi)+i\sin(n\phi)]\cdot [\cos(\phi)+i\sin(\phi)]\\ &=&\cos(n\phi)\cos(\phi)+i\sin(\phi)\cos(n\phi)-i\sin(n\phi)\cos(\phi)-\sin(n\phi)\sin(\phi)\\ &=&\cos((n+1)\phi)+i\sin((n+1)\phi) \end{array}$$

- In particular, if $z=r\exp(i\phi)$ is a complex number written in its polar coordinates, then $z^n=r^n\exp(in\phi),$ by Euler's formula.∎

**Modler, F.; Kreh, M.**: "Tutorium Analysis 1 und Lineare Algebra 1", Springer Spektrum, 2018, 4th Edition