Question Number 66421 by Rio Michael last updated on 14/Aug/19
$${show}\:{that}\:{for}\:{a}\:{given}\:{complex}\:{number}\:{z} \\ $$$$\:{z}^{{n}} \:=\:{r}^{{n}} \:\left({cosn}\theta\:+\:{isinn}\theta\right)\: \\ $$
Answered by MJS last updated on 14/Aug/19
$${z}={r}\left(\mathrm{cos}\:\theta\:+\mathrm{i}\:\mathrm{sin}\:\theta\right)={r}\mathrm{e}^{\mathrm{i}\theta} \\ $$$${z}^{{n}} =\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)^{{n}} ={r}^{{n}} \mathrm{e}^{\mathrm{i}{n}\theta} ={r}^{{n}} \left(\mathrm{cos}\:{n}\theta\:+\mathrm{i}\:\mathrm{sin}\:{n}\theta\right) \\ $$
Commented by Rio Michael last updated on 15/Aug/19
$${thanks}\:{sir} \\ $$$$ \\ $$