Question Number 29049 by NECx last updated on 03/Feb/18
$${Prove}\:{that} \\ $$$$\:\:\:\:\:{e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$
Answered by Penguin last updated on 03/Feb/18
$${e}^{{i}\theta} =\mathrm{cos}\left(\theta\right)+{i}\mathrm{sin}\left(\theta\right) \\ $$$$\theta=\pi \\ $$$$\therefore{e}^{{i}\pi} =−\mathrm{1}+{i}\left(\mathrm{0}\right) \\ $$$$=−\mathrm{1} \\ $$