Question Number 8838 by tawakalitu last updated on 31/Oct/16
$$\mathrm{Show}\:\mathrm{that}\::\:\:\mathrm{e}^{\mathrm{i}\pi\:+\:\mathrm{1}} \:=\:\mathrm{0} \\ $$
Answered by FilupSmith last updated on 31/Oct/16
$$\mathrm{do}\:\mathrm{you}\:\mathrm{mean}\:{e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$$${e}^{{ix}} =\mathrm{cos}\left({x}\right)+{i}\mathrm{sin}\left({x}\right) \\ $$$${x}=\pi \\ $$$${e}^{{i}\pi} =\left(−\mathrm{1}\right)+{i}\left(\mathrm{0}\right) \\ $$$${e}^{{i}\pi} =−\mathrm{1} \\ $$$$\therefore{e}^{{i}\pi} +\mathrm{1}=\left(−\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$
Commented by tawakalitu last updated on 31/Oct/16
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$