Question Number 227063 by fantastic2 last updated on 29/Dec/25
$${is}\:\left(−{e}\right)^{\pi} \:\:{real}\:{number}? \\ $$
Answered by Frix last updated on 29/Dec/25
$$\left.\begin{matrix}{\mathrm{abs}\:\left(−\mathrm{e}\right)\:=\mathrm{e}}\\{\mathrm{arg}\:\left(−\mathrm{e}\right)\:=\pi}\end{matrix}\right\}\:\Rightarrow\:−\mathrm{e}=\mathrm{e}^{\mathrm{i}\pi} \mathrm{e} \\ $$$$\left(−\mathrm{e}\right)^{\pi} =\mathrm{e}^{\mathrm{i}\pi^{\mathrm{2}} } \mathrm{e}^{\pi} =\mathrm{e}^{\pi} \left(\mathrm{cos}\:\pi^{\mathrm{2}} \:+\mathrm{i}\:\mathrm{sin}\:\pi^{\mathrm{2}} \right)\:\notin\mathbb{R} \\ $$