Question Number 71806 by ahmadshahhimat775@gmail.com last updated on 20/Oct/19
Answered by MJS last updated on 20/Oct/19
$$\left(−\mathrm{2}\right)^{\sqrt{\mathrm{2}}} =\left(\mathrm{2e}^{\mathrm{i}\pi} \right)^{\sqrt{\mathrm{2}}} =\mathrm{2}^{\sqrt{\mathrm{2}}} \mathrm{e}^{\mathrm{i}\pi\sqrt{\mathrm{2}}} = \\ $$$$=\mathrm{2}^{\sqrt{\mathrm{2}}} \mathrm{cos}\:\pi\sqrt{\mathrm{2}}\:+\mathrm{2}^{\sqrt{\mathrm{2}}} \mathrm{i}\:\mathrm{sin}\:\pi\sqrt{\mathrm{2}}\approx−.\mathrm{709609}−\mathrm{2}.\mathrm{56894i} \\ $$