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Question Number 71806 by ahmadshahhimat775@gmail.com last updated on 20/Oct/19

Answered by MJS last updated on 20/Oct/19

$${(-2)}^{\sqrt{2}}={\left({2\mathrm{e}}^{\mathrm{i}\pi }\right)}^{\sqrt{2}}=2^{\sqrt{2}}{\mathrm{e}}^{\mathrm{i}\pi \sqrt{2}}=$$$$=2^{\sqrt{2}}\cos \pi \sqrt{2}+2^{\sqrt{2}}\mathrm{i}\sin \pi \sqrt{2}\approx -.709609-2.56894\mathrm{i}$$

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