Question-71806

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}} = {({2 e}^{i π})}^{\sqrt{2}} = 2^{\sqrt{2}} e^{i π \sqrt{2}} =

= 2^{\sqrt{2}} \cos π \sqrt{2} + 2^{\sqrt{2}} i \sin π \sqrt{2} \approx - . 709609 - 2 . 56894 i

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