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Question Number 122854 by Study last updated on 20/Nov/20

Answered by mathmax by abdo last updated on 20/Nov/20

$${(-2)}^{\sqrt{2}}={\mathrm{e}}^{\sqrt{2}\ln (-2)}={\mathrm{e}}^{\sqrt{2}(\ln (-1)+\ln \left(2\right))}={\mathrm{e}}^{\sqrt{2}(\mathrm{i}\pi +\ln 2)}$$$$={\mathrm{e}}^{\sqrt{2}\ln \left(2\right)+\mathrm{i}\sqrt{2}\pi }={\mathrm{e}}^{\sqrt{2}\ln \left(2\right)}\{\cos \left(\pi \sqrt{2}\right)+\mathrm{isin}\left(\pi \sqrt{2}\right)\}$$

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