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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)^(√2) =e^((√2)ln(−2))  =e^((√2)(ln(−1)+ln(2)))  =e^((√2)(iπ+ln2))   = e^((√2)ln(2)+i(√2)π)  =e^((√2)ln(2)) {cos(π(√2)) +isin(π(√2))}

$$\left(−\mathrm{2}\right)^{\sqrt{\mathrm{2}}} =\mathrm{e}^{\sqrt{\mathrm{2}}\mathrm{ln}\left(−\mathrm{2}\right)} \:=\mathrm{e}^{\sqrt{\mathrm{2}}\left(\mathrm{ln}\left(−\mathrm{1}\right)+\mathrm{ln}\left(\mathrm{2}\right)\right)} \:=\mathrm{e}^{\sqrt{\mathrm{2}}\left(\mathrm{i}\pi+\mathrm{ln2}\right)} \\ $$$$=\:\mathrm{e}^{\sqrt{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{i}\sqrt{\mathrm{2}}\pi} \:=\mathrm{e}^{\sqrt{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)} \left\{\mathrm{cos}\left(\pi\sqrt{\mathrm{2}}\right)\:+\mathrm{isin}\left(\pi\sqrt{\mathrm{2}}\right)\right\} \\ $$

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