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Question Number 199165 by hardmath last updated on 28/Oct/23

Answered by Frix last updated on 28/Oct/23

$$-1={\mathrm{e}}^{\mathrm{i}\pi }$$$${(-1)}^{\pi }={\mathrm{e}}^{\mathrm{i}{\pi }^2}=$$$$=\cos {\pi }^2+\mathrm{i}\sin {\pi }^2$$$$\approx -.902685362+-.430301217\mathrm{i}$$

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