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Question-199165




Question Number 199165 by hardmath last updated on 28/Oct/23
Answered by Frix last updated on 28/Oct/23
−1=e^(iπ)   (−1)^π =e^(iπ^2 ) =  =cos π^2  +i sin π^2   ≈−.902685362+−.430301217i
$$−\mathrm{1}=\mathrm{e}^{\mathrm{i}\pi} \\ $$$$\left(−\mathrm{1}\right)^{\pi} =\mathrm{e}^{\mathrm{i}\pi^{\mathrm{2}} } = \\ $$$$=\mathrm{cos}\:\pi^{\mathrm{2}} \:+\mathrm{i}\:\mathrm{sin}\:\pi^{\mathrm{2}} \\ $$$$\approx−.\mathrm{902685362}+−.\mathrm{430301217i} \\ $$

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