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Prove-i-i-e-pi-2-i-2-1-




Question Number 4556 by FilupSmith last updated on 07/Feb/16
Prove i^i =e^(−π/2) ,  i^2 =−1
$$\mathrm{Prove}\:{i}^{{i}} ={e}^{−\pi/\mathrm{2}} ,\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$
Answered by Yozzii last updated on 07/Feb/16
i=0+i×1=cos0.5π+isin0.5=e^(0.5πi)   ∴i^i =(e^(0.5πi) )^i =e^(0.5πi^2 ) =e^(−0.5π)
$${i}=\mathrm{0}+{i}×\mathrm{1}={cos}\mathrm{0}.\mathrm{5}\pi+{isin}\mathrm{0}.\mathrm{5}={e}^{\mathrm{0}.\mathrm{5}\pi{i}} \\ $$$$\therefore{i}^{{i}} =\left({e}^{\mathrm{0}.\mathrm{5}\pi{i}} \right)^{{i}} ={e}^{\mathrm{0}.\mathrm{5}\pi{i}^{\mathrm{2}} } ={e}^{−\mathrm{0}.\mathrm{5}\pi} \\ $$

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