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i-i-




Question Number 80926 by 20092104 last updated on 08/Feb/20
i^i
$${i}^{{i}} \\ $$
Commented by mr W last updated on 08/Feb/20
i=cos (π/2)+i sin (π/2)=e^(i(π/2))   i^i =(e^(i(π/2)) )^i =e^(i^2 (π/2)) =e^(−(π/2)) =(1/( (√e^π )))≈0.2079
$${i}=\mathrm{cos}\:\frac{\pi}{\mathrm{2}}+{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{2}}={e}^{{i}\frac{\pi}{\mathrm{2}}} \\ $$$${i}^{{i}} =\left({e}^{{i}\frac{\pi}{\mathrm{2}}} \right)^{{i}} ={e}^{{i}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}} ={e}^{−\frac{\pi}{\mathrm{2}}} =\frac{\mathrm{1}}{\:\sqrt{{e}^{\pi} }}\approx\mathrm{0}.\mathrm{2079} \\ $$
Commented by mr W last updated on 08/Feb/20
we can also get  (i)^(1/i) =(√e^π )
$${we}\:{can}\:{also}\:{get} \\ $$$$\sqrt[{{i}}]{{i}}=\sqrt{{e}^{\pi} } \\ $$

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