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