Is it REAL?

Similar solution as Renz Jimwel Mina , using Euler Identity:

$e^{i\theta} = \cos {\theta} + i \sin{\theta}$

$\Rightarrow i^i = (0 + i)^i = (\cos{\frac{\pi}{2}} + i \sin{\frac{\pi}{2}})^i = (e^{i\frac{\pi}{2}})^i = e^{-\frac{\pi}{2}} = 0.207879576$

e^iπ=-1 =>e^(iπ/2)=i =>e^((i^2 π)/2)=i^i =>e^((-π)/2)=i^i And so this is it.

i^i = e^(iΠ/2)^i = e^(-Π/2)