+9519 \(\text{Let }x=i^i\\ \ln x = i \ln i\)
Now we need to find the value of ln(i).
\(e^{i\pi} = -1\\ \ln(-1)=i\pi\\ \ln i = \dfrac{1}{2}\ln(-1)= \dfrac{i\pi}{2}\)
Substitute the result into the equation:
\(\ln x = i\left(\dfrac{i\pi}{2}\right)\\ \quad \;\;=-\dfrac{\pi}{2}\\ e^{\ln x}=e^{-\pi /2}\\ x = e^{-\pi/2}\\ \therefore i^i = e^{-\pi /2}\)
e^(-pi /2) approximately equals 0.207879.
