Question

What's the value of i^i, where i = (-1)^1/2 ?

Answer 1

This is quite surprising and beautiful:

In polar form, i has infinity many representations:
\[i = e^{i(\pi/2 + 2k\pi)}, \ \ k \in \mathbb{Z}\]

Thus
\[i^i = e^{i^2(\pi/2 + 2k\pi)} = e^{-\pi/2(1 + 4k)} , \ \ \ k \in \mathbb{Z}\]