Question

(a) Find all the possible values of i^i

Answer 1

a. Since $i=e^{i\pi/2}$, we have

$i^i=(e^{i\pi/2})^i=e^{i^2\pi/2}=e^{-\pi/2}$

b. Since $1=e^0$, we have

$-1^{-i}=-(e^0)^{-i}=-(e^0)=-1$

c. Yes, for the reason illustrated in part b. $1=e^0$, and raising this to any power $z\in\mathbb C$ results in $e^{0z}=e^0=1$.