(a) Find all the possible values of i^i (b) Find all the possible values of -1^(-i) (c) Is 1 to every power (real or complex) necessarily equal to 1?

Question

asked Apr 14, 2020 147k views

1 Answer

Jondlm · Answer 1 · 2020-04-19T18:59:16+0000

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 .