Question

How to find Re: z, when `z=i^(i+1)` ?

Answer 1

`Re(i^(i+1)) = 0`

Explanation:

Note that `i = e^(pi/2i)`

Hence:

`i^(i+1) = i^i*i^1 = (e^(pi/2i))^i*i = e^((pi/2i)i)*i = e^(-pi/2)*i`

...which is pure imaginary.

So:

`Re(i^(i+1)) = 0`