Question

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

Answer 1

The real part of `z=i^(i+1)` is zero `0`

Because `z=i^(i+1)=i*i^i`

But `i^i=e^-(pi/2)` hence

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

which means that the real part is zero and the imaginary part

is `e^(-pi/2)`

Footnote Why `i^i=e^-(pi/2)` ?

Because `i=e^(pi/2*i)` hence `i^i=e^((pi/2)*i*i)=e^(pi/2*i^2)= e^(-pi/2)`