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

The real part of $z = {i}^{i + 1}$ is zero $0$

Because $z = {i}^{i + 1} = i \cdot {i}^{i}$

But ${i}^{i} = {e}^{-} \left(\frac{\pi}{2}\right)$ hence

$z = {i}^{i + 1} = i \cdot {i}^{i} = i \cdot {e}^{- \frac{\pi}{2}}$

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

is ${e}^{- \frac{\pi}{2}}$

Footnote Why ${i}^{i} = {e}^{-} \left(\frac{\pi}{2}\right)$ ?

Because $i = {e}^{\frac{\pi}{2} \cdot i}$ hence i^i=e^((pi/2)*i*i)=e^(pi/2*i^2)= e^(-pi/2)