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

##### 1 Answer

Mar 18, 2017

#### Explanation:

Note that

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#