How to find Re: z, when z=ii+1 ?

1 Answer
Mar 18, 2017

Re(ii+1)=0

Explanation:

Note that i=eπ2i

Hence:

ii+1=iii1=(eπ2i)ii=e(π2i)ii=eπ2i

...which is pure imaginary.

So:

Re(ii+1)=0