How to find Re: z, when z=ii+1 ?
1 Answer
Mar 18, 2017
Explanation:
Note that
Hence:
ii+1=ii⋅i1=(eπ2i)i⋅i=e(π2i)i⋅i=e−π2⋅i
...which is pure imaginary.
So:
Re(ii+1)=0
Note that
Hence:
ii+1=ii⋅i1=(eπ2i)i⋅i=e(π2i)i⋅i=e−π2⋅i
...which is pure imaginary.
So:
Re(ii+1)=0