A Complex Easy Problem

Since ${ i }^{ 2 }=-1$

it can be generalised that ${ i }^{ 4n }=1$ for n is element of integer.

Since $8\equiv 0\quad (mod\quad 4)\\$ and $4008\equiv \quad 0\quad (mod\quad 4)\\$

${ i }^{ 8 }-{ i }^{ 4008 }=1-1\\ \qquad \qquad =0$