Question 134532
First, it would be helpful to have a table of values for {{{i^n}}}.  You can do this quite easily:
{{{i^0 = 1}}}
{{{i^1 = i}}}
{{{i^2 = -1}}} because {{{i^2 = (sqrt(-1))(sqrt(-1))}}}= {{{sqrt(-1)^2 = -1}}}
{{{i^3 = -i}}}
{{{i^4 = 1}}}
{{{i^5 = i}}}
{{{i^6 = -1}}}
{{{i^7 = -i}}}
{{{i^8 = 1}}}
{{{i^9 = i}}}
Do you see the pattern?
Simplify:
{{{(i^7+i^2+i^12+i^9)/(1-i)^2 = (-i+(-1)+1+i)/(1-i)^2}}}={{{0/(1-i)^2 = 0}}}