Question 134532
{{{(i^7+i^2+i^12+i^9) / (1-i)^2}}}
Note that
{{{i^1 = i}}} 
{{{i^2 = -1}}}
{{{i^3 = -1*i}}}
{{{i^4 = -1*-1}}}
{{{i^5 = -1*-1*i}}}
{{{i^6 = -1*-1*-1}}}
{{{i^7 = -1*-1*-1*i}}}
{{{i^8 = -1*-1*-1*-1}}}
So, the sequence is
i, -1, -i, 1, i, -1, -i, 1
If the exponent is 1,5,9,13. . . every other odd #, then
{{{i^n = i}}}
If the exponent is 3,7,11,15 . . .every other odd #, then
{{{i^n = -i}}}
If the exponent is 2,6,10,14 . . .every other even #, then
{{{i^n = -1}}}
If the exponent is 4,8,12,16 . . .every other even #, then
{{{i^n = 1}}}
{{{(i^7 + i^2 + i^12 + i^9) / (1-i)^2}}}
{{{(-i + (-1) + 1 + i) / (1 -2i + (-1))}}}
{{{0 / (-2i)}}}
Unless I goofed, the answer is {{{0}}}