How to break $\frac{1}{z^2}$ into real and imaginary parts? $$ \frac{1}{(x+iy)^2}=\frac{1}{x^2+i2xy-y^2}=\frac{x^2}{(x^2+y^2)}-\frac{2ixy}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
So I thought I could just say:
$$ Re(\frac{1}{z^2})=\frac{x^2}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
and
$$ Im(\frac{1}{z^2})=-\frac{2ixy}{(x^2+y^2)}$$
But I know that is wrong because it looks nothing like the graph of the real part of 1/z^2 on wolfram alpha found here: http://www.wolframalpha.com/input/?i=1%2F(x%2Bi*y)%5E2
Then I thought I could must multiply $1/z^2$ by $z/z$ to get $\frac{x}{z^3}$ and $\frac{iy}{z^3}$ however graphing these again shows that they are not the real and complex parts of $\frac{1}{z^2}$.

\begin{eqnarray} 
f(z)=1/(z^2)=1/((x+iy)^2)&=&1/(x^2-y^2+2ixy) \newline
                         &=&(1/(x^2-y^2+2ixy))\cdot((x^2-y^2-2ixy)/(x^2-               y^2-2ixy)) \newline
                         &=&((x^2-y^2-2ixy)/((x^2-y^2)^2+4x^2y^2)) \newline
                         &=&(x^2-y^2-2ixy)/(x^2+y^2)^2
\end{eqnarray}
therefore $\Re(1/z^2)=(x^2-y^2)/(x^2+y^2)^2$ and $\Im(1/z^2)=-2xy/(x^2+y^2)^2.$
