Show that fourier transformation is a circle I've calculated the fourier transformation of a function and got $\hat{f}(w) = \frac{1}{\rho+iw}$.
Now I must show, that this is equal to a circle with radius $r$ and middlepoint $m$, where $r = m = \frac{1}{2\rho}$
When I plot the function, I see the circles, but I'm not sure how to show that. 
I thought that I must split it into the real and imaginary part and got:
$Re(\hat{f}(w))=\frac{1}{2\rho+2iw}+\frac{1}{2\rho-2iw}$
$Im(\hat{f}(w))=\frac{1}{2i\rho-2w}-\frac{1}{2i\rho+2w}$
Am I on the right way? Can anyone help me to get to $\frac{1}{2\rho}$ for the radius and middlepoint ?

Note that
$$
\hat{f}(\omega) = \frac{1}{\rho + i \omega}\frac{\rho - i \omega}{\rho - i \omega} = \frac{\rho - i\omega}{\rho^2 + \omega^2}
$$
Therefore
$$
x = \Re(\hat{f}(\omega)) = \frac{\rho}{\rho^2 + \omega^2} ~~~\mbox{and}~~~
y = \Im(\hat{f}(\omega)) = -\frac{\omega}{\rho^2 + \omega^2}
$$
We then have
\begin{eqnarray}
\left(x - \frac{1}{2\rho} \right)^2 + y^2 &=& \left(\frac{\rho}{\rho^2 + \omega^2} - \frac{1}{2\rho} \right)^2 + \left(- \frac{\omega}{\rho^2 + \omega^2} \right)^2 \\
&=& \frac{1}{(\rho^2 + \omega^2)^2} \left[ \left(\rho - \frac{\rho^2 + \omega^2}{2\rho} \right)^2 + \omega^2\right] \\
&=& \frac{1}{(\rho^2 + \omega^2)^2} \left[ \left(\frac{2\rho^2 - \rho^2 - \omega^2}{2\rho} \right)^2 + \omega^2\right] \\
&=& \frac{1}{4\rho^2(\rho^2 + \omega^2)^2} \left[ (\rho^2 - \omega^2) + 4\rho^2 \omega^2\right] \\
&=& \frac{1}{4\rho^2(\rho^2 + \omega^2)^2} \left[ \rho^4 - 2\rho^2\omega^2 + \omega^4 + 4\rho^2 \omega^2\right] \\
&=& \frac{1}{4\rho^2(\rho^2 + \omega^2)^2}  (\rho^2 + \omega^2)^2 \\
&=& \frac{1}{4\rho^2}
\end{eqnarray}
So the locus of $\hat{f}(\omega)$ on the complex plane is a circle of radius $\frac{1}{2\rho}$ centered at $x = \frac{1}{2\rho}$, $y = 0$
