Show that fourier transformation is a circle

Question

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 ?

Answer 1

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 $x = \frac{1}{2\rho}$, $y = 0$

Answer 2

This can be expressed nicely in complex variables. Let $z=\hat{f}(\omega)$. Then we want to show that

$$\left|z-\frac{1}{2\rho} \right|=\frac{1}{2\rho}$$

So

$$ \begin{align} \left|z-\frac{1}{2\rho} \right| & =\left|\frac{1}{\rho+i\omega}-\frac{1}{2\rho} \right|\\ & =\left|\frac{2\rho-(\rho+i\omega)}{2\rho(\rho+i\omega)} \right|\\ & =\left|\frac{\rho-i\omega}{2\rho(\rho+i\omega)} \right|\\ & = \frac{1}{2\rho} \end{align} $$

This seems more direct. You were in the complex plane, so why not stay there?