Since \(\omega = \frac {1}{z}\), we have \(z = \frac {1}{\omega }\). Hence, \begin {align*} \frac {1}{\omega } \frac {1}{\omega ^*} - a \frac {1}{\omega ^ *} - \frac {1}{\omega } a^* + a a^* & = r^2 \\ 1 - \omega a - \omega ^* a^* + a a^* \omega \omega ^* & = r^2 \omega \omega ^* \\ (r^2 - a a^*) \omega \omega ^* + \omega a + \omega ^* a^* & = 1 \\ \omega \omega ^* + \frac {a}{r^2 - a a^*} \omega + \frac {a^*}{r^2 - a a^*} \omega ^* & = \frac {1}{r^2 - a a^*} \\ \left (\omega + \frac {a^*}{r^2 - a a^{*}}\right ) \left (\omega + \frac {a^*}{r^2 - a a^{*}}\right )^* & = \frac {1}{r^2 - a a^*} + \frac {a a*}{\left (r^2 - a a^*\right )^2} \\ \abs *{\omega - \frac {a^*}{a a^{*} - r^2}}^2 & = \frac {r^2}{\left (r^2 - a a^*\right )^2} \\ \abs *{\omega - \frac {a^*}{a a^{*} - r^2}} & = \frac {r}{\abs *{r^2 - a a^*}}, \end {align*}
so \(\omega \) is on a circle \(C'\) with centre \(\frac {a^*}{a a^{*} - r^2}\) and radius \(\frac {r}{\abs *{r^2 - a a^*}}\).
If \(C\) and \(C'\) are the same circle, we have \[ a = \frac {a^*}{a a^{*} - r^2}, r = \frac {r}{\abs *{r^2 - a a^*}}. \]
The second equation gives \(\abs *{r^2 - a a^*} = 1\), which means \(r^2 - a a^* = \pm 1\).
\begin {align*} r^2 - a a^* & = \pm 1 \\ r^2 - \abs *{a}^2 & = \pm 1 \\ \left (\abs *{a}^2 - r^2\right )^2 & = 1, \end {align*}
as desired.
When \(r^2 - a a^* = 1\), \(a = - a^*\), and hence \(a\) is pure imaginary. Since \(r^2 = 1 + \abs *{a}^2\) in this case, \(r > \abs *{a}\), so the circle must contain the origin. The diagrams are as below, with the case \(\im (a) > 0\) on the left, \(\im (a) = 0\) in the middle, and \(\im (a) < 0\) on the right:
When \(r^2 - a a^* = -1\), \(a = a^*\), and hence \(a\) is real. Since \(r^2 = -1 + \abs *{a}^2\) in this case, \(r < \abs *{a}\), so the circle cannot contain the origin, and \(\abs *{a} > 1\). The diagrams are as below, with the case \(\re (a) > 1\) on the left, and \(\re (a) < -1\) on the right:
In the case where \(\omega = \frac {1}{z^*}\), we have \(z = \frac {1}{\omega ^*}\), and hence similar to the previous one, \begin {align*} \omega \omega ^* + \frac {a}{r^2 - a a^*} \omega ^* + \frac {a^*}{r^2 - a a^*} \omega & = \frac {1}{r^2 - a a^*} \\ \left (\omega + \frac {a}{r^2 - a a^{*}}\right ) \left (\omega + \frac {a}{r^2 - a a^{*}}\right )^* & = \frac {1}{r^2 - a a^*} + \frac {a a*}{\left (r^2 - a a^*\right )^2} \\ \abs *{\omega - \frac {a}{a a^{*} - r^2}}^2 & = \frac {r^2}{\left (r^2 - a a^*\right )^2} \\ \abs *{\omega - \frac {a}{a a^{*} - r^2}} & = \frac {r}{\abs *{r^2 - a a^*}}, \end {align*}
so \(\omega \) is on a circle \(C'\) with centre \(\frac {a}{a a^{*} - r^2}\) and radius \(\frac {r}{\abs *{r^2 - a a^*}}\).
If they are the same circle, we have \[ a = \frac {a}{a a^* - r^2}, r = \frac {r}{\abs {r^2 - a a^*}}. \]
We still have \(r^2 - a a^* = \pm 1\).
When \(r^2 - a a^* = 1\), we have \(a = -a\), and \(a = 0\).
When \(r^2 - a a^* = -1\), we have \(a = a\), and \(a\) can be any complex number satisfying \(\abs *{a} = \sqrt {r^2 + 1}\).
It is not the case that \(a\) is either real or pure imaginary.