Given a Möbius Transformation $w=f(z)$ find $f(-\bar{z})$ Let $w=f(z)$ be a Möbius  Transformation, and let $\gamma,\Gamma\subset\mathbb{C}$ be the following curves:
$$\gamma=\{z\in\mathbb{C}\mid \Re(z)=0\}\\ \Gamma=\{w\in\mathbb{C}\mid |w-w_0|=r\}$$ 
Where $w_0\in\mathbb{C}$ and $r\in\mathbb{R^+}$. Given that $f(\gamma)=\Gamma$, find the value of $h(z)=f(-\bar{z})$ using $w,w_0$ and $r$.

I know that $f(z)$ is a Möbius Transformation, therefore:
$$f(z)=\frac{az+b}{cz+d}$$
For some $a,b,c,d\in\mathbb{C}$. Now because $f(\gamma)=\Gamma$, I receive:
$$f(0)=\frac bd\\f(\bar\infty)=\frac ac$$
Therefore $\frac ac, \frac bd\in\Gamma$. Choosing any other $z\in\gamma$ and trying to work out the equations led me to some ugly algebra which I couldn't manage to handle. I am sure there is an easier way to handle this but I don't know how.
Thanks

$z$ and $-\bar z$ are symmetric about the line $\gamma$.  Such points are  conjugate.
Mobius transformations happen to preserve conjugate points. Thus $h(z)$ is symmetric to $f(z)$ relative to the circle $\Gamma$.  This means they are the images of each other under circle inversion.  
So we get $h(z)=w_0+\dfrac{r^2(
w-w_0)}{|w-w_0|^2}$.
