Consider the map from unit disk to upper half plane $f(\omega) = i \frac{\omega + 1}{1-\omega}$ for $|\omega| <1$. Let $r<1$ and $\Gamma = \{\omega: |\omega|=r\}$, what is the image $f(\Gamma)$?
Clearly $f$ sends $1$ to $\infty$ hence $f(\Gamma)$ does not contain the point $\infty$, therefore $f(\Gamma)$ should be a circle. The symmetric point of $1$ w.r.t $\Gamma$ is the point $2r-1$, hence the center of the circle $f(\Gamma)$ should be the point $f(2r-1) = i \frac{r}{1-r}$ the we compute
$$|f(\omega) - f(2r-1)| = \frac{|\omega+1-2r|}{|1-\omega| \ |1-r|}$$
which doesn't seem to be constant. What am I doing wrong?