Let $C$ be the circle with center $0$ and radius $1$. Find a Möbius transformation which transforms $C$ onto $C$ and transforms $0$ to $1/2$.
Notes: consider $h(z)= az+\dfrac{b}{c}z+d$ then $h(0) = 1/2$ so $2b = d$. Check this works by confirming $\left|h(z)\right| = 1$ when $\left|z\right| = 1$ (along the contour $C$) so that
\begin{equation} \left|az+b\right|^{2} = \left|cz+2b\right|^2 \end{equation}
when
\begin{equation} \left|z\right| = 1 \rightarrow \left|a\right|^{2}\left|z\right|^2+2\Re(a\bar{b}z + \left|b\right|^2) = \left|c\right|^{2}\left|z\right|^2+4\Re((\bar{b}cz) + 4\left|b\right|^2) \end{equation}
The problem is I cannot get them equal, so I'm not sure if this method will satisfy.
I also tried another method considering the case $i \cdot \dfrac{z+i}{z-i}$ with formula $h(z)=f^{-1}(rf(z))$ this yielded an imaginary answer for $r$ which is wrong.