I have trouble finding a Möbius (conformal) transformation $w=f(z)$ that fulfills the criteria that $w=f(0)=\frac{1}{2}$ for the following case:
Find $w=f(z)$ so that $|z|<1$ maps onto $|w|<1$ so that $f(0)=\frac{1}{2}$.
How can one go about this problem?
While, a bit more generally, $SU(1,1)$ acts transitively on the disk, let's consider a specific map:
Fix $q\in D_1(0),$ and consider the map $$\varphi_q(z)=\frac{z-q}{1-\bar{q}z}.$$ It's not difficult to show that $\varphi_q:D_1(0)\rightarrow D_1(0)$, and that $\varphi_q(q)=0.$ Also, $\varphi_q(0)=-q$ (you should check all of these properties!). With this in mind, we consider $\varphi_{-1/2}$. It's easy to see from the above facts that $\varphi_{-1/2}(0)=1/2,$ and $\varphi_{-1/2}$ is a conformal map from the disk to itself.