Conformally Map Region between tangent circles to Disk Suppose we are given two circles, one inside the other, that are tangent at a point $z_0$. I'm trying to map the region between these circles to the unit disc, and my thought process is the following:
I feel like we can map $z_0$ to $\infty$, but I'm not really sure about this. If it works, then I get a strip in the complex plane, and I know how to handle strips via rotation, translation, logarithm, power, and then I'm in the upper half-plane (if I've thought this through somewhat properly). My problem really lies in what point $z_0$ can go to, because I thought the point symmetric to $z_0$ (which is $z_0$ in this case) had to be mapped to $0$. Is this the right idea, and if what other details should I make sure I have? Thanks!

I just recently solved this myself.
Let $z_1$ be the center of the inner tangent circle such that $|z_1 - z_0| =r$ and let $z_2$ be the center of the larger circle with $|z_2 - z_0| = R$ with $R>r$. Rotate and translate your circles so that $z_0$ lies on the real axis (as so does $z_1$) and $z_2$ is 0. You can map this a vertical strip by sending $z_0$ to $\infty$, 0 to 0 and the antipodal point on the inner circle of $z_0$ to 1.
To get from the upper half plane to the vertical strip, you'll need a logarithm and a reflection.
E.g., Consider $B(0,2) \cap \overline{B(3/2,1/2)}$ as the domain. The map as above (sending the balls to the vertical strip) is $\Phi(z) = \frac{z}{z-2} \frac{-1}{1}$ (this is a Möbius transformation). You'll need to invert this to go from the strip to the balls.
