# 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!

As it turns out, writing down an explicit formula is very long, but describing the general process is easy (and is similar to Chris C's, although I still feel his solution has a hole). Take $z_0$ to be the point of tangency, $z_1$ to be the antipodal point on the inner circle, and $z_2$ to be the antipodal point on the outer circle (in particular, these points are on the boundaries of the circles). Assuming we're given equations for the circles, we can explicitly calculate these antipodal points. From here, rotate (if necessary) so that these points are all on the real axis and translate so that $z_2$ is at $0$. Now, given the equations of the circles, we can ensure $z_0$ and $z_1$ lie on the positive real axis. From here, we can use a Moebius transformation to send $z_2\mapsto 0$ (so $0$ is fixed), $z_1\mapsto 1$, and $z_0\mapsto\infty$. Since angles are preserved and this maps the circles to lines, this maps the region to a vertical strip in the complex plane of width $1$. Now rotate this by $\pi/2$, dilate it by $\pi$, and apply the exponential map. This brings us to the upper half-plane, and now we can use the standard map to go the unit disc.

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.

• This is roughly what I was thinking, but is there any sort of an explicit map we can write down? – Clayton Apr 28 '13 at 16:15
• I've edited my answer to include an example transformation. The transformations of this type are also known as fractional linear transformations. – Chris C Apr 28 '13 at 20:53
• There seems to be a small problem with this solution. How do we know the antipodal point isn't $z_2$? Take for example, $B(0,2)$ and $B(1,1)$. In more generality, it seems feasible that $r=R/2$. – Clayton Apr 29 '13 at 5:49
• Another problem is that lines intersect at $\infty$, but they don't intersect at $z_1$. That is, if anything goes to $\infty$, it has to be the point of tangency. – Clayton Apr 29 '13 at 12:21
• The antipodal point can be the center. You'll just have to send another point to 0. As in your case, you could send -1/2 to 0 by $\Phi(z) = \frac{z+1/2}{z-1}\frac{1/2}{-1}$ – Chris C Apr 30 '13 at 14:50