Möbius transform calculation, over an annulus I started learning about Möbius transformations in my Complex Analysis textbook. This question appeared as an exercise (no solutions are provided, sadly):
Let's say you have a Möbius transform that maps the annulus $r<|z|<1$ to a region bounded by two circles (for the sake of example take $|z|=1$ and $|z-1/4|=1/4$). Is this enough information to figure out $r$?
What I've tried: Trying to figure out what different compositions of Möbius transforms will affect in here. I was thinking taking the inverse somehow would be a good idea but I couldn't find my way around the details of how I'd do this.
I'm stumped, so an explanation of how this is possible would be great. Thanks!
 A: Yes, $r$ is determined by the other doubly-connected region $\Omega$ (even if we do not restrict ourselves to Möbius transformations or to domains bounded by circles). A high-tech explanation involves the concept of conformal modulus, see Conformal maps of doubly connected regions to annuli. 
Here is a low-tech approach which still does not rely on solving for the Möbius map. Let $L$ be a line of symmetry of $\Omega$; in your example it's the real line. Its image under  a Möbius transformation onto $r<|z|<1$ is a line (or circle) $L'$. Since $L$ is invariant under inversion with respect to either boundary component of $\Omega$, $L'$ must be invariant under inversion in the circle $|z|=1$ and under inversion in $|z|=r$. Since the composition of two latter inversions is the scaling map $z\mapsto r^2z$, it follows that $L'$ is a line through the origin $z=0$. Composition with a rotation ensures that $L'$ is the real line.
Now recall that the cross-ratio is preserved by Möbius transformations. The line $L$ crosses $\partial \Omega$ at four points: in your example they are $-1,0,1/2,1$. The line $L'$ crosses the boundary of $r<|z|<1$  at the points $-1,-r,r,1$. Therefore, the cross-ratios of these quadruples are equal: 
$$(-1,0;1/2,1)=\frac{-1-1/2}{0-1/2}\cdot\frac{0-1}{-1-1}= \frac32 \tag1$$
must be equal to  $$(-1,-r;r,1)=\frac{-1-r}{-r-r}\cdot\frac{-r-1}{-1-1}=\frac{(1+r)^2}{4r}\tag2$$ 
Hence $r=2-\sqrt{3}$.
Added: One can do without the knowledge of cross-ratio. The  Möbius map sending $0,1/2,1$ to $-r,r,1$ is described by the equation
$$\frac{z-1/2}{0-1/2}\cdot\frac{0-1}{z-1} = \frac{w-r}{-r-r}\cdot\frac{-r-1}{w-1}\tag3$$
Since we also want $w=-1$ at $z=-1$, equation (3) yields the equality of (1) and (2). Incidentally, (3) also gives a formula for our Möbius map, should we want to have it.
A: First solve this exercise: Given two non-intersecting circles you can find an inversion that makes them concentric. 
Use for this that if two points are inverse of each other with respect to a circle and some other inversion is applied, then their images remain inverses of each other with respect to the image of the circle.
This will allow you to make the region in the target an annulus. Then bring it to the origin by a translation, and by a dilation overlap it to the known boundary of the annulus in the source. If they don't completely overlap, invert with center the origin and radious the known boundary of the annulus in the target to exchange the order of the boundary (which one is inside and which one is outside). The radius of the other boundary gives you the value of $r$.
Still, computing the transformation is not the most efficient way if you are interested only in computing $r$.
A: Do you know any hyperbolic geometry? My first thought with such a question would be to first map the circle of radius $1$ to the vertical line going through the origin and then note that a Möbius transformation which fixes this line is a dilation (ie the original Möbius transformation is conjugate to a dilation). It is then easy to determine what this dilation must be to map the image of the circle of radius $r$ to the image of the corresponding circle of radius $\frac{1}{4}$.
