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!
Best Answer
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.