Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 11F / Solution

Inversion in a circle of center $c$ and radius $R$ is $K(z)=c+R^2/(\bar{z}-\bar{c})$ on the extended plane. It sends infinity to $c$, so it interchanges zero and infinity exactly when $c=0$.

For the unit-circle inversion $J$, both $J$ and $K$ are involutions. If $KJ{z}_0=z_0$, applying $K$ gives $J{z}_0=K{z}_0$. Writing $w=J{z}_0$, we have $Jw=z_0$ and $Kw=z_0$, and therefore $KJw=K{z}_0=w$. This proves both assertions about the fixed points.

We now show that one fixed point is inside the disk and use it as the hyperbolic center. Rotate coordinates to make the Euclidean center $c=a\ge 0$. The hypothesis is $a+R<1$. If $a=0$ the conclusion follows immediately from the radial hyperbolic distance. If $a>0$, thenfor its fixed points. Since $1+a^2-R^2>2a$, this quadratic has two positive roots with product one, one $p\in (0,1)$ and the other $1/p>1$.

The disk automorphism $h(z)=(z-p)/(1-pz)$ is a Poincare disc automorphism. It sends the two fixed points to zero and infinity and commutes with $J$ on the extended plane. Consequently $K^{\prime }=hK{h}^{-1}=(hT{h}^{-1})J$ interchanges zero and infinity. Conjugation takes circle inversion to inversion in the image circle, so the first part says that $h(\Gamma )$ is centered at zero. Its Euclidean radius is some $s<1$.

In the Poincaré disk metric, integrating the radial length element givesSince $h$ preserves distance, the original circle is exactlyUndoing the preliminary rotation gives the center for arbitrary complex $c$.

This identifies the hyperbolic circle in the Poincare disc intrinsically.