Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 3F (Geometry)

The Poincare disc model is the open unit disc $D=\{z=x+iy:|z|<1\}$ equipped with the Riemannian metricThis normalization gives Gaussian curvature $-1$. In particular, a Euclidean curve has hyperbolic length $\int 2|dz|/(1-|z{|}^2)$; the boundary circle is infinitely far away in hyperbolic distance.

A suitable Poincare disc automorphism is the Möbius transformationIt sends $a$ to zero and has inverse $(w+a)/(1+\overline{a}w)$. To verify the isometry directly, note thatThese identities show both that $f$ maps the unit disc onto itself and that $2|df|/(1-|f{|}^2)=2|dz|/(1-|z{|}^2)$, preserving the Riemannian metric.

The radial segment is a geodesic of the Poincare disc model. Its hyperbolic length isOne can also verify that it minimizes length of a curve: along any piecewise smooth curve from zero to radius $r$, the Euclidean speed is at least the absolute radial speed. Integrating $2|dr|/(1-r^2)$ is at least the absolute change in $\log ((1+r)/(1-r))$, with equality on the radial segment. Thus the answer is independent of $\theta$.

Use an isometry of the Poincare disc model to move the centre to zero. A hyperbolic disc of radius $\rho$ then has Euclidean radius $R=\tanh (\rho /2)$ by the hyperbolic distance formula. The area element of a surface of the Riemannian metric is $4rdrd\theta /(1-r^2{)}^2$, soTherefore the area is independent of the centre, and