Poincare disc model

The Poincare disc carries the metric $4|dz{|}^2/(1-|z{|}^2{)}^2$ on $|z|<1$.

For $|a|<1$, the Möbius transformation $f_a(z)=(z-a)/(1-\overline{a}z)$ is an isometry of the Poincare disc model sending $a$ to zero. Rotations combined with these transformations describe its orientation-preserving isometries.

In curvature minus one, a hyperbolic disc of radius $R$ has areaIn the Poincare disc this follows by integrating $4r(1-r^2{)}^{-2}drd\theta$ up to Euclidean radius $\tanh (R/2)$.

Hyperbolic discs of area $\pi /2$ have radius $\log 2$. Starting at the origin and placing their centers successively along one radial geodesic gives center distance $n\log 4$ and Euclidean coordinate

An isometry sends the centers of a collinear family of equal hyperbolic discs to points on one hyperbolic geodesic. In the upper-half-plane model no geodesic contains three distinct points on a horizontal Euclidean line, since geodesics are vertical lines or semicircles orthogonal to the boundary.

For a piecewise smooth curve $z(t)$ in the Poincare disc, its hyperbolic length is

The geodesics are diameters and arcs of circles orthogonal to the unit circle. A radial segment is length minimizing becausewith equality for constant angle and monotone radius. Hence the distance from the origin to $z$ is $2\mathrm{artanh}|z|$.

For $z_1,z_2$ in the Poincare disc model, putTheir hyperbolic distance is

If $P$ is at hyperbolic distance $\rho$ from the origin, the Poincare-disc geodesic through $P$ perpendicular to $OP$ is represented by a Euclidean circle of radius $\mathrm{csch}\rho$ whose centre is at Euclidean distance $\coth \rho$ from the origin.