The <Poincare disc model> is
$$
D=\{z\in\mathbb C:|z|<1\},
\qquad
ds^2=\frac{4|dz|^2}{(1-|z|^2)^2}.
$$
This <Riemannian metric> has constant Gaussian curvature $-1$.

For $a\in D$ and $|\eta|=1$, the <Möbius transformation>
$$
T_{a,\eta}(z)=\eta\frac{z-a}{1-\overline a z}
$$
is an <isometry> of the disc, and every orientation-preserving disc isometry has this form. In particular, $T_{z_1,1}$ sends $z_1$ to zero and $z_2$ to a point of modulus
$$
\rho=\left|\frac{z_2-z_1}{1-\overline{z_1}z_2}\right|.
$$
Rotations are also isometries, so it remains only to integrate the metric along a radial geodesic from $0$ to $\rho$:
$$
d(0,\rho)=\int_0^\rho\frac{2\,dr}{1-r^2}
=2\operatorname{artanh}\rho.
$$
Thus the <Hyperbolic distance in the Poincare disc> is
$$
\boxed{d(z_1,z_2)=2\operatorname{artanh}\left|\frac{z_2-z_1}{1-\overline{z_1}z_2}\right|
=\log\frac{1+\rho}{1-\rho}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 1 / 3E / Solution

{c}

For $z_1,z_2$ in the <Poincare disc model>, put
$$
\rho=\left|\frac{z_2-z_1}{1-\overline{z_1}z_2}\right|.
$$
Their <hyperbolic distance> is
$$
d(z_1,z_2)=2\operatorname{artanh}\rho
=\log\frac{1+\rho}{1-\rho}.
$$

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