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The Cayley transform gives a conformal bijection between an unbounded half-plane and the unit disc. Two conventions are useful. For the right half-plane, useFor the upper half-plane, useRotating the upper half-plane onto the right half-plane by $z\mapsto -iz$ explains the relationship $C(z)=R(-iz)$. The factors of $i$ distinguish which half-plane is mapped onto the unit disc. ^[ref]

In particular, $C(i)=0$, $C(2i)=1/3$ and $C(1+i)=(1-2i)/5$. The whole positive imaginary axis becomes the real diameter: $C(iy)=(y-1)/(y+1)$ for $y>0$. Horizontal grid lines become circles tangent to the unit circle at $1$ when completed by their endpoint at infinity; they are not hyperbolic geodesics.

On the Riemann sphere, $C$ maps the extended real line onto the unit circle:Every finite real $x$ has $|C(x)|=1$; the remaining point $w=1$ corresponds to infinity. The pole $z=-i$ is outside the upper half-plane, and the pole of $C^{-1}$ at $w=1$ is on the boundary of the unit disc. Neither pole obstructs the interior conformal bijection.

For a short verification, subtract the squared moduli:These identities put each interior image in the required domain; solving $w(z+i)=z-i$ gives the stated inverse function. Both maps are holomorphic, with $C^{\prime }(z)=2i/(z+i{)}^2\ne 0$, so the conformal bijection preserves Euclidean angles between tangent directions.

Euclidean distance is not preserved: $i$ and $2i$ are one unit apart, whereas $0$ and $1/3$ are only one third of a unit apart. The striking feature is that $C$ does preserve length for the hyperbolic metrics of the Poincaré half-plane model and Poincaré disk model:Consequently it is an isometry for these hyperbolic metrics, including their induced hyperbolic distances. For the same example, the corresponding hyperbolic geodesic segments have equal length:Thus the Cayley transform lets us choose the most convenient picture: a straight boundary for calculations, or a bounded unit disc for visualization. It also transports automorphisms of the unit disk into upper half-plane automorphisms by function composition $C^{-1}\circ g\circ C$.

The curvature-minus-one Poincaré disk model has Riemannian metricThe geodesics through zero are Euclidean diameters. Indeed, for a path written in polar coordinates, its length is at least $\int 2|\dot{r}|/(1-r^2)dt$, and hence at least $2\mathrm{artanh}r$ between zero and a point of modulus $r$. The radial segment attains this lower bound. By uniqueness of a geodesic with given initial tangent, these radial paths describe all geodesics through the origin. The radial hyperbolic distance isThus a hyperbolic circle in the Poincare disc is the indicated Euclidean circle.

An isometry from the Poincaré half-plane model to the Poincaré disk model is the Cayley transform between the half-plane and disk,Writing $z=x+iy$, one has $1-|w{|}^2=4y/|z+i{|}^2$ and $|dw{|}^2=4|dz{|}^2/|z+i{|}^4$. Substitution into the disk Riemannian metric gives $d{s}^2=|dz{|}^2/y^2$, the half-plane metric, and $i$ maps to zero.

Put $r=\tanh (\rho /2)$. The inverse image of $|w|=r$ obeysSince $(1+r^2)/(1-r^2)=\cosh \rho$, completing the square yieldsThis proves that the hyperbolic circle in the upper half-plane has Euclidean centre $i\cosh \rho$ and radius $\sinh \rho$.

If two hyperbolic lines intersect, move their intersection to the centre of the Poincaré disk model. Lines through the centre are diameters. Reflection in a diameter making Euclidean angle $\alpha$ with the real axis is $z\mapsto e^{2i\alpha }\overline{z}$. The composition of two such reflections is the rotation $z\mapsto e^{2i(\alpha -\beta )}z$. The Poincaré disk model is conformal, so the Euclidean angle difference is the hyperbolic intersection angle $\theta$, up to sign and replacing it by $\pi -\theta$, neither of which changes rationality of $\theta /\pi$.

The rotation has finite order exactly when $m\theta \in \pi \mathbb{Z}$ for some positive integer $m$, equivalently $\theta /\pi \in \mathbb{Q}$. If $\theta /\pi =r/s$ in lowest terms, its order is $s$. Conversely, the nonintersecting cases in part(a) have infinite order. Therefore