= Cayley transform between the half-plane and disk
{c}
{title2=$C(z)=(z-i)/(z+i)$}
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>, use
$$
R(z)=\frac{z-1}{z+1},\qquad R^{-1}(w)=\frac{1+w}{1-w},\qquad 1-|R(z)|^2=\frac{4\operatorname{Re}z}{|z+1|^2}.
$$
For the <upper half-plane>, use
$$
C(z)=\frac{z-i}{z+i},\qquad C^{-1}(w)=i\frac{1+w}{1-w},\qquad 1-|C(z)|^2=\frac{4\operatorname{Im}z}{|z+i|^2}.
$$
Rotating 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>. https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/ef46ddd9941227b8700c945bb7475ae9_MIT18_04S18_topic10.pdf[Orloff, MIT Topic 10, Example 10.14 and section 10.11]{ref}
\Image[/cayley-half-plane-grid.png]
{title=The upper half-plane Cayley transform bends a rectangular grid into the unit disc while preserving angles at intersections}
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>:
$$
C(-1)=i,\qquad C(0)=-1,\qquad C(1)=-i,\qquad C(\infty)=1.
$$
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>:
$$
|z+i|^2-|z-i|^2=4\operatorname{Im}z,\qquad \operatorname{Im}C^{-1}(w)=\frac{1-|w|^2}{|1-w|^2}.
$$
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 $C'(z)=2i/(z+i)^2\ne0$, 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>:
$$
ds_{\mathbb H}=\frac{|dz|}{\operatorname{Im}z},\qquad ds_{\mathbb D}=\frac{2|dw|}{1-|w|^2},\qquad \frac{2|C'(z)|\,|dz|}{1-|C(z)|^2}=\frac{|dz|}{\operatorname{Im}z}.
$$
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:
$$
\int_1^2\frac{dy}{y}=\log2=\int_0^{1/3}\frac{2\,dr}{1-r^2}.
$$
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> $C^{-1}\circ g\circ C$.
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