\begin{align*} \Psi: {\mathbb{H}}&\rightleftharpoons{\mathbb{D}}\\ z &\mapsto {z-i \over z+i} \\ i \qty{1+w \over 1-w} &\mapsfrom w .\end{align*}

Mnemonic: every \(z\in {\mathbb{H}}\) is closer to \(i\) than \(-i\). This restricts to a map \begin{align*} \Psi: Q_1 &\rightleftharpoons{\mathbb{D}}\cap{\mathbb{H}} .\end{align*}

\begin{align*} F: {\mathbb{D}}&\mapsto Q_{12} \coloneqq\left\{{\Re(z) > 0}\right\} \\ z &\mapsto {1+z \over 1-z} \\ {w-1\over w+1} &\mapsfrom w .\end{align*}

This satisfies \begin{align*} {\left[ {-1, 0, 1} \right]} \mapsto {\left[ {0, 1, \infty} \right]} .\end{align*}

Note that \(\Psi\) inverse from above can be recovered by post-composing with a rotation by \(\pi/2\): \begin{align*} \Psi^{-1}(z) = i\qty{1+z\over1-z} = i \cdot F(z) && {\mathbb{D}}\xrightarrow{F} Q_{12} \xrightarrow{\cdot i} {\mathbb{H}} ,\end{align*} and up to a negative sign, we can recover \(\Psi\) by recomposition with a rotation by \(-\pi/2\): \begin{align*} F(-iz) = {1+ iz \over 1-iz} = {-i + z \over -i-z} = -{z-i\over z+i} = -\Psi(z) && {\mathbb{H}}\xrightarrow{\cdot -i} Q_{12} \xrightarrow{F} {\mathbb{D}} .\end{align*}

This restricts to a map \(F: {\mathbb{D}}\cap{\mathbb{H}}\to Q_1\):

• Why this lands in the first quadrant:
  • Use that squares are non-negative and \(z=x+iy\in {\mathbb{D}}\implies x^2 + y^2 < 1\): \begin{align*} f(z)=\frac{1-\left(x^{2}+y^{2}\right)}{(1-x)^{2}+y^{2}}+i \frac{2 y}{(1-x)^{2}+y^{2}} .\end{align*}
• Why the inverse lands in the unit disc:
  • For \(w\) in Q1, the distance from \(w\) to 1 is smaller than from \(w\) to \(-1\).
  • Check that if \(w=u+iv\) where \(u, v>0\), the imaginary part of the image is positive:

` \begin{align*} {w-1 \over w+1} &= { (w-1) \overline{(w+1)} \over {\left\lvert {w+1} \right\rvert}^2}\ &={ \qty{u-1 + iv} \qty{u+1-iv} \over (u+1)^2 + v^2 } \ &= {u^2 + v^2 + 1 \over (u+1)^2 + v^2}

• i\qty{ 2v \over (u+1)^2 + v^2} .\end{align*} `{=html}

Boundary behavior:

• On the upper half circle \(\left\{{ e^{it } {~\mathrel{\Big\vert}~}t\in (0, \pi) }\right\}\), write \begin{align*} f(z)=\frac{1+e^{i \theta}}{1-e^{i \theta}}=\frac{e^{-i \theta / 2}+e^{i \theta / 2}}{e^{-i \theta / 2}-e^{i \theta / 2}}=\frac{i}{\tan (\theta / 2)} ,\end{align*} so as \(t\) ranges \(0\to \pi\) we have \(f(z)\) ranging from \(0\to i\infty\) along the imaginary axis.

\begin{align*} F: \left\{{z{~\mathrel{\Big\vert}~}\operatorname{Arg}(z) \in \qty{0, {\pi \over n}} }\right\} &\to {\mathbb{H}}\\ z &\mapsto z^n \\ w^{1\over n} &\mapsfrom w .\end{align*}

More generally, for \(0 < \alpha < 2\), \begin{align*} F: {\mathbb{H}}&\to \left\{{z{~\mathrel{\Big\vert}~}\operatorname{Arg}(z) \in \qty{0, \alpha} }\right\} \\ z &\mapsto z^{\alpha\over \pi} \\ w^{\pi \over \alpha} &\mapsfrom w .\end{align*}

\begin{align*} F: {\mathbb{H}}&\to {\mathbf{R}}\times i(0, \pi) \\ z &\mapsto \operatorname{Log}(z) \\ e^w &\mapsfrom w .\end{align*}

• Why this lands in a strip: use that \(\arg(z) \in (0, \pi)\) and \(\log(z) = {\left\lvert {z} \right\rvert} + i\arg(z)\).

\begin{align*} F: {\mathbf{C}}\setminus{\mathbf{R}}_{\leq 0} &\to {\mathbf{R}}\times i(-\pi, \pi) \\ z &\mapsto \operatorname{Log}(z) .\end{align*}

• Circles of radius \(R\) are mapped to vertical line segments connecting \(\ln(R) + i\pi\) to \(\ln(R) - i\pi\), and rays are mapped to horizontal lines.
• Inverse is useful: \(z\mapsto e^z\) sends \(\left\{{ {\left\lvert { \Re(z) } \right\rvert} < \pi/2 }\right\}\) to \(Q_{14}\), the right half-plane.

\begin{align*} F: {\mathbb{D}}\cap{\mathbb{H}}&\to {\mathbf{R}}_{<0} \times i (0, i\pi) \\ z &\mapsto \operatorname{Log}(z) \\ e^w &\mapsfrom w .\end{align*}

In general,

• \({\mathbb{D}}\cap{\mathbb{H}}\) maps to a half-strip in \(Q_2\)
• \({\mathbb{D}}\cap Q_{34}\) maps to a half-strip in \(Q_1\).
• \({\mathbb{D}}^c \cap{\mathbb{H}}\) maps to a half-strip in \(Q_1\)
• \({\mathbb{D}}^c \cap Q_{34}\) maps to a half-strip in \(Q_4\)

\begin{align*} F: \qty{-{\pi \over 2}, {\pi \over 2}} \times i{\mathbf{R}}&\to {\mathbb{D}}\cap\left\{{\Re(z) > 0}\right\} \\ z &\mapsto e^{iz} \\ -i\operatorname{Log}(w) &\mapsfrom w .\end{align*}

This is essentially polar coordinates: write \(e^z = e^{-y} e^{ix}\), then \(x\in (-\pi/2, \pi/2)\) and \(y\in (0, \infty)\) so this fills out a half-disc as \(x,y\) vary.

\begin{align*} F: {\mathbb{D}}^c \cap{\mathbb{H}}&\to {\mathbf{C}}\setminus[-2, 2] \\ z &\mapsto z+ z^{-1} .\end{align*}

\begin{align*} F: {\mathbb{D}}\cap{\mathbb{H}}&\to {\mathbb{H}}\\ z & \mapsto -{1\over 2}\qty{ z + z^{-1}} .\end{align*}

• It additionally maps \({\mathbb{D}}^c\to {\mathbf{R}}\setminus[-1, 1]\).
• A similar variant: \(z\mapsto {1\over 2i}(z+z^{-1})\) sends \(Q_{14}\) to \({\mathbf{C}}\setminus\left\{{t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}{\left\lvert {t} \right\rvert} \geq 1}\right\}\), the plane with a slit on \({\mathbf{R}}\) through infinity (i.e. a doubly slit plane).

This is sometimes referred to as a Joukowski map.o The inverse is a bit complicated.

\begin{align*} F: {\mathbb{H}}&\to \qty{-{\pi \over 2}, {\pi \over 2}} \times i{\mathbf{R}}\\ z &\mapsto \sin(z) .\end{align*}

The mapping \(z\mapsto \sin(z)\):

• As \(z\) travels from \(i\infty \to i0\), \(\sin(iz) = i\sinh(z)\) also traverses \(i\infty\to i0\)
• For \(z\in[-\pi/2, \pi/2]\), \(\sin(z)\) is real and in \([-1, 1]\).
• As \(z\) travels along \(\pi/2 + it\) for \(t\in [0, \infty)\), \(\sin(\pi/2 + it) = \cosh(t)\) traverses \(1\to \infty\) along \({\mathbf{R}}\)

Note that this isn’t new: set \(w \coloneqq e^{iz}\), then \begin{align*} \sin(z) = -{1\over 2}\qty{iw + {1\over iw}} ,\end{align*} which is the composition \begin{align*} \qty{z \mapsto e^{z} } \circ \qty{z\mapsto iz} \circ \qty{z\mapsto {1\over 2}(z+z^{-1})} .\end{align*}