# Standard Examples: Conformal Maps

A summary:

• $$z\mapsto -z$$ is a reflection about $$y=x$$, so for example sends $${\mathbb{H}}\to Q_{34}$$ and $${\mathbb{D}}\cap{\mathbb{H}}\to {\mathbb{D}}\cap Q_{34}$$ and vice-versa.
• $$z\mapsto 1/z$$: write as $$Re^{it} \mapsto R^{-1}e^{-it}$$, which is reflection about $${\mathbf{R}}$$ and inversion through $$S^1$$.
• $${\mathbb{D}}\to {\mathbb{D}}$$: $$\lambda {z-a\over 1-\overline{a}z}$$ for rotations $$\lambda \in S^1$$
• $${\mathbb{H}}\to {\mathbb{D}}$$: the Cayley map $$z\mapsto {z-i\over z+i}$$
• Horizontal strips to $${\mathbb{H}}$$: use $$z\mapsto e^z$$ for $$0<\Im(z) < \pi \to {\mathbb{H}}$$
• These send $$\left\{{\Im(z) \in (-\pi, \pi) }\right\}\longrightarrow{\mathbf{C}}\setminus{\mathbf{R}}_{\leq 0}$$.
• Sectors to $${\mathbb{H}}$$: for $$0<\operatorname{Arg}(z) < {\pi \over n}$$, use $$z\mapsto z^{n}$$ to get $${\mathbb{H}}$$.
• Some variants:
• Unfolding a half-plane: Generally, \begin{align*}\left\{{\operatorname{Arg}(z) \in (-\theta_0, \theta_0) }\right\} \xrightarrow{z^a} \left\{{\operatorname{Arg}(z) \in (-a\theta_0, a\theta_0) }\right\}\end{align*}

• Unfolding a symmetric sector:

\begin{align*}z\mapsto z^{\pi \over 2\theta_0}: \left\{{\operatorname{Arg}(z) \in (-\theta_0, \theta_0)}\right\}\to \left\{{\operatorname{Arg}(z) \in (-\pi/2, \pi/ 2)}\right\}.\end{align*}

• Half-discs to planes: the Joukowski maps $$z\mapsto z+z^{-1}$$
• Bigons/lunar domains: map the intersection points $$z_0\to 0, z_1\to\infty$$ to get strips.
• In general, take the line through the centers and intersection points. Call the intersection of the two circles $$a$$, the line with the small circle $$b$$, and the last $$c$$, and take $$(z; c,b,a)$$ to get $$0<\Re(z) < 1$$.

Some tips:

• A computational shortcut: $$z^{-1}= {\overline{z}\over {\left\lvert {z} \right\rvert}^2}$$. Use this to quickly compute images, e.g. for $${\left\lvert {z-1} \right\rvert} = 1$$ under $$f(z) = 1/z$$, write $$f(1+i)={1\over 1+i} = {1-i \over 2}$$.
• The locus of points equidistant to two fixed points is the perpendicular bisector.
• Seems obvious, but use that conformal maps preserve angles. You can use tangent vectors to reason about angles of intersection (even at $$z=\infty$$). E.g. Since $${\left\lvert {z-i} \right\rvert}=1$$ intersects $${\mathbf{R}}$$ in a parallel way, since the tangent vectors at zero will line up. So any conformal map must send them to parallel lines or circles with intersection angle zero.
• Similarly if two circles intersect orthogonally, they must go to orthogonal lines or a line orthogonally intersecting a circle.
• $$z\mapsto 1/z$$ corresponds to a rotation of $${\mathbf{CP}}^1$$ around the $$x{\hbox{-}}$$axis by $$\pi$$.

# The Big 9 Conformal Maps

## $${\mathbb{H}}$$ and $${\mathbb{D}}$$

\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.

## Sectors

\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*}

## Logs and Exponentials

The exponential generally sends boxes to sectors, so \begin{align*} \left\{{z {~\mathrel{\Big\vert}~}\Re(z) \in [a, b], \Im(z)\in [c, d]}\right\}\mapsto \left\{{Re^{i\theta} {~\mathrel{\Big\vert}~}R \in [e^a, e^b], \theta\in [c, d]}\right\} .\end{align*}

Pictures of the situation:

\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.

## Joukowski Maps

A nice resource: https://complex-analysis.com/content/joukowsky_airfoil.html

In general, $$z\mapsto z+z^{-1}$$ has the following effects:

• $${\left\lvert {z} \right\rvert} = 1$$ is mapped onto $$[-2, 2]$$
• $${\mathbb{D}}\cap{\mathbb{H}}$$ is mapped to $$Q_{34}$$
• $${\mathbb{D}}^c \cap{\mathbb{H}}$$ is mapped to $${\mathbb{H}}$$
• $${\mathbb{D}}^c$$ is mapped to $${\mathbf{C}}\setminus[-2, 2]$$

\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*}