# Exercises: Conformal Maps

Notation:

• Almost everything is an open set, so don’t include boundaries in definitions.
• $${\mathbb{D}}\coloneqq\left\{{z {~\mathrel{\Big\vert}~}{\left\lvert {z} \right\rvert} < 1}\right\}$$ is the open unit disc.
• $${\mathbb{H}}\coloneqq\left\{{z {~\mathrel{\Big\vert}~}\Im(z) > 0}\right\}$$ is the open upper half-plane.
• $$Q_i$$ is the $$i$$th quadrant, e.g. $$Q_1 \coloneqq\left\{{z {~\mathrel{\Big\vert}~}\Re(z), \Im(z) > 0}\right\}$$.
• $$Q_{ij} \coloneqq Q_i \cup Q_j$$ is the union of two quadrants. E.g. $${\mathbb{H}}= Q_{23}$$, or $$Q_{14}$$ is the right half-plane.

Tips:

• If just mapping the disc to itself, use the hypberolic translations \begin{align*} \psi_a \coloneqq{z-a\over 1-\overline{a} z} .\end{align*}

• For lunes (regions bounded by arcs): map the cusps to $$0$$ and $$\infty$$ to get a sector.

• For discs with slits: aim for $${\mathbf{C}}\setminus[0, \infty) \xrightarrow{\sqrt z} {\mathbb{H}}$$.

• For circles with tangencies: send the tangent point to $$\infty$$ to get parallel lines.

• Remembering the cross ratio: the order $$1,0,\infty$$ is very important (as images of $$a, b, c$$).

• Send $$b\to 0$$ by including $$z-b$$ in the numerator.
• Send $$c\to \infty$$ by including $$z-c$$ in the denominator.
• Send $$a\to 1$$ by canceling the terms just added:
• Cancel $$z-c$$ in the denominator with $$a - c$$ in the numerator.
• Cancel $$z-b$$ in the numerator with a $$a - b$$ in the denominator.
• Inverting conformal maps: just set $$f(z) = w$$ and solve for $$w$$.

• Conformal maps preserve generalized spheres, i.e. circles get mapped to circles (which could be lines on $${\mathbf{CP}}^1$$).

• Orthogonal circles must go to orthogonal circles.

• Arcs between two points must go to arcs between their images

• $${\mathbf{R}}=\left\{{\tan(t) {~\mathrel{\Big\vert}~}t\in (-\pi/2, \pi/ 2)}\right\}$$.

## Cross-Ratios

Find a Mobius transformation sending

• $$1\to 3$$
• $$i\to 0$$
• $$2\to -1$$

Use cross ratios: set $$T(z) \coloneqq(z;,1,i,2)$$ and $$S(w) = (w;,3,0,-1)$$ and solve $$T(z) = S(w) \implies w = (S^{-1}T)(z)$$: \begin{align*} {z-i \over z-2}{1-2\over 1-i} &= {w-0\over w+1}{3+1 \over 3-0} \\ \implies -\frac{\left(i + 1\right) \, {\left(z - i\right)}}{2 \, {\left(z - 2\right)}} &= \frac{4 \, w}{3 \, {\left(w + 1\right)}} \\ \implies w &= -\frac{3 \, {\left(\left(i + 1\right) \, z - i + 1\right)}}{\left(3 i + 11\right) \, z - 3 i - 13} \\ &= - {3z - 3i \over {3i+11\over i+1}z + {-3i-13 \over 3i+11}} \\ &= - {3z - 3i \over (7-4i) z + {-8+5i} } \\ &= \frac{-3 z+3 i}{(7-4 i) z+(-8+5 i)} .\end{align*}

Find a conformal map from $${\mathbb{D}}^c \cap{\mathbb{H}}$$ to $${\mathbb{H}}$$ using cross-ratios.

Idea: all cross-ratios send the complement of a positively oriented region $$(a,b,c)$$ to the half-hemisphere $$(1,0,\infty)$$ on $${\mathbf{CP}}^1$$. So take $$(i, -1, 1)\to (1,0,\infty)$$:

This is the cross-ratio \begin{align*} f(z) = {z+1\over z-1} {i-1\over i+1} .\end{align*}

The image of $${\mathbb{D}}^c \cap{\mathbb{H}}$$ is the 1st quadrant $${\mathbb{H}}\cap\Re(z) > 0$$. Send this to $${\mathbb{H}}$$ with $$z\mapsto z^2$$.

## Discs and Planes

### $${\mathbb{H}}\to{\mathbb{D}}$$#complex/exercise/completed

Find a fractional linear transformation $$T$$ which maps $${\mathbb{H}}$$ to $${\mathbb{D}}$$, and explicitly describe the image of the first quadrant under $$T$$.

Unclear to me how to motivate this formula, but choose $$f(z) = {z-i\over z+i}$$. Note that

• $$f(-1) = i$$
• $$f(0) = -1$$
• $$f(1) = -i$$,

so $${\mathbf{R}}$$ oriented from $$-\infty\to\infty$$ is sent to $$S^1$$ oriented counterclockwise. Since this is conformal, it preserves handedness – noting that $${\mathbb{H}}$$ is on the left with respect to $${\mathbf{R}}$$, it gets mapped to the left of $$S^1$$ with its induced orientation, i.e. the interior of $${\mathbb{D}}$$. How to remember: $${\left\lvert {z-i} \right\rvert}<{\left\lvert {z+i} \right\rvert}$$ in $${\mathbb{H}}$$, since points are closer to $$i$$ than $$-i$$.

The image of the first quadrant: the claim is that this is $${\mathbb{D}}\cap Q_{34}$$. Note that parameterizing seems hard! The naive idea would be to check the image of horizontal lines $$t + ci$$ for $$c$$ fixed heights and $$t\in (0, \infty)$$ the parameterization. Instead consider handedness and where sub-regions go:

Noting that $$Q_1$$ is the bigon enclosed by $$0, \infty$$, this maps to a bigon spanned by $$-1, 1$$. By handedness, since $$Q_1$$ is to the left of $${\mathbf{R}}$$, it gets mapped to the left of the image of $${\mathbf{R}}_{>0}$$, which is the lower half of the circle.

### $${\mathbb{H}}\to{\mathbb{D}}$$, cross-ratio #complex/exercise/completed

Find a conformal map $${\mathbb{H}}\to {\mathbb{D}}$$ using cross-ratios.

Idea: rotate the upper hemisphere $$({\mathbb{H}})$$ of $${\mathbf{CP}}^1$$ to make the equator $${{\partial}}{\mathbb{D}}$$, “zoom” by placing $$i$$ at the center so $$0\to i\mapsto -1\to 0$$ and $$i\to \infty\mapsto 0\to 1$$. Accomplish this by sending

• $$\infty\to 1$$
• $$i\to 0$$
• $$-i\to \infty$$

Use the cross-ratio \begin{align*} R(z) \coloneqq(z, \infty, i, -i) = {z-i \over z-(-i)} {\infty - (-i) \over \infty - i} = {z-i\over z+i} .\end{align*}

Checking that this works:

• If $$z\in {\mathbf{R}}$$ then $${\left\lvert {z-i} \right\rvert} = {\left\lvert {z+i} \right\rvert}$$ so $${\left\lvert {F(z)} \right\rvert} = 1$$.
• If $$z\in {\mathbb{H}}$$ then $${\left\lvert {z-i} \right\rvert}\leq {\left\lvert {z+i} \right\rvert}$$ so $${\left\lvert {F(z)} \right\rvert}< 1$$.

### $${\mathbb{D}}\to{\mathbb{H}}$$#complex/exercise/completed

Find a conformal map from $${\mathbb{D}}$$ to $${\mathbb{H}}$$.

Note that the standard Cayley map $$f(z)\coloneqq{z-i\over z+i}$$ sends $${\mathbb{H}}\to {\mathbb{D}}$$. Why this is true: $${\left\lvert {f(z)} \right\rvert} < 1$$, since $${\left\lvert {z-i} \right\rvert} < {\left\lvert {z+i} \right\rvert}$$ for $$z\in {\mathbb{H}}$$. Finding an explicit inverse: \begin{align*} w &= {z-i\over z+i} \\ \implies w(z+i) - (z-i) &= 0 \\ \implies z &= -i {w+1\over w-1} ,\end{align*} which is the desired map. Why the image is in $${\mathbb{H}}$$: it suffices to show that $$\Im(f(z)) > 0$$ for all $$z\in {\mathbb{D}}$$. Write $$z = x+iy$$ and note that $$\Im(iz) = \Re(z)$$, then \begin{align*} \Im(f(z)) &= \Re\qty{1-z\over 1+z} \\ &= \Re\qty{1-x-iy \over 1+x+iy} \\ &= \Re\qty{1-x^2-y^2 - i2y \over 1+x^2 + y^2} \\ &= {1-(x^2+y^2) \over 1+(x^2+y^2) } \\ &> 0 ,\end{align*} since $$x^2+y^2<1$$ for $$x+iy \in {\mathbb{D}}$$.

### Upper half-disc to $${\mathbb{D}}$$#complex/exercise/completed

Find a conformal map from $$\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}{\left\lvert {z} \right\rvert} < 1, \Im(z) > 0}\right\} = {\mathbb{D}}\cap{\mathbb{H}}$$ to $${\mathbb{D}}$$.

Note that $$z\mapsto z^2$$ doesn’t actually work, because the image is $${\mathbb{D}}\setminus{\mathbf{R}}_{\geq 0}$$ and has a slit deleted. Instead compose:

• $$z\mapsto i{z-1\over z+1}$$, which maps $${\mathbb{D}}\to {\mathbb{H}}$$ and restricts to map $${\mathbb{D}}\cap{\mathbb{H}}\to Q_1$$.
• $$z\mapsto z^2$$, which maps $$Q_1\to {\mathbb{H}}$$
• $$z\mapsto {z-i\over z+i}$$ which maps $${\mathbb{H}}\to {\mathbb{D}}$$.

In parts:

• Use $$z\mapsto z^{-1}$$ to send $${\mathbb{H}}\cap{\mathbb{D}}$$ to $$Q_{34} \cap{\mathbb{D}}^c$$.
• Use $$z\mapsto -z$$ to map this to $${\mathbb{H}}\cap{\mathbb{D}}^c$$
• Use $$z\mapsto {1\over 2}(z+z^{-1})$$ to map $${\mathbb{H}}\cap{\mathbb{D}}^c$$ to $${\mathbb{H}}$$
• Then use the Cayley map $$z\mapsto {z-i\over z+i}$$ to map $${\mathbb{H}}\to {\mathbb{D}}$$.

### Upper half-disc to $${\mathbb{H}}$$#complex/exercise/completed

Find a conformal map from the upper half-disc to the upper half-plane.

• $$z\mapsto {1+z\over 1-z}$$ is a standard map $${\mathbb{D}}\to Q_{14}$$ which restricts to $${\mathbb{D}}\cap{\mathbb{H}}\to Q_1$$
• $$z\mapsto z^2$$ unwraps $$Q_1\to {\mathbb{H}}$$.

### $${\mathbb{D}}^c \cap{\mathbb{H}}\to{\mathbb{H}}$$#complex/exercise/completed

Find a conformal map $${\mathbb{D}}^c \cap{\mathbb{H}}\to {\mathbb{H}}$$.

Claim: the map $$f(z) \coloneqq z+z^{-1}$$ works. Consider the images of circles $$\gamma_r(t) \coloneqq rei^{t}$$ where $$t\in [-\pi, \pi]$$. For $$r=1$$, \begin{align*} f(\gamma_1(t)) = e^{it} + e^{-it} = 2\cos(t) ,\end{align*} which sweeps out $$[-2, 2]$$ twice. For arbitrary $$r$$, \begin{align*} f(\gamma_r(t)) = re^{it} + r^{-1}e^{-it} = (r+r^{-1})\cos(t) +i(r-r^{-1})\sin(t) ,\end{align*} which sweeps out an ellipse with horizontal radius $$r+r^{-1}$$ and vertical radius $$r-r^{-1}$$. For $$1<r<\infty$$, these sweep out all of $${\mathbf{C}}\setminus{\mathbb{D}}$$. Restricting $$t\in [0, \pi]$$, the $$\gamma_r(t)$$ are top halves of circles which cover all of $${\mathbb{H}}\setminus{\mathbb{D}}$$, and the images $$f(\gamma_r(t))$$ are top halves of ellipses which sweep out all of $${\mathbb{H}}$$. This includes points inside of $${\mathbb{D}}\cap{\mathbb{H}}$$ – this is because for any $$t\in (0, \infty)$$, there is always a solution $$r$$ to $$t=r-r^{-1}$$: \begin{align*} t = r-r^{-1}\implies r^2-tr-1 \implies r = {t \pm \sqrt{t^2+4}\over 2} .\end{align*} So there is an image ellipse at that vertical height. Since every point $$z_0\in {\mathbb{H}}$$ is on an ellipse of some vertical height $$t$$, $${\mathbb{H}}$$ is in the image.

That this map is conformal: a computation shows $$f'(z) = 1 + {1\over r^2}$$, which vanishes only at $$z=\pm 1$$. Since these are not in the domain, the derivative is nonvanishing, making $$f$$ conformal.

## Slits

Find a conformal map \begin{align*} {\mathbb{D}}\setminus\left[ {1\over 2}, 1\right) \to {\mathbb{D}} .\end{align*}

The picture:

In steps:

• $$f_1$$: send $$1/2\to 0$$ and $$1\to 1$$ in order to lengthen the slit. Mobius transformations preserve lines, so take $$f_1(z) = {z-1/2 \over 1-1/2}$$. New domain: $${\mathbb{D}}\setminus[0, 1)$$.

• $$f_2$$: send $${\mathbb{D}}\to {\mathbb{H}}$$ and keep track of the slit. Use the standard inverse to the Cayley transform, $$f_2(z) = i{1-z\over 1+z}$$. New domain: $${\mathbb{H}}\setminus[i, i\infty)$$, noting that $${\mathbb{H}}$$ is the open half-plane.

• $$f_3$$: rotate the slit so it becomes a line segment from $$-1$$ to $$1$$ passing through $$\infty$$. Use $$f_3(z) = z^2$$. New domain: $${\mathbf{C}}\setminus\qty{(-\infty, 1] \cup[1, \infty) }$$

• $$f_4$$: move the slit off to infinity to be left with a ray, so send $$-1\to \infty$$ and $$0\to 0$$. Take $$f_4(z) = {z\over z+1}$$, the new domain is $${\mathbf{C}}\setminus[0, \infty)$$.

• $$f_5$$: fold it back. Branch cut log along $$[0, \infty)$$ to define $$f_5(z) = z^{1\over 2}$$, so the new domain is $${\mathbb{H}}$$.

• $$f_6$$: apply the standard Cayley transform $$f_6(z) = {i-z\over i+z}$$

### 8 #complex/exercise/work

Let $$D$$ be the region obtained by deleting the real interval $$[0, 1)$$ from $${\mathbb{D}}$$; find a conformal map from $$D$$ to $${\mathbb{D}}$$.

### 9 #complex/exercise/work

Find a conformal map from $${\mathbf{C}}\setminus\left\{{x\in {\mathbf{R}}{~\mathrel{\Big\vert}~}x\leq 0}\right\}$$ to $${\mathbb{D}}$$.

### 10 #complex/exercise/work

Find a conformal map from $${\mathbf{C}}\setminus\left\{{x\in {\mathbf{R}}{~\mathrel{\Big\vert}~}x\geq 1}\right\}$$ to $${\mathbb{D}}$$.

## 11 #complex/exercise/work

Find a bijective conformal map from $$G$$ to $${\mathbb{H}}$$, where \begin{align*} G \coloneqq\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}{\left\lvert {z-1} \right\rvert} < \sqrt 2,\, {\left\lvert {z+1} \right\rvert} < \sqrt 2}\right\} \setminus [0, i) .\end{align*}

## Strips

### Horizontal strip to $${\mathbb{H}}$$#complex/exercise/completed

Find a conformal map from the strip $$\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}0 < \Im(z) < 1}\right\}$$ to $${\mathbb{H}}$$.

In steps:

• Dilate by $$z\mapsto \pi z$$ to get $$0<\Im(z) < \pi$$.
• Exponentiate by $$z\mapsto e^z$$ to get $${\mathbb{H}}$$.

Why $$e^z$$ works: apply $$\operatorname{Log}$$ to $${\mathbb{H}}$$, use polar coordinates to write $$w=re^{i\theta}$$ with $$0<\theta<\pi$$ and note \begin{align*} \operatorname{Log}(w) = \ln{\left\lvert {w} \right\rvert} +i\operatorname{Arg}(w) = \ln(r) + i\theta ,\end{align*} and noting that the image of $$\ln({-})$$ is all of $${\mathbf{R}}$$.

## Lunes

### Intersection of circles

Find a conformal map $$L\to {\mathbb{D}}$$ where \begin{align*} L\coloneqq\left\{{{\left\lvert {z - i } \right\rvert} < \sqrt 2}\right\} \cap\left\{{{\left\lvert {z+i} \right\rvert} < \sqrt 2}\right\} ,\end{align*} i.e. a lune with vertices $$-1$$ and $$1$$.

The key insight: for lunes, map the corners to $$0$$ and $$\infty$$; this yields a sector. Here we want $$-1\mapsto 0$$ and $$1\mapsto \infty$$, so $$f(z) = {z+1\over z-1}$$ gets things started.

In steps:

• $$z\mapsto {1+z\over 1-z}$$ sends the lune to the sector $$\operatorname{Arg}(z) \in (-\pi/4, \pi/4)$$, since it fixes and must be symmetric about the real axis, and preserves the right angle between the circles at $$z=-1$$.
• $$z\mapsto e^{i\pi/4}z$$ to rotate this into $$Q_1$$.
• $$z\mapsto z^2$$ to dilate into $${\mathbb{H}}$$.
• $$z\mapsto {z-i\over z+i}$$ the standard Cayley map $${\mathbb{H}}\to{\mathbb{D}}$$.

### Lune with one intersection point

Find a conformal map: \begin{align*} {\mathbb{D}}\setminus\left\{{{\left\lvert {z - {1\over 2}} \right\rvert} = {1\over 2} }\right\} \to {\mathbb{D}} .\end{align*}

The picture:

• Key insight: send the point of tangency to $$\infty$$ to get parallel lines. Send $$z\mapsto {1+z \over 1-z}$$ to send $$-1\to 0, 0\to 1, 1\to \infty$$ to get a vertical strip.

• Rotate and dilate: $$z\mapsto i\pi z$$ to get a strip $${\mathbf{R}}\times i(0, \pi)$$.

• Standard exponential $$z\mapsto e^{iz}$$ sends this to $${\mathbb{H}}$$.

• Standard Cayley $$z\mapsto {z-i \over z+i}$$ sends this to $${\mathbb{D}}$$.

### 4 #complex/exercise/work

Find a conformal map from $$\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}{\left\lvert {z-i} \right\rvert} > 1,\, \Re(z) > 0}\right\}$$ to $${\mathbb{H}}$$.

### 5 #complex/exercise/work

Find a conformal map from $$\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}{\left\lvert {z} \right\rvert} < 1,\, {\left\lvert {z - {1\over 2}} \right\rvert} > {1\over 2} }\right\}$$ to $${\mathbb{D}}$$.

### 6 #complex/exercise/work

Find a conformal map from $$\left\{{{\left\lvert {z-1} \right\rvert} < 2}\right\} \cap\left\{{{\left\lvert {z+1} \right\rvert} < 2}\right\}$$ to $${\mathbb{H}}$$.

### 13 #complex/exercise/work

Find a conformal map from $$D = \{z :\ |z| < 1,\ |z - 1/2| > 1/2\}$$ to the unit disk $$\Delta=\{z: \ |z|<1\}$$.

## Sectors

Find a conformal map from the sector $$\left\{{\operatorname{Arg}(z) \in (0, \alpha)}\right\} \to {\mathbb{D}}$$.

The picture:

In steps:

• Map the sector to $${\mathbb{H}}$$ using $$z\mapsto z^{\pi/\alpha}$$, choosing a branch cut for $$\operatorname{Log}$$ along $${\mathbf{R}}_{\leq 0}$$.
• Map $${\mathbb{H}}\to {\mathbb{D}}$$ using the standard $$z\mapsto {z-i\over z+i}$$.

## Joukowski-Type Regions

Map $${\mathbf{C}}\setminus[-1, 1]$$ to $${\mathbb{D}}$$.

In steps:

• Send $$-1\to 0$$ and $$1\to \infty$$ with $$z\mapsto {z+1\over z-1}$$. Checking that $$f(0) = -1$$, this yields $${\mathbf{C}}\setminus{\mathbf{R}}_{\leq 0}$$.

• Unwrap with $$z\mapsto \sqrt{z}$$ to obtain the right half-plane $$-\pi/2<\operatorname{Arg}(z) < \pi/2$$.

• Apply the rotated Cayley map $$z\mapsto {z-1\over z+1}$$ to map this to $${\mathbb{D}}$$.

Note that the composition is \begin{align*} {\sqrt{z+1\over z-1} -1 \over \sqrt{z+1\over z-1} +1} &= { \qty{ \sqrt{z+1} - \sqrt{z-1}}^2 \over (z+1)-(z-1) } \\ &= z - \sqrt{z^2-1} ,\end{align*} which has inverse $$z\mapsto {1\over 2}\qty{z+{1\over z}}$$.

## Misc

Find a conformal map that sends $$i{\mathbf{R}}$$ to $${\left\lvert {z-{1\over 2}} \right\rvert} = {1\over 2}$$.

• The double angle formulas: \begin{align*} \sin(2t) = {2\tan(t) \over 1+\tan^2(t)} && \cos(2t) = {1-\tan^2(2t) \over 1+\tan^2(t)} .\end{align*}

• Parameterize a line by $$x=\tan(t)$$ for $$t\in (-\pi/2, \pi/2)$$

• The methods in SS Theorem 1.2: particularly the boundary behavior of $$F(z) \coloneqq{i-z\over i+z}$$, where $$F({\mathbf{R}}) = \left\{{\cos(2t)+i\sin(2t) = e^{2it} {~\mathrel{\Big\vert}~}t\in (-\pi/2, \pi/2)}\right\}$$.

• Nyquist plots: a common applied problem, essentially computing the image $$F(i{\mathbf{R}})$$.

Idea: need the line $$-i\infty\to 0 \to i\infty$$ to get mapped to a circle $$0\to 1\to 0$$:

Take the cross ratio $$R(z) = (z, 0, \infty, -1)$$ to send

• $$0\to 1$$
• $$\infty\to 0$$
• $$-1\to \infty$$

This yields \begin{align*} R(z) = {z-\infty \over z+1}{0+1\over 0-\infty} = {1\over z+1} .\end{align*} Some deductions:

• $$R$$ preserves circles, so $${\mathbf{R}}\mapsto {\mathbf{R}}$$.
• The segment $$0\to\infty\mapsto 1\to 0$$
• The ray $$-1\to 0 \to \infty \mapsto \infty\to 1\to 0$$
• The ray $$-\infty \to -1 \mapsto 0\to -\infty$$
• $$R$$ preserves angles, to at $$w=0$$, the image of $$i{\mathbf{R}}$$ must be orthogonal to $${\mathbf{R}}$$, so it’s either a circle or $$i{\mathbf{R}}$$ itself.
• Check $$-i\infty \to 0 \to -\infty\mapsto 0\to 1\to 0$$, so the image is a circle.
• Parameterize $$i{\mathbf{R}}= \left\{{it{~\mathrel{\Big\vert}~}t\in {\mathbf{R}}}\right\}$$, then compute the image \begin{align*} f(i{\mathbf{R}}) &= \left\{{{1\over 1+it}}\right\} \\ &= \left\{{1+t\over 1+t^2}\right\} \\ &= \left\{{{1\over 1+t^2} + i {t\over 1+t^2}}\right\} \\ &= \left\{{{1\over 2} \qty{ 1 + {1-t^2\over 1+t^2} } + i{1\over 2}\qty{2t\over 1+t^2}}\right\} \\ &= \left\{{{1\over 2}\qty{1 + \cos(2t) + i\sin(2t)}}\right\} \\ &= \left\{{{1\over 2} + {1\over 2}e^{2it} }\right\} ,\end{align*} which is a circle of radius $$1/2$$ about $$1/2$$.

Conclusion:

• $$i{\mathbf{R}}\to \left\{{{\left\lvert {z-{1\over 2}} \right\rvert} = {1\over 2} }\right\}$$ by $$z\to {1\over 1+z}$$.
• The reverse map: $$w\mapsto {1-w\over w}$$.

Map $${\mathbb{D}}^c \cap{\mathbb{H}}$$ to $${\mathbb{H}}$$, sending

• $$-1\to -1$$
• $$i\to 0$$
• $$1\to 1$$

Compose to get: \begin{align*} {1\over z}{z + z^{-1}} .\end{align*}

### 7 #complex/exercise/work

Let $$\Omega$$ be the region inside the unit circle $${\left\lvert {z} \right\rvert} = 1$$ and outside the circle $${\left\lvert {z-{1\over 4}} \right\rvert} = {1\over 4}$$.

Find an injective conformal map from $$\Omega$$ onto some annulus $$\left\{{r < {\left\lvert {z} \right\rvert} < 1}\right\}$$ for constant $$r$$.

### 12 #complex/exercise/work

Prove that TFAE for a Möbius transformation $$T$$ given by $$T(z) = {az + b \over cz + d}$$:

• $$T$$ maps $${\mathbf{R}}\cup\left\{{\infty}\right\}$$ to itself.
• It is possible to choose $$a,b,c,d$$ to be real numbers.
• $$\overline{T(z)} = T(\overline{z})$$ for every $$z\in {\mathbf{CP}}^1$$.
• There exist $$\alpha\in {\mathbf{R}}, \beta \in {\mathbf{C}}\setminus {\mathbf{R}}$$ such that $$T(\alpha) = \alpha$$ and $$T(\overline{\beta}) = \overline{T(\beta)}$$.
#complex/exercise/completed #complex/exercise/work