Exercises: Conformal Maps

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

Idea: all cross-ratios send the complement of a positively oriented region $(a,b,c)$ to the half-hemisphere $(1,0,\infty)$ on ${\mathbb{CP}}^1$. 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 $z\mapsto z^2$.

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 ${\mathbb{R}}$ 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 ${\mathbb{R}}$, 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 ${\mathbb{R}}$, it gets mapped to the left of the image of ${\mathbb{R}}_{>0}$, which is the lower half of the circle.

Idea: rotate the upper hemisphere $({\mathbb{H}})$ of ${\mathbb{CP}}^1$ 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 $z\in {\mathbb{R}}$ 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 ${\left\lvert {F(z)} \right\rvert}< 1$.

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 $x+iy \in {\mathbb{D}}$.

Note that $z\mapsto z^2$ doesn’t actually work, because the image is ${\mathbb{D}}\setminus{\mathbb{R}}_{\geq 0}$ 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 ${\mathbb{H}}\to {\mathbb{D}}$.

• $z\mapsto {1+z\over 1-z}$ 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 $Q_1\to {\mathbb{H}}$.

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 ${\mathbb{C}}\setminus{\mathbb{D}}$. 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.

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: ${\mathbb{C}}\setminus\qty{(-\infty, 1] \cup[1, \infty) }$

• $f_4$: 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 ${\mathbb{C}}\setminus[0, \infty)$.

• $f_5$: 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 $f_6(z) = {i-z\over i+z}$

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 ${\mathbb{R}}$.

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 ${\mathbb{H}}\to{\mathbb{D}}$.

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 ${\mathbb{R}}\times i(0, \pi)$.

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

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

The picture:

In steps:

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

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 ${\mathbb{C}}\setminus{\mathbb{R}}_{\leq 0}$.

• 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 $z\mapsto {1\over 2}\qty{z+{1\over z}}$.

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

Conclusion:

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

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