On my complex analysis prelim this morning I was asked to give a conformal map from the region $L=\{z\in\mathbb{C}:|z-i|<\sqrt{2},|z+i|<\sqrt{2}\}$, a lune with vertices at $-1$ and $1$ to the unit disc $\mathbb{D}=\{z:|z|<1\}$. I tried to send $L$ to the upper half plane by the Möbius transform sending $(-1,0,1)$ to $(0,i,\infty)$. Then I composed with the Cayley transformation to get to the unit disc.

My question is: does my first map do what I want it to(presuming I calculated it correctly)?

To be brief, does the Möbius transform which takes $(-1,0,1)$ to $(0,i,\infty)$ send $L$ to the upper half plane?

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    $\begingroup$ Your first map should send your "lune" to the quadrant $\{\operatorname{Im}{z} \gt \operatorname{Re}{z}\}$ (since the circles intersect orthogonally and the boundaries of the "lune" are sent to rays), so you need to square before you get a half-plane. $\endgroup$ – t.b. Aug 23 '11 at 18:06

Your map sending $(-1,0,1)$ to $(0,i,\infty)$ is given by $i = e^{i\pi/2}$ times the cross ratio $[z,0,-1,1] = \frac{z+1}{z - 1} : \frac{1}{-1}$, so it is $$\tilde{\phi}(z) = e^{i\pi/2}\cdot \frac{1+z}{1-z}.$$ It is easy to see geometrically (or by a direct calculation) that $\tilde\phi(L)$ is the quadrant $\{z:\,\operatorname{Im}{z} \gt |\operatorname{Re}{z}|\}$ since the circles $|z-i|=\sqrt{2}$ and $|z+i|=\sqrt{2}$ are sent to the lines $\{\operatorname{Im}{z} = \operatorname{Re}{z}\}$ and $\{\operatorname{Im}{z} = -\operatorname{Re}{z}\}$. Note that the angles at the vertices of the lune are equal to $\pi/2$.

It is more convenient to work with $\displaystyle\phi(z) = e^{i\pi/4}\cdot \frac{1+z}{1-z}$ which sends $L$ to the quadrant $\{z:\operatorname{Re}{z}, \, \operatorname{Im}{z}\gt 0\}$, then square to get to the upper half plane and apply the Cayley-transfom $\displaystyle\kappa(z) = \frac{z-i}{z+i}$ to get to the unit disk.

The solution to your problem then is $\kappa((\phi(z))^2)$, which you can compute yourself if needed.

The general procedure is succinctly explained in GEdgar's answer.

  • $\begingroup$ Do you mean the quadrant $\{z : \operatorname{Im} z > \lvert \operatorname{Re} z \rvert\}$? $\endgroup$ – Rahul Aug 23 '11 at 19:03
  • $\begingroup$ Oh, yes. Thanks. Hm... It's really not my day. It should be fixed now. $\endgroup$ – t.b. Aug 23 '11 at 19:03
  • $\begingroup$ Thanks for the explanation! $\endgroup$ – RHP Aug 24 '11 at 2:26

Just to put everyone's answers together and finish things off, $f:z\mapsto \frac{1+z}{1-z}$ sends $-1$ to $0$ and $1$ to $\infty$. Because $f:(-1,1)\to(0,+\infty)$, the sector is symmetric about $\mathbb{R}^+$. Since $f$ is conformal, it preserves the right angle at which the circles cross, so the sector has a right angle. Therefore, $g:z\mapsto z^2$ maps the sector to the right half-plane. Then, $h:z\mapsto\frac{z-1}{z+1}$ maps the right half-plane to the unit disk. Thus, $$ h\circ g\circ f:z\mapsto\frac{2z}{1+z^2} $$ should map the given lune to the unit disk.

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    $\begingroup$ Just to clarify: $f$ sends $(-1,1)$, a segment of $\mathbb{R}$, to $(0,+\infty)$, another segment of $\mathbb{R}$. $\endgroup$ – robjohn Aug 24 '11 at 14:07
  • $\begingroup$ Ah, that looks simpler than what I suggested. I didn't actually do the computation, but I trust you :) $\endgroup$ – t.b. Aug 24 '11 at 14:57
  • $\begingroup$ @Theo: it is a nice map in that it maps $[-1,1]$ to itself, fixing the endpoints. Of course, it then has to map the top and bottom of the lune to the top and bottom of the disk. $\endgroup$ – robjohn Aug 24 '11 at 17:49
  • $\begingroup$ I saw that when I finally made the short computation and now I'm even more convinced by the superiority of your solution than before. Thanks for contributing it. $\endgroup$ – t.b. Aug 24 '11 at 20:52

Theo is right. For a general lune: First apply a linear fractional transformation mapping the two corners (vertices, horns, whatever you call them) to $0$ and $\infty$. The result is a sector. Then apply a power to get a half-plane. Then appy a linear fractional transformation to get a disk.


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