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I would like to know how one can find a conformal map that maps a given set onto another (not $\mathbb D\rightarrow\mathbb D$ or $\mathbb H\rightarrow \mathbb H$ since those are clear).

For example, if the question is to

  1. find a conformal map that maps the upper half of $\mathbb D$ to the second quadrant of the plane.
  2. find a conformal map that maps the upper half of $\mathbb D$ to $\mathbb H$

How should one proceed? I've heard one can fix 3 points and this determines the mapping but how to find to where they should be mapped

NB: $\mathbb D=D(0,1)$ the unit disk and $\mathbb H=\{z\in\mathbb C:Im(z)>0\}$

David Clyde
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PNM
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    I don't think there is a general method. In general, given any open set $U$ in the plane that is homeomorphic to the disk, there is a conformal map $f: \mathbb D \to U$ but unless $U$ is very special you won't be able to compute $f$. in practice, for such questions you need to have in mind a catalogue of such functions, involving Möbius maps, $\exp$, and $z \mapsto z^2$ for instance – Albert Dec 05 '22 at 10:13
  • Yes is see. But in the case of the two examples above which are quite specific, how would one proceed? – PNM Dec 05 '22 at 10:26
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    it would be nice to know why the question was closed... – PNM Dec 05 '22 at 12:08

1 Answers1

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Strategies

There is not a single general method for finding these mappings. Instead, we can learn several "building blocks" and then assemble these to construct most mappings we might want. The following tricks are enough to cover your example questions:

  1. Mapping from $\mathbb{H}$ to $\mathbb{D}$: $T(z) = \frac{z-i}{z+i}$.
  2. Mapping onto $\mathbb{D}$ from any region bounded between two rays with endpoints at 0: First apply a rotation if necessary so that the region becomes $\{r e^{i \theta} : 0 < r, 0 < \theta < 2 \pi \alpha\}$ for some real number $0 < \alpha <= 1$. Next, apply $f(z) = z^{1/2\alpha}$ which maps the region onto $\mathbb{H}$. Finally, use the previous trick to get $\mathbb{D}$.
  3. Mapping onto $\mathbb{D}$ from any region bounded between two circles (lines count as circles): Let the two circles' intersection points be $w_1$ and $w_2$. Choose a Mobius transformation $T$ such that $T(w_1) = 0$ and $T(w_2) = \infty$. $T(z) = \frac{z-w_1}{z-w_2}$ works. Now applying $T(z)$ gives a region bounded between two rays with endpoints at 0, so we can apply the method above.
  4. Mapping between two regions that both aren't $\mathbb{D}$: Use some combination of other tricks to come up with conformal bijections $f: \Omega_1 \to \mathbb{D}$ and $g: \Omega_2 \to \mathbb{D}$. Then $g^{-1} \circ f$ is a conformal mapping from $\Omega_1 \to \Omega_2$.

A couple of bonus tricks just for fun:

  1. Mapping a "strip" (region between two parallel lines) onto $\mathbb{D}$: First translate/rotate/scale so that the strip becomes $\{a + bi : a, b \in \mathbb{R}, 0 < b < \pi\}$. Next apply $f(z) = e^z$, ending up with $\mathbb{H}$. Finally, map $\mathbb{H} \to \mathbb{D}$ as above.
  2. Mapping a "half strip" onto $\mathbb{D}$: After translation/rotation/scaling, suppose the starting region is $\{a + bi: a < 0, 0 < b < \pi\}$. Apply $f(z) = e^z$, ending with the upper half disc. Now apply the method from trick (3) above to map from the upper half disc onto $\mathbb{D}$.

Solving your specific proposed examples

Upper half disc to 2nd quadrant of the plane: Using trick (4), we see it's sufficient to find mappings from each region to $\mathbb{D}$. Trick (3) shows how to map from upper half disc to $\mathbb{D}$, since the upper half disc is bounded between the unit circle and the line $\text{Im}(z) = 0$. Trick (2) shows how to map from the 2nd quadrant to $\mathbb{D}$.

Upper half disc to upper half plane: Use trick (3) in the same way as above.

David Clyde
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    Thank you for your answer. I still have one question. For (1), how can one guarantee that that the mapping $T^{-1}$ you propose does the trick? It works for $\mathbb D\rightarrow\mathbb H$ but how can one know that this works when restricted to the particular parts $\mathbb D\cap\mathbb H\rightarrow\mathbb H\cap{z\in\mathbb C:Re(z)<0}$ ? – PNM Dec 05 '22 at 11:55
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    I misread part of your question. Edited the answer and now it should be fully addressed. – David Clyde Dec 05 '22 at 12:32
  • Thank you so much. Have a great day – PNM Dec 05 '22 at 15:23