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How can I construct a conformal mapping from the union of three discs $$D = \{|z-i|<\sqrt 2\} \cup \{|z+i|<\sqrt 2\} \cup \{|z-\sqrt 2|<1\}$$ to the upper half plane?

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Can someone give me a hint which transformation to use? I am trying to solve this task already couple of hours, I tried using 1/z, mobius ... but I couldn't achieve any desirable results.

David Clyde
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Johnaaa11
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  • please stop editing this question unless you post a reason here. So far your edits seem to keep removing important information (the image) and also undoing my changes that made the title clearer. – David Clyde Dec 16 '22 at 13:20
  • @amWhy It looks to me like you rolled back to a version with lots of typos and unclear title and also it doesn't even show what region is being discussed. My edited version included the region in mathjax plus restored OP's original image of the region. (OP edited to remove the image after the question had been answered.) Overall, I don't understand why you reverted and I will re-revert later today unless you post some explanation - like maybe there's some very strange website malfunction happening?? – David Clyde Dec 29 '22 at 14:49
  • Your job is not to change it to the question you want to answer, the moment you answer it. – amWhy Dec 29 '22 at 14:53
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    @amWhy I didn't do that. I first answered a question that was not perfectly explained but at least quite comprehensible. I made a tiny edit at that time, just adding an appropriate tag. The OP accepted my answer and probably also upvoted my answer and then a few hours later they edited the question to make it incomprehensible (removed all reference to the region in question). I assumed that was a mistake and I fixed the image and also added the same information in mathjax alongside. – David Clyde Dec 29 '22 at 15:06
  • I answered the original question and OP accepted that answer. My edits since then have only been re-adding information that the OP removed for some reason. (I still mostly expect that they were doing it accidentally somehow; if they really don't want the question to be usable by future readers then they could have just deleted it.) – David Clyde Dec 29 '22 at 15:07

1 Answers1

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Let $D$ denote the starting domain.

The first step is pretty tricky! Denote the three starting circles as $C_1 = \{z : |z-i|=\sqrt 2\}$, $C_2 = \{z : |z+i|=\sqrt 2\}$, and $C_3 = \{z:|z-\sqrt 2|=1\}$. I want a Mobius transformation $T$ that satisfies:

  1. $T$ maps $C_3$ to the unit circle $|z|=1$
  2. $T(-1) = \infty$

I'll do this in two steps. First, apply $T_1(z) = z - \sqrt 2$ to shift the whole diagram sideways so that $C_3$ is mapped to $|z|=1$. Then, apply $T_2(z) = \frac{z-a}{1-\bar a z}$ with $a = \frac{-1}{1+\sqrt 2}$. Mobius transformations in this form always map the unit disc to itself (source) and we can check by hand that $T_2(-1-\sqrt 2) = \infty$. Now $T = T_2 \circ T_1$ is the desired Mobius transformation that fulfills both conditions (1) and (2).

Now let's see what $T$ does to our domain $D$. First, we note $T(\sqrt 2 + i) = e^{i \pi/4}$ and $T(\sqrt 2 - i) = e^{- i \pi / 4}$. The circles $C_1$ and $C_2$ must have mapped to lines because the images contain $T(-1) = \infty$. The original circles intersected at $z=-1$ but also at $z=1$, so both images will contain $T(1)$. It turns out $T(1) = 0$, so our image domain is $$T(D) = \left\{r e^{i \theta} : r > 1, -\frac \pi 4 < \theta < \frac \pi 4 \right\}$$

Phew! From here, the rest of the mapping problem is pretty straightforward. Let $f(z) = z^2$; then $$f(T(D)) = \left\{r e^{i \theta} : r > 1, -\frac \pi 2 < \theta < \frac \pi 2 \right\}$$ which is an important step because our boundary is now made up of only two circles/lines instead of 3. Next, let $S(z)$ be the Mobius transformation satisfying $S(i) = \infty$, $S(-i) = 0$, and $S(1) = 1$. That gives $S(z) = \frac{z+i}{i(z-i)}$ and our domain becomes a quadrant: $$S(f(T(D))) = \{x+iy:x>0,y<0\}$$

Finally, square the quadrant to get the lower half plane, and then negate that to get the upper half plane. We can do these steps using $g(z) = -z^2$. And we're done! The final conformal map is the composition $g \circ S \circ f \circ T$.

David Clyde
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