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Let $H = \{z\in \mathbb{C}: Im(z) > 0$ and $|z-2i|>1\}$.

How can we find a conformal mapping from $H$ onto an annulus $r < |z| < 1$?

zhw.
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Rachel.Lee
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    You should have included some thoughts of your own on this. – zhw. Aug 08 '18 at 19:50
  • @zhw. I considered $f(z) = \frac{z-\beta}{z-\bar{\beta}}$($\beta$ is in the upper half plane) which is conformal from the upper half plane onto the unit disk. Since f sends the imaginary axis to the real line, we have $f(i)+f(3i)=0$ where $0$ is center of a circle and $f(3i) = r$. Then, find $\beta$ in terms of $r$. But I find $\beta$ implicitly. I would like to know whether it is right and if there's another way. – Rachel.Lee Aug 08 '18 at 20:30
  • @Rachel.L: Good. Usually, though, those comments should be placed in the body of the question. – Brian Tung Aug 08 '18 at 21:24

2 Answers2

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There is a Mobius transform $T$ that satisfies the conditions. The points $0$ and $\infty$ are conjugate wrt both circles bounding the annulus, therefore $T$ maps points that are conjugate wrt both the real axis and the circle $\mathcal C = \{z: |z - 2 i| = 1\}$ to $0$ and $\infty$. The only such points are $\pm i \sqrt 3$, and $T(z) = a(z \pm i \sqrt 3)/(z \mp i \sqrt 3)$.

$T$ maps the real axis to $\{z: |z| = |a|\}$ and maps $\mathcal C$ to $\{z: |z| = (2 \pm \sqrt 3) |a|\}$. The requirement $r < 1$ gives $r = 2 - \sqrt 3$, while $|a| = 1$ or $|a| = 2 - \sqrt 3$. There is no conformal mapping for other values of $r$, because the ratio of the radii is preserved.

Maxim
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One thought: A Moebius transformation will take this region to the open unit disc minus a smaller closed disc. Now you have the well known automorphisms of the unit disc to play with. They all keep the unit circle fixed, but one of them should map that smaller disc to a disc centered at the origin.

zhw.
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