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$\textbf{Exercise}$ Find a conformal mapping which maps the domain $D$ onto the open unit disc, where $D$ is the intersection of $\vert z \vert <1 $ and $\vert z-1 \vert <1 $.

I knew that Mobius Transformation is a self-conformal map on the open unit disc. However, I don't know how to find a conformal map when two regions intersect...

Any help is appreciated...

Thank you!

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Note that $D$ has two corners that somehow have to be straightened by the map. Construct your map in several steps:

  • The two circles intersect in two points $p$, $q$. Find a Moebius transformation $T$ that maps these two points to $0$ and $\infty$. Such a $T$ will map the two circles to a pair of lines through $0$, hence $D$ to a wedge. Determine the opening angle of this wedge.
  • A second map of the form $w\mapsto w^\lambda$ with a suitable $\lambda>0$ will map the wedge to a half plane.
  • Map this half plane to the unit disc by a suitable Moebius transformation.
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  • $\begingroup$ I found two points $1/2 + i \sqrt {3}/2 , 1/2 - i\sqrt{3}/2 $ . Thus, If T be $(az+b)/(cz+d)$, $-b/a = 1/2 - i \sqrt {3} /2 , -d/c = 1/2 + i \sqrt {3} /2 $. Then, $T(1) \neq 0 $.... How to find a Moebius transformation??.. $\endgroup$ – w.sdka Jun 19 '18 at 23:46
  • $\begingroup$ I thought the opening angle of this wedge is $ 2 / 3 \pi $ $\endgroup$ – w.sdka Jun 19 '18 at 23:53
  • $\begingroup$ Ah.... Sorry... I found $ T = \frac{(1-\sqrt{3} i) +1+\sqrt{3}i}{(1+\sqrt{3}i)z+1-\sqrt{3}i}$. $\lambda = 3/2 $ and we knew $F(z)=\frac{i-z}{i+z}$ map from the upper half plane to the unit disk. I got it! Thank you! $\endgroup$ – w.sdka Jun 20 '18 at 0:06

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