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Find a complex function that maps the region $$D =\left\{|z|<1, \left|z- \frac{1}{2}\right|>\frac {1}{2}\right\}$$ conformally on to the upper half plane.

Can somebody help me to find the conformal map of the above domain with step by step explanation of the map? I think this is something about Lunar domain and I need to be able to map the first intersection of the domain in to the vertical strip passing through $\frac {1}{2}$ and $1$, and then map the vertical strip to the unit disk by exponential. Am I thinking this right? Please help.

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3 Answers 3

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Hint: Moving away the exceptional point $z=1$ to infinity seems a good idea. What does the region look like after applying $z\mapsto \frac1{z-1}$?

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$$z\mapsto\exp\left(\pi i\frac{1+z}{1-z}\right)$$ This is the composition of the following functions:

  1. $z\mapsto 1-z$
  2. $z\mapsto \frac{1}{z}$
  3. $z\mapsto z-\frac{1}{2}$
  4. $z\mapsto 2\pi i \,z$
  5. $z\mapsto \exp(z)$
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$$L_1(z)=\frac{1}{z-1}$$ will send the domain into the region between the vertical lines $x=-1$ and $x=-\frac{1}{2}$. Then you translate with $$L_2(z)=z+1$$ Now you have the region between $x=0$ and $x=\frac{1}{2}$. Next, you rotate $90$ degrees with $$ L_3(z)=iz$$

Finally, you take $$L_4(z)=e^{4\pi z}$$ which sends the strip onto the uppper half plane.

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