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As stated in the title, what is the starting point in finding a conformal map between doubly-slit domain to the open unit disk? I know how to deal with a single-slit domains, but have trouble trying to modify them to work on doubly slit domains.

For instance, if $D = \Bbb{C}\setminus_{\{(-\infty, -1]\cup[1, \infty)\}}$, then $f(z) = \log(1-z^2)$ conformally maps $D$ onto $V$, where $V = \{z: |\Im(1-z^2)|<\pi\}$. But I find it hard to explicitly construct a conformal map from $V$ to the open unit disk so I can compose it with $f$ to obtain my desired map. So I am guessing there must be a different approach to this problem that I have not seen before.

Any hint is appreciated.

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If you think of the domain $D=\mathbb C \setminus {(-\infty,1]\cup[1,\infty)}$, this is not a doubly slit domain on the sphere, but it only has a single slit, passing through infinity (note that $\infty \notin D$). So you will have to move this slit to a slit from $0$ to $\infty$. This can be done with a Mobius transformation that preserves the real line, and maps $1$ to $0$ and $-1$ to $\infty$, say. It is easy to see that $(z-1)/(z+1)$ works.

Now you can unfold $\mathbb C \setminus [0,\infty)$ using $z^{1/2}$ (the branch that maps $-1$ to $i$). The image is the upper half plane, and it is easy now to map it to the unit disk with the Cayley transform $(z-i)/(z+i)$.

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  • $\begingroup$ That makes perfect sense. Thanks! $\endgroup$
    – dezdichado
    Commented May 8, 2016 at 21:30

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