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Is there a direct proof, using the Riemann mapping theorem for the Jordan domain, than every doubly connected domain in the complex plane can be mapped conformaly onto a rounded annulus.

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    $\begingroup$ What is the rounded annulus? $\endgroup$ – Michael Albanese Feb 6 '17 at 22:04
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Yes, here is a sketch. Let $A$ be your doubly connected domain, wlog $0$ and $\infty$ are in different components of the complement. Consider the preimage of $A$ under $\exp(z)$. This is an unbounded simply connected domain $D$ with two boundary points at infinity. This domain will be periodic: $z\mapsto z+2\pi i$ will map it onto itself. Now using the Riemann mapping theorem, map $D$ by some function $g$ onto the vertical strip $0<\Re z<1$, so that the infinite points on the boundary of $D$ correspond to the infinite points on $\partial D$. By the uniqueness in the Riemann theorem, your map $g$ will satisfy $g(z+2\pi i)=g(z)+c$ with some constant $c\neq 0$. Multiply $g$ on $k=2\pi i/c$ to obtain a new function $g_1=kg$ which satisfies such an equation with $c=2\pi i$. Function $g_1$ will map $D$ on the strip $0<\Re z<k$. Now the function $\exp\circ g_1\circ\log$ will map $A$ onto the round ring $1<|z|<\exp(k)$.

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  • $\begingroup$ Yes, afterwards I realised that this indeed done by Golusin.Thanks $\endgroup$ – djole Feb 8 '17 at 16:12

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