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Prove the existence of a conformal map from a double connected domain, in extended complex plane, where none of the components is singleton, onto a domain whose boundary consists of two analytic Jordan Curve.

It's a homework exercise that I find really hard. I know I should use the Riemann Mapping Theorem. But I don't know how.

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If $D$ is a double-connected domain in the extended complex plane $\hat{\Bbb C}$, let $A$ and $B$ be the connected components of $\hat{\Bbb C} \setminus D$.

Then $D \cup A$ is simply-connected (and its complement is not a single point), according to the Riemann mapping theorem there is a conformal map $f$ from $D \cup A$ to the unit disk $\Bbb D$.

$f(A)$ is connected, so the Riemann mapping theorem can be applied again to $\hat{\Bbb C} \setminus f(A)$.

The composition of these two mappings (restricted to $D$) has the desired property.

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  • $\begingroup$ I didn't get it. How do we end up with a domain that it's boundary consists of two Jordan Curves? If you restrict f to D then D is mapped onto {Unit Disk}\f(A)? ( If correct) g maps from {Unit Disk}\f(A) to where? $\endgroup$ – Mathitis Jan 7 '16 at 12:48
  • $\begingroup$ @Mathitis: $g$ maps $\hat{\Bbb C} \setminus f(A)$ to the unit disk. The restriction of $g$ to $\Bbb D \setminus f(A)$ maps to a subset of the unit disk, bounded by two analytic Jordan curves (one if them is the circle $|z|=1$). $\endgroup$ – Martin R Jan 7 '16 at 12:56
  • $\begingroup$ I can see why it's mapped to a subset of {Unit Disk}, but can't see why this subset is like an annulus $\endgroup$ – Mathitis Jan 7 '16 at 13:18
  • $\begingroup$ @Mathitis: $g(\{|z|=1\})$ is an analytic Jordan curve and that is one component of the boundary of the image. $\endgroup$ – Martin R Jan 7 '16 at 13:24
  • $\begingroup$ And the other component is the boundary of f(A)? I think I got it know, thank you very much for your time. $\endgroup$ – Mathitis Jan 7 '16 at 13:42

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