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I'm working on the conformal mappings of complex analysis. I can find conformal mappings from&onto simple domains but cannot find one from the abnormal domains.

Can any one give me a hint? Is there a rule that I can apply to find such mappings?

($D(z;r)$ means the open disk centered at $z$ with radius $r$.)

  • 1
    I just realized your question has a little problem after noticing the answer given by @pritam. Actually $D(0;1)\setminus D(1;1)$ is not conformal to $D(0;1)$, and my answer gives a conformal map from $D(0;1)\setminus \overline{D(1;1)}$ to $D(0;1)$.2012-12-14
  • 0
    I think the correct region should be the open region inside the unit disc and outside the circle centered at 1 with radius 1 i.e. $\{ |z|<1 \}$ and $\{|z-1|>1 \}$.2012-12-14

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