I remember seeing this statement, I don't remember where (maybe in Lang's Complex). Is this true or do I have a faulty memory. It was always somewhere in the back of my mind but I never believed it. Is it true that there is a conformal map from the punctured unit disc onto the unit disc?
Conformal map from the punctured unit disc onto the unit disc?
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complex-analysis
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0Ok let me edit that. – 2011-07-12
2 Answers
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There is a surjective conformal map, though (in opposition to bijective, which is what Soarer has in mind in his answer). Just compose an holomorphic bijection from the disc to the upper half-plane with $z\mapsto e^{i z}$.
(I made a video showing the images under the map $z\mapsto\exp\frac{z-1}{z+1}$, which is a surjection from the unit disk to the punctured unit disk, of the circles centered at $0$. You can see in it how it manages to avoid the origin.) (The file will not last forever in that location... if someone can upload it to some site or another, it would be great!)
Later In any case, this is the Mathematica code I used:
Animate[ ParametricPlot[ With[{z = r Exp[I theta]}, Through[{Re, Im}[Exp[(z - 1)/(z + 1)]]] ], {theta, 0, 2 \[Pi]}, ImageSize -> Medium, PlotRange -> {{-1, 1}, {-1, 1}}, PlotPoints -> 1000, Ticks -> False ], {r, 0, 0.999, 0.001} ]
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0@J.M.: great! Can you upload it and add a link to my answer? – 2011-07-16
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No, because punctured disc is not simply connected.
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0I had in mind a bijective conformal map, which is just a biholomorphism. Sorry for the confusion. – 2011-07-13