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Does there exist a conformal bijection/Mobius transformation from the open unit disk to the whole complex plane?

Does there exist a conformal bijection/Mobius transformation from the annulus $\{z\in \mathbb{C}\mid 1<|z|<2\}$ to $\mathbb{C}\setminus \overline{D(0,1)}$, the whole complex plane removing the closed unit disk?

I guess there does not exist such Mobius transformations. But I do not know how to prove... Such homeomorphisms really exist. But under the extra condition "conformal map", I do not know how to solve... Thank you a lot.

Shiquan
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2 Answers2

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Does there exist a conformal bijection/Mobius transformation from the open unit disk to the whole complex plane?

No. The inverse of such a map would be a bounded function holomorphic in $\mathbb C$. Such a function must be a constant by Liouville's theorem.

Does there exist a conformal bijection/Mobius transformation from the annulus to the whole complex plane removing the closed unit disk?

No. The plane minus a disk is biholomorphic to punctured unit disk, via $z\mapsto 1/z$. If we had a holomorphic bijection from $\{0<|z|<1\}$ to $\{1<|z|<2\}$, it would extend continuously to $z=0$ (removable singularity theorem). On the other hand, topological considerations require the cluster set of $f$ at $z=0$ to be a boundary component of $\{1<|z|<2\}$, and neither of those is a single point.

Also, see Non-existence of a bijective analytic function between annulus and punctured disk.

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user127096
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The Riemann mapping theorem specifically excludes the case of the whole complex plane. Investigating where that is important in the proofs might be useful in disprooving the existence of the first conformal map.

Both annulus and plane minus disk are open subsets of $\mathbb C$, but not simply connected. So again Riemann doesn't apply, although in this case I doubt whether that can be used as a strong argument against the existence of a conformal map. When I first wrote this answer, I had missed the simply connected precondition, perhaps I wouldn't have posted otherwise since I don't know a more detailed answer here.

A Möbius transformation can always be extended to $\mathbb{CP}^2$ where it is a bijection. Both boundaries of the annulus are circles, whereas the point at infinity is a one-point boundary of the complex plane, as embedded into $\mathbb{CP}^2$. So there can be no bijection between the boundaries components, which implies that there can be no Möbius transformation which satisfies your requirements.

MvG
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