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.
Best Answer
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.
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.