Let $\mathbb{H}$ denote the complex upper half plane (not including the real axis). Let $A$ be a bounded subset with $A = \mathbb{H} \cap \overline{A}$ and $\mathbb{H} \backslash A$ simply connected. Why does it follow from the Riemann mapping theorem that there are conformal maps $g: \mathbb{H} \backslash A \rightarrow \mathbb{H}$ with $|g(z)| \rightarrow \infty$ as $z \rightarrow \infty$?
Consequence of Riemann mapping theorem
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complex-analysis
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0I'm not sure why it should even matter for the problem at hand, where the closure is taken....... – 2011-09-25
1 Answers
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By the Riemann Mapping Theorem you can find a conformal map $f:\mathbb{H}\setminus A \to \mathbb{H}$. Assuming that $A$ is contained in the disk $|z| \le R$, you can extend $f$ by Schwarz reflection meromorphically to $|z| > R$, with a pole or removable singularity at $\infty$. If $f(\infty) = \infty$ for this extension, you are done (set $g = f$), otherwise $g(z) = \frac{1}{f(\infty) - f(z)}$ works.