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I want to show that if $f:\mathbb{C} \to \mathbb{C}$ is a holomorphic automorphism then $\vert f(z) \vert \to \infty$ as $\vert z \vert \to \infty$.

I need this result to show that $f$ extends uniquely to a biholomorphim of the Riemann sphere $\mathbb{C}_{\infty}$ so that $f$ is an affine map.

I would appreciate any hints or references. Thank you.

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    $\begingroup$ My first thought is to use Jordan Curve Theorem sequentially applied to larger and larger circles centered at 0 and their images. The bounded set must correspond to the bounded set. Might be a bit overkill. $\endgroup$ – Robert Wolfe Feb 23 '18 at 18:59
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    $\begingroup$ Every homeomorphism extends continuously to the one-point compactifications, which translates to your requirement. Am I missing something? $\endgroup$ – Aloizio Macedo Feb 23 '18 at 19:24
  • $\begingroup$ @AloizioMacedo You are right. Thank you! $\endgroup$ – GouldBach Feb 24 '18 at 11:58
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$f$ is a homeomorphism. Thus, it extends continuously to the one-point compactifications, which is a translation of what you want to arrive at.

To purge away the general topology terminology if desired, note that "$|f(z)| \to \infty$ as $|z| \to \infty$" is equivalent to "for every compact $K$, there exists a compact $L$ such that if $x \notin L$, then $f(x) \notin K$".

Thus, pick a compact $K$. Since $f$ is a homeomorphism, $f^{-1}(K)$ is compact. Letting $L:=f^{-1}(K)$ we have the result.

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Any holomorphic function $f:\Bbb C \to \Bbb C$ has an isolated singularity at $\infty$, so there are three possibilities:

  1. $f$ has a removable singularity at $\infty$. Then $f$ may be extended to a holomorphic map $f: \Bbb C_\infty \to \Bbb C$. But every such map is constant (you can also use Liouville for this). So $f$ can't be an automorphism.
  2. $f$ has an essential singularity at $\infty$. Then by Casorati-Weierstraß, the image $f(\{ z \in \Bbb C \mid |z| > 1 \})$ is dense in $\Bbb C$, so it intersects every non-empty open subset of $\Bbb C$. By the open mapping theorem $f(\{ z \in \Bbb C \mid |z| < 1 \})$ is open in $\Bbb C$, so $f(\{ z \in \Bbb C \mid |z| > 1 \}) \cap f(\{ z \in \Bbb C \mid |z| < 1 \}) \neq \varnothing$. But if we take $x \in f(\{ z \in \Bbb C \mid |z| > 1 \}) \cap f(\{ z \in \Bbb C \mid |z| < 1 \})$, then $x$ has both a preimage $z$ under $f$ with $|z|>1$ and a preimage with $|z| < 1$, so $f$ is not injective.
  3. $f$ has a pole at $\infty$. This is the only case that remains and that $\lim_{|z| \to \infty}\vert f(z) \vert \to \infty$ follows from a well-known characterization of poles.
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