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What kind of singularity must a univalent function have at $\infty$?

I do not know how to be rigorous for this problem. I would appreciate if someone can prove this rigorously?

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Starting point: $f$ is univalent in a neighborhood of $\infty$ if and only if $g(z)=f(1/z)$ is univalent in a neighborhood of $0$. (Reason: $z\mapsto 1/z$ is a bijection.) – user53153 Dec 23 '12 at 21:56
You can use the Casorati-Weierstrass theorem plus open mapping theorem (or even easier, the great Picard theorem) to rule out an essential singularity. – Robert Israel Dec 23 '12 at 22:13
@RobertIsrael, How do I rule out pole? Please help. I am lost totally. – Deepak Jan 8 '13 at 1:43

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