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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.)2012-12-23
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    You can use the Casorati-Weierstrass theorem plus open mapping theorem (or even easier, the great Picard theorem) to rule out an essential singularity.2012-12-23
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    @RobertIsrael, How do I rule out pole? Please help. I am lost totally.2013-01-08

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