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Apr 29, 2023 at 2:19 comment added Alexandre Eremenko Very simple. The map $f$ has finitely many zeros and poles in $C$. Let $g$ be the rational function with the same zeros and poles in $C$ (counting multiplicity). Then $h=f/g$ has no zeros or poles in $C$. If $h(\infty)$ is infinite, consider $g/f$ instead. A bounded holomorphic function on the sphere must be constant. (Both the Maximum Principle and Liouville theorem which can be used to make this conclusion have simple algebraic proofs).
Apr 28, 2023 at 16:42 comment added Vít Tuček How do you prove, using only essentially elementary algebra, that any holomorphic map from sphere to itself is meromorphic? :)
Apr 28, 2023 at 12:29 history edited Alexandre Eremenko CC BY-SA 4.0
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Apr 28, 2023 at 12:18 history answered Alexandre Eremenko CC BY-SA 4.0