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if f is an analytical injective on a disc beside 0 then f has a pole or removeable point on z=0.

My try: there can't be an essential singularity, as otherwise, near the pole we would have all the values on the entire plane (but 1). Now it must be removeable or a simple pole, otherwise we can take the Laurent series, of this function, and there will be at least one value for which we get 2 different roots.

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    I think your question is answered here: https://math.stackexchange.com/q/53303/42969. – Martin R Jan 24 '21 at 10:22
  • @MartinR nope, I can see why this is not essential singularity, but why can't be 2nd order pole for example? –  Jan 24 '21 at 10:28
  • That is answered here: https://math.stackexchange.com/q/566422/42969. (But actually your question does not ask that explicitly.) – Martin R Jan 24 '21 at 10:30
  • Your question is unclear, it is mostly trivial to show that if $f$ is meromorphic then it must have either a simple pole or $f'$ must have a simple zero at $0$, the hard part is dealing with the essential singularity case that Martin answered to. – reuns Jan 24 '21 at 10:33

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