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This question already has an answer here:

Let $f$ be a meromorphic function on $\mathbb C$ such that $|f(z)|\ge|z|$ at each $z$ where $f$ is holomorphic. Then which of the following is/are true?

  1. The hypothesis are contradictory.

  2. Such an $f$ is entire.

  3. There is a unique $f$ satisfying the given conditions.

  4. There is an $A\in\mathbb C$ with $|A|\ge1$ such that $f(z)=Az$ for each $z\in\mathbb C.$

I think 2 and 4 are correct. But I can't proceed rigorously. Neither do the discussion https://math.stackexchange.com/questions/631244/f-be-a-meromorphic-function-on-mathbbc-ni-fz-ge-z-forall-z is complete.

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marked as duplicate by Henrik, John B, Servaes, colormegone, Daniel W. Farlow Jun 5 '16 at 0:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ well 1 and 3 are obviously wrong... (where do people come up with these weird multiple choice questions?) $\endgroup$ – Seth Mar 22 '14 at 0:32
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Note that $f$ has no zeros, except possibly at the origin. Define $$g(z) = \frac z {f(z)}$$

Since $g$ is bounded near $0$ (even if $f$ has a pole at zero, $g$ is still bounded), we can extend $g$ to be holomorphic in a neighborhood of zero. Anywhere else $f$ has a pole, $g$ has a zero; thus, $g$ is holomorphic everywhere, or entire.

Finally, $g$ is bounded on $\mathbb{C}$, so must be constant. The correct choices then follow immediately.

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