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Clearly $f$ has zeroes everywhere $g$ does, and since $|f/g| < 1$ in a deleted neighborhood of each of $g$s zeroes any singularities are at worst removable, but I haven't got a clue where to go from there.

Hunter
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  • Well, you said it. The singularities are removable. – Friedrich Philipp Mar 20 '16 at 11:51
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    You should interpret the statement as “the meromorphic function $f/g$ has only removable singularities, so it can be extended to an entire function”. The statement is not really precise, because it is assumed $g$ is not the constant zero function. With this implicit assumption, you can observe that $g$ only has isolated zeros. – egreg Mar 20 '16 at 11:52
  • @egreg Well, I really want to show that $f = c * g$, and for that I believe I need to show that $f/g$ is entire and not that it can be extended, unless thats all that can be said. – Hunter Mar 20 '16 at 11:57
  • @Hunter $f/g$ can be entire only if $g$ has no zero. But is the fact it only has removable singularities really of a concern? – egreg Mar 20 '16 at 12:02
  • Actually I suppose thats enough to show its true, because I can create a function that shows $f = c * g$ except at the zeroes, and at the zeroes f is 0. – Hunter Mar 20 '16 at 12:02

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