# What results are known If $f,g$ are both analytic in $\mathbb{C}$, having infinitely many poles or zeros for the same $z$'s?

What results are known If $f,g$ are both analytic in $\mathbb{C}$, having infinitely many poles (or zeros) that all coincide? Where each pole from $f$ has has the same order pole of $g$ for the same $z$.

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## 2 Answers

Hint:

What happens if you multiply by an entire function without zeros - such as $h(z)=e^z$?

Do you know of any factorization theorems?

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OK! Anything else is out there besides Weierstrass and Mittag-Leffler's ? +1 for reminding me the obvious – Arjang Apr 29 '11 at 8:21

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Yes, can you prove this: If $f$ is meromorphic, then there are entire functions $g$ and $h$ such that $f= g/h$. In particular $g=fh$. – AD. Apr 29 '11 at 9:06

Theorem: Let $U\subseteq \mathbb{C}$ be open and connected, let $f:U\rightarrow \mathbb{C}$ be holomorphic, and let $A\subseteq U$ have an accumulation point in $U$. Then, if $f(A)=\{ 0\}$, then $f$ is identically $0$ on all of $U$.

In other words, the values a holomorphic function takes on an inifnite set with an accumulation point uniquely determine the function (on a certain connected component).

As for when $f$ and $g$ both have the same poles, I don't think you can say much. For example, if $h$ is holomorphic, then $f+h$ has exactly the same poles as $f$ and $g$. I guess you could try to apply the above theorem to $1/f$; however, if the set of points where $f$ has a pole has an accumulation point, then $f$ is identically equal to $\infty$ (follows from the above theorem). . .probably isn't a particularly useful fact.

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While this is an important theorem, it's not directly relevant to the question because those infinitely many points at which $f$ and $g$ coincide need not have an accumulation point. – lhf Apr 29 '11 at 13:20

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True, but he did not mention this one way or the other. He just asked what results are known when $f$ and $g$ share infinitely many zeroes, and this theorem is such a result. – Jonathan Gleason Apr 29 '11 at 15:33
you're right. Sorry for the noise. – lhf Apr 29 '11 at 15:58
@lhf No problem =) – Jonathan Gleason Apr 29 '11 at 16:20
@lhf and @GleasSpty : Thank you both for the discussion, lhf what you said was not noise it was a good point, GleasSpty's reply to you clarified how it related back to the question ( I wasn't sure myself when I read the answer ). – Arjang Apr 30 '11 at 0:00
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