Let $f: \mathbb{C} \rightarrow \overline{\mathbb{C}}$ be a meromorphic function such that all its poles are simple with itegral residues. Then there exists a meromorphic function $h: \mathbb{C} \rightarrow \overline{\mathbb{C}}$ such that $f(z) = h'(z)/h(z)$.

In my attempts at this, I write $f$ as a function that has simple integral residues, so we want it to be of the form

$$ f(z) = \frac{p(z)}{g(z)} = \frac{p(z)}{\prod_{k=0}^n (z - a_k)} $$

Where $p(z) = g'(z)$. Since $g(z) = \prod_{k=0}^n (z - a_k)$, we can take the log of each side and differentiate, giving

$$ Log(g(z)) = Log\left( \prod_{k=0}^n (z - a_k) \right) = \sum_{k=0}^n Log ((z - a_k)) \\ g'(z) = g(z) \left( \sum_{k=0}^n \frac{1}{z - a_k} \right) = \left( \prod_{k=0}^n (z - a_k) \right) \left( \sum_{k=0}^n \frac{1}{z - a_k} \right) $$

Using this, I should have that

$$ f = \frac{g'(z)}{g(z)} = \frac{\left( \prod_{k=0}^n (z - a_k) \right) \left( \sum_{k=0}^n \frac{1}{z - a_k} \right)}{\left( \prod_{k=0}^n (z - a_k) \right)} = \sum_{k=0}^n \frac{1}{z - a_k} $$

Clearly, here $g$ is meromorphic, as it has a discrete set of poles (actually, none), and the construction of $g$ matches the requirements for our definition of $f$. My first question relates to the general correctness of the proof. Is this a unique solution? Are there alternative methods for constructing something that looks like this? Additionally, I was a little bit hesitant to state that $g(z)$ was composed of a finite number of linear terms. Is at an issue to assume that they are infinite?

Lastly, I feel that I'm missing something because I never really used the fact that $f$ or $g$ was meromorphic. In fact, each is analytic in punctured neighborhoods around the poles, and I feel that that would have been a sufficient condition for solving this problem. The fact that I didn't use it makes me suspicious that I am leaving out a detail of importance somewhere.

  • $\begingroup$ Do you assume that $f$ has only finitely many poles? – Here are some other proofs: Which meromorphic functions are logarithmic derivatives of other meromorphic functions?. $\endgroup$
    – Martin R
    Aug 15 '16 at 20:36
  • $\begingroup$ Actually your proof seems to be the wrong way around. You assume that $f$ has the desired representation. Also you don't treat the case of infinitely many poles. $\endgroup$
    – Martin R
    Aug 15 '16 at 20:43
  • $\begingroup$ @MartinR, I haven't finished going through the first link, but it does look more like what I thought it would. With regards to my assumptions, I understand what you are saying about it being backwards. I think that at the very least I am OK with writing $f$ with a product of linear terms in the bottom, as we know that the poles are all simple and with integral residues (which I guess I didn't specify). But if I am understanding you correctly, that should be as far as I am able to go down that line of reasoning $\endgroup$
    – cnolte
    Aug 15 '16 at 20:58
  • $\begingroup$ @MartinR Actually, now that I think about it, yeah, I feel like this is a mess. Clearly not all meromorphic functions are the sum of reciprocal linear terms. I'm going to pursue the link and see if I can't get a more comprehensive answer. Thank you for finding it for me. I guess the way one asks a question really determines the search results $\endgroup$
    – cnolte
    Aug 15 '16 at 21:00

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