Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f$ be a meromorphic function in a domain $D$. The set of zeros $Z_f$ and the set of poles $P_f$ are both discrete in $D$; it means that doesn't exist a sequence of zeros (risp. sequence of poles) that converges to a zero of $f$ (risp. to a pole of $f$). My question is the following:

Let $a\in D$ such that $a\not\in Z_f $ and $a\notin P_f$; does exist a sequence $\{b_n\}$ with $b_n\in Z_f\cup P_f$ such that $b_n\rightarrow a$? Roughly speaking, can $a$ be an accumulation point for $Z_f\cup P_f$?

I've tried to give myself an answer but I don't know if it is correct:

If $f$ is continuous then $$\lim_{b_n\to a} f(b_n)=f\big(\lim_{b_n\to a}b_n\big)$$ but in the above case $$\lim_{b_n\to a}f(b_n)=0,\infty$$ or it doesn't exist, and $$f\big(\lim_{b_n\to a}b_n\big)=f(a)$$ Contradiction!

share|cite|improve this question
You will need the definition of "meromorphic" for this. Then answer questions: Is $f$ analytic at $a$? Is $a$ a pole of $f$? – GEdgar Sep 14 '12 at 15:37
I said that $a\notin P_f$ – Dubious Sep 14 '12 at 15:51
The argument looks good to me. – Gerry Myerson Sep 17 '12 at 12:58
up vote 1 down vote accepted

CW answer to push it from unanswered queue:

Argument looks good.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.