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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!

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  • $\begingroup$ You will need the definition of "meromorphic" for this. Then answer questions: Is $f$ analytic at $a$? Is $a$ a pole of $f$? $\endgroup$ – GEdgar Sep 14 '12 at 15:37
  • $\begingroup$ I said that $a\notin P_f$ $\endgroup$ – Dubious Sep 14 '12 at 15:51
  • $\begingroup$ The argument looks good to me. $\endgroup$ – Gerry Myerson Sep 17 '12 at 12:58
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CW answer to push it from unanswered queue:

Argument looks good.

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