# Any Meromorphic function can at most have a countable number of poles.

I can prove this: Any Meromorphic function $$f$$ can at most have a countable number of poles.

Any pole a is contained in $$B(a,r)$$ some $$r>0$$ small. For all $$a,b$$ poles, $$B(a,r)\cap B(b,r)=\emptyset$$ some $$r>0$$ small. if i could prove that $$\bigcup_{a\, pole} B(a,r)\subset B(0,R)$$ some $$R>0$$ then $$Area(\bigcup_{a\, pole} B(a,r))=\sum_{a\, pole} Area(B(a,r))<\infty$$. Then, $$Area(B(a,r))=0$$ except for some countable many with $$Area(B(a,r))>0$$. Therefore the poles of $$f$$ can at most have a countable number of poles. This is correct?

Question 1.Why exists $$B(0,R)$$ such that $$\bigcup_{a\, pole} B(a,r)\subset B(0,R)$$ some $$R>0$$ big?

Question 2. If $$f:G\to\mathbb{C}$$ meromorphic function has an infinite countable number of poles. The poles accumulate on the border of region $$G$$? Why?

• Why so complicated? Each of the (disjoint) $B(a,r)$ contains a point with rational coordinates, and there are at most countably many rational pairs. Commented Jul 28, 2020 at 17:09
• oh right! it is more simply Commented Jul 28, 2020 at 17:10

## 1 Answer

Question 1.Why exists $$B(0,R)$$ such that $$\bigcup_{a\, pole} B(a,r)\subset B(0,R)$$ some $$R>0$$ big?

That is not true. If the domain is unbounded then the poles can accumulate at infinity. Example: $$f(z) = 1/\sin(z)$$.

But each of the (disjoint) $$B(a,r)$$ contains a point with rational coordinates, and there are at most countably many rational pairs.

Question 2. If $$f:G\to\mathbb{C}$$ meromorphic function has an infinite countable number of poles. The poles accumulate on the border of region $$G$$? Why?

First note that the poles can not accumulate at a point $$a \in G$$: Otherwise there is a sequence $$(z_n)$$ of poles with $$z_n \ne a$$ and $$\lim_{n\to \infty} z_n = a$$. But $$f$$ is either holomorphic at $$z_0$$ or has an (isolated) pole at $$a$$. In both cases there is a $$r > 0$$ such that $$f$$ is holomorphic in $$B(a,r)\setminus \{a\}$$, i.e. $$B(a,r)$$ contains no poles except possibly $$a$$. This is a contradiction.

Now assume that $$f$$ has infinitely many poles:

• If the set of poles is bounded then it must have an accumulation point in $$\Bbb C$$ (that is the Bolzano–Weierstrass theorem). As we saw above, the accumulation point cannot be in $$G$$, and therefore must be on the boundary of $$G$$.

• If the set of poles is unbounded then the poles accumulate at $$\infty$$. This is also boundary point of $$G$$ in the extended complex plane.

One can also consider $$G$$ as a subset of the extended complex plane $$\hat{ \Bbb C}$$ with the chordal metric. That is a compact metric space so that every sequence has a convergent subsequence. In particular, the sequence of poles has an accumulation point in the closure $$\overline G$$, and since the poles can not accumulate in $$G$$, the every accumulation point is on $$\partial G$$.

• But is it possible that the poles never accumulate? or specifically will they accumulate on the bundary of G? Commented Jul 28, 2020 at 17:49
• If the function has infinitely many poles the they must accumulate (at the boundary). Commented Jul 28, 2020 at 18:09
• If $f$ has an pole at $z_0$ then $f$ is holomorphic in $B(z_0,r)\setminus \left\{z_0\right\}.$ This implies that $B(z_0,r)\setminus \left\{z_0\right\}$ contain poles differents of $z_0$ because $z_0$ is a accumulation point but this is a contradiction because the poles are aislated. This is correct? Same argument when $f$ is analytic in $z_0$.In any case, I cannot see the contradiction clearly with the fact $z_0\in G$ Commented Jul 28, 2020 at 18:13
• If the function has infinitely many poles the they must accumulate (at the boundary). This is a knows results? Commented Jul 28, 2020 at 18:30
• @eraldcoil: I have extended the answer. Please let me know if something is still unclear. Commented Jul 29, 2020 at 7:02