# Proving that meromorphic on the extended plane implies polynomial.

As the title says, I am trying to prove that if $$f(z)$$ is meromorphic on the extended plane, then $$f(z)$$ is rational. Is my proof below complete?

My proof so far is as follows: We can enumerate the poles of $$f(z)$$ as $$z_1, z_2, \dots, z_n$$ $$(1)$$. Then, we see that $$f(z) = \frac{h(z)}{\prod_{i = 1}^{n} (z-z_i)^{h_i}}$$, where $$h(z)$$ is an analytic function. Since $$f(z)$$ is meromorphic, $$h(z)$$ has a non-essential singularity at $$\infty$$ $$(2)$$. However, by a previous exercise, we know any function that is analytic everywhere in the non-extended plane and has a non-essential singularity at $$\infty$$ must be a polynomial. So, $$h(z)$$ is a polynomial and hence, $$f(z)$$ is a rational function.

My worries are regarding $$(1)$$ and $$(2)$$. To prove $$(1)$$ we know if there are infinitely many poles in a bounded region, then there must be a limit point of poles. However, this can never happen as $$f(z)$$ is meromorphic. Now, suppose the poles are the natural numbers $$1, 2, \dots \in \mathbb{C}$$. Then every neighborhood of $$\infty$$ contains infinitely many naturals so we again have an accumulation of poles. Therefore, meromorphic functions only have finitely many poles.

For $$(2)$$, how would I show that $$f(z)$$ doesn't have an essential singularity at $$\infty$$?

• By definition, $f(z)$ does not have an essential singularity at $\infty$ as it is meromorphic. The same holds for $h(z)$ as it equals $f(z) \prod (z-z_i)^{h_i}$. – hellHound Dec 3 '18 at 22:39
• @hellHound Oh yes, essential singularities result in non-analytic behavior in some neighborhood around the point. – Yunus Syed Dec 3 '18 at 23:37
• Meromorphic at $\infty$ implies for some $A,n,B,r$ : $|f(1/z) - A z^n| < B |z^{n+1}|$ for $|z| < r$ so $f(z)$ has no poles (nor zeros) for $|z| > 1/R$. Thus you are looking at $z_i$ in the compact set $|z| \le 1/R$. – reuns Dec 3 '18 at 23:53

(1) a meromorphic function $$f$$ on a riemann surface $$X$$ is a olomorphic function $$f: X/S \to \mathbb{C}$$ such that $$S$$ is a closed and discrete subset of $$X$$ and each point of $$S$$ is a non essential pole of $$f$$

In your case the extended plane $$\mathbb{C}_\infty$$ is a compact Riemann surface and so every closed and discrete subset of $$\mathbb{C}_\infty$$ is finite.. then the set $$S$$ of non essential pole of a meromorphic function $$f$$ on $$\mathbb{C}_\infty$$ is a finite set $$S=\{a_1,\dots, a_n\}$$;

(2) every meromorphic function $$f$$ on a compact riemann surface $$X$$ verify the follow identity:

$$\sum_{p\in X} ord_p(f)=0$$

So in the case $$X=\mathbb{C}_\infty$$ you have that

$$\sum_{k=1}^n ord_{a_k}(f)+ord_\infty (f)=0$$

so

$$ord_\infty (f)=-\sum_{k=1}^n ord_{a_k}(f)$$

Now you can prove your theorem:

Let $$g(z):=\prod_{k=1}^n(z-a_k)^{l_k}$$ where $$l_k:=ord_{a_k}(f)$$. Then you have that $$g(z)$$ is a non null meromorphic function of $$\mathbb{C}_\infty$$ so $$\frac{f}{g}$$ is a meromorphic function of $$\mathbb{C}_\infty$$ but for every $$a_k\neq \infty$$ you have that

$$ord_{a_k}(\frac{f}{g})= ord_{a_k}(f)-ord_{a_k}(g)=l_k-l_k=0$$

And

$$ord_{\infty}(\frac{f}{g})= ord_{\infty}(f)-ord_{\infty}(g)=$$

$$=ord_{\infty}(f)-ord_{0}(\prod_{k=1}^n (\frac{1}{w}-a_k)^{l_k} =$$

$$=ord_{\infty}(f)-ord_{0}(w^{-\sum_{k=1}^n l_k}(\prod_{k=1}^n (1-a_kw)^{l_k})=$$

$$=ord_{\infty}(f)+\sum_{k=1}^n l_k= 0$$

So $$\frac{f}{g}$$ is a olomorphic function on the compact Riemann surface $$\mathbb{C}_\infty$$ then it is costant.

In other words $$f$$ is a rational function