1
$\begingroup$

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 $\infty$?

$\endgroup$
3
  • 1
    $\begingroup$ 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} $. $\endgroup$
    – hellHound
    Commented Dec 3, 2018 at 22:39
  • $\begingroup$ @hellHound Oh yes, essential singularities result in non-analytic behavior in some neighborhood around the point. $\endgroup$
    – Yunus Syed
    Commented Dec 3, 2018 at 23:37
  • $\begingroup$ 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 $|z| \le 1/R$. $\endgroup$
    – reuns
    Commented Dec 3, 2018 at 23:53

1 Answer 1

1
$\begingroup$

(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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .