Jtdylktuy

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 $infty$?

(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:

$sum_{pin X} ord_p(f)=0$

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 $a_kneq infty$ 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