Weierstrass Factorization Theorem for meromorphic functions. On Wikipedia page about Weierstrass factorization theorem one can find a sentence which mentions a  generalized version so that it should work for meromorphic functions. I mean:

We have sets of zeros and poles of function $f$. How could we use that sets to find formula for $f$.

I think that it should be in the form of quotient of two entire functions.
 A: Given a meromorphic function $f$ with poles $(p_i)_{i\in I}$ and zeros $(z_i)_{i\in J}$  repeated according to multiplicity we have the corresponding weierstrass product  for the poles $$\Pi(s)=s^{m}\prod_{i\in I} E_{n_i}\left(\frac{s}{p_i}\right)$$
where $m$ is the order of the pole at $s=0$ and $(n_i)_{i\in I}$ is a sequence of positive integers such that the weierstrass sum $$\sum_{i \in I} \left(\frac{r}{\vert p_i \vert}\right)^{1+n_i}$$ and thus the product converges. This product then defines a holomorphic function on the whole of $\mathbb{C}$ that has a zero of order $m$ at the point $s$ if and only if $f$ has a pole of order $m$ at $s$. Thus the function $\Pi(s)\cdot f$ can be continued to a holomorphic function on all of $\mathbb{C}$ with exactly the zeros $(z_i)_{i\in J}$ of $f$ (again according to multiplicity). Due to the fact that this function is holomorphic, by Weierstrass we have a sequence of positive integers $(m_i)_{i\in J}$ and a zerofree holomorphic function $g$ such that: $$\Pi(s)\cdot f(s)=g(s)s^{m'} \prod_{i\in J} E_{m_i}\left(\frac{s}{z_i}\right)$$ where $m'$ denotes the order of the zero at $s=0$ of $f$. All together we have that: $$f(s)=g(s) s^{m'-m}\frac{\prod_{i\in J} E_{m_i}\left(\frac{s}{z_i}\right)}{\prod_{i\in I}E_{n_i} \left(\frac{s}{p_i}\right)}$$
A: First,  a meromorphic function in the plane is the quotient of two entire functions: Say $f$ is entire except for poles at $p_j$. Say the  pole at $p_j$ has order $n_j$.  There is an entire function $h$ such that $h$ has a zero of  order $n_j$ at $p_j$ (and no other zeroes); in fact you can construct such a function $h$ as a product of the sort you see in the Weierstrass factorization theorem. Now all the singularities of $g=hf$ are removable, so $g$ is entire, and $f=g/h$.
I don't know precisely what factorization that Wikipedia  page is referring to, but if $f=g/h$ then a factorization of $g$ plus a factorization of $h$ give a factorization of $f$. This "must" be the result they're talking about...
(And in fact we've obtained $f=e^\varphi\Pi_1/\Pi_2$, where $\Pi_2$ is a product depending only on the poles of $f$ and $\Pi_1$ is a product depending only on the zeroes of $g$, which are the same as the zeroes of $f$. So, in the strongest possible sense, we have in fact given a factorization of $f$ that depends only on the zeroes and poles of $f$.)
