# Can we choose a meromorphic function's poles arbitrarily?

By the Weierstrass theorem, if $\{z_n\}$ is any set of points in a domain $\Omega$ with no limit point in $\Omega$, then there is a holomorphic function vanishing exactly at the $z_n$ and nowhere else. Furthermore, we can prescribe the multiplicities. Is it true that we can also find a meromorphic function with poles exactly at the $z_n$ and nowhere else? I think yes, just by inverting the function obtained by the Weierstrass theorem, but I want to be certain.

Furthermore, what happens if I allow the $z_n$ to have a limit point inside $\Omega$, for instance $\Omega$ is the unit disc and $z_n = 1/(n+1)$? By Rudin's definition of meromorphic, such a function is not meromorphic, but what can be said about a function with simple poles at $1/(n+1)$ for all $n$?

• Looks like you want Mittag-Leffler here. Basically you can prescribe the principal parts at all the points $z_n$. – Jyrki Lahtonen Jun 7 '17 at 5:18

You mean a function a bit like $$\frac1{\sin(\pi/z)}?$$ This is not meromorphic at $0$; it's like an "essential singularity" just slightly worse.
The usual Weierstrass theorems, applied to the domain $U=D-\{0\}$ where $D$ is a unit disc give meromorphic functions with poles at the points of any sequence tending to $0$. But these functions have nothing like a Laurent series at $0$.
• Yes, but $\sin(\pi/z)$ is not a holomorphic function that one could construct with Weierstrass' theorem if your domain is including $0$. It's not even meromorphic. So the fact that $\frac{1}{\sin(\pi/z)}$ is not meromorphic either doesn't seem like a useful observation. Also it doesn't seem relevant whether a function will have a Laurent series about $0$. – user452093 Jun 7 '17 at 6:35