What can the multiset of zeros of a meromorphic function look like? Suppose I have a multiset $S$ of complex numbers. Under what conditions is there a meromorphic function $f$ whose zeros are precisely the elements of $S$, and have the same multiplicities?
I know that $S$ must be a discrete set (unless it is the entire complex plane with infinite multiplicity, for which $f(x) = 0$ has $S$ as its multiset of zeros), but as far as I know, there are no other conditions. 
 A: If you have a discrete closed set with non-zero rational integers attached to them, you can find a meromorphic function with zeros and poles dictated by your integers.
It will have  zeros exactly at the points with positive integers with the given multiplicity and poles at the points with negative integers, again with the right multiplicity. This is the Weierstrass/Mittag-Leffler theorem.
The same result is true in much greater generality: Given any divisor on any non-compact Riemann surface, there exists a meromorphic function whose divisor is the given one.    
There are generalizations to higher-dimensional manifolds, but this would take us too far. Suffice it to say that the key-word is Stein manifold.
Caveat A common misconception is that the right concept in those questions is discrete subset of $\mathbb C$. It is however indispensable to add closed. It is clearly impossible, for example, to find a holomorphic (or meromorphic)  function on $\mathbb C$ which  vanishes exactly on $1,1/2,1/3,1/4,...$
