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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.

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Mittag-Leffler's theorem gives a complete answer to your question: No further restrictions. See also Weierstrass's product theorem. – t.b. Jul 23 '11 at 16:41
Theo's comment should be answer instead. – Omar Antolín-Camarena Jul 23 '11 at 16:43
up vote 6 down vote accepted

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 $1,1/2,1/3,1/4,...$

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Dear Theo, sorry for answering: your fine comment wasn't yet there when I started to (slowly as usual!) compose my answer. – Georges Elencwajg Jul 23 '11 at 17:09
Dear Georges, No worries! In fact, I was thinking that you'd answer this question with pleasure, so I just left the comment. I'll do my usual job and add a reference I like: Remmert, Classical topics in complex function theory, Chapter 3, pages 73ff, see also Chapter 4, pages 89ff. Best wishes, – t.b. Jul 23 '11 at 17:15
I'm always amazed that this is not only "existence" but produces the function for you, too! – Matt Jul 23 '11 at 17:24
Dear Theo, I have never seen such a friendly and selfless attitude on the Internet! But, please, don't do that in the future: although I indeed like holomorphic functions, I don't own them! And I like reading your answers. By the way, Remmert is my favourite reference too, but I can't quote it here because my copy is in German! (I'm surprised yours isn't...) – Georges Elencwajg Jul 23 '11 at 17:29

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