determination of a holomorphic function by its poles and zeros While reading a text about the application of complex analysis to elasticity, I thought about the following problem:

Let $f$ be a holomorphic function in all $\mathbb{C}$. Is $f$ uniquely determined by the list of its poles and zeros (and their orders, of course)?

EDIT: By "the list of its poles and zeros" I include also the point at $\infty$. I assume that $f$ has a proper limit at infinity.
I guess that if that was true it was an undergrad theorem that I'm supposed to know. 
 A: Let me summarize the comments: I'm assuming that we talk about meromorphic functions.


*

*As long as we don't control the singularity at infinity, poles and zeroes and their multiplicities are far from enough to determine the function.
Indeed, if $h:\mathbb{C} \to \mathbb{C}$ is any entire function then $\exp{h}$ has no zeroes and no poles and thus $f(z)$ and $g(z) = f(z) \cdot \exp{(h(z))}$ have the same poles and zeroes.
This is the “only” ambiguity: if $f(z)$ and $g(z)$ are meromorphic functions with the same set of poles and zeroes and with the same multiplicities then $k(z) = \frac{f(z)}{g(z)}$ only has removable singularities and thus it extends to an entire function: $k : \mathbb{C} \to \mathbb{C}$. It is not hard to show that $k$ has no zeroes, and a standard fact from complex analysis tells us that a function $k: \mathbb{C} \to \mathbb{C} \smallsetminus \{0\}$ can be written as $k(z) = \exp{(h(z))}$ for some entire function $h$. (a)

*There is the surprising fact that for every closed discrete subset $D \subset \mathbb{C}$ (which may well be infinite) and any assignment $o: D \to \mathbb{Z}$ there exists a meromorphic function $f: \mathbb{C} \to \mathbb{D}$ such that $f$ has a zero of order $k$ at $z_0$ if and only if $z_0 \in D$ and $o(z_0) = k \gt 0$, and a pole of order $k$ at $z_0$ if and only if $z_0 \in D$ and $k = -o(z) \gt 0$. Of course, if $D$ is finite, this is easy to achieve (and a good exercise to do), but if $D$ is infinite you need to work a bit. This is Weierstrass's factorization theorem.
My favorite reference is Remmert's Classical topics in complex function theory, see chapter 3, p.73ff and also chapter 4, p.89ff for more on this.

*Finally, if we do control the singularity of the meromorphic function $f$ at infinity and decide that it should be non-essential (that is to say $f(1/z)$ has either a pole or a removable singularity at $0$), then it follows that $f$ is a rational function.
This is proved in detail e.g. in Theorem 4.7.7 on page 144 of Greene-Krantz, Function theory of one complex variable.

(a) Indeed, $h(z)$ can be chosen to be 
$$
h(z) = C \cdot \int_{1}^z \frac{k'(w)}{k(w)}\,dz
$$
where $C$ is such that $e^C = k(1)$ and a straightforward computation shows that $k(z) \cdot \exp{(-h(z))}$ has zero derivative while $k(1)e^{-h(1)} = 1$ so that $k(z) = \exp{(h(z))}$ as desired (note that $h$ is holomorphic and well-defined precisely because $k$ has no poles and zeroes).
