# 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.

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You can always multiply $f$ with a function without poles or zeroes, i.e. if $g$ is an entire function then $f$ and $e^g f$ have the same poles and zeroes with the same multiplicities. See also Weierstrass's factorization theorem. –  t.b. Apr 3 '12 at 13:46
@t.b. But if you include infinity too? The pole/zero at infinity of $e^g f$ will not be the same as that of $f$, would it? –  yohBS Apr 3 '12 at 13:49
Unless $f$ is a rational function, $f$ has an essential singularity at $\infty$ and so has $e^g f$. –  t.b. Apr 3 '12 at 13:50
Not sure about the $e^g f$ (because that does in general change the location of zeroes), but consider a constant function. It is holomorphic and has no poles or zeroes. This works for any constant though, so it can clearly not be unique. Other examples include all polynoms (??) eg. $x$ vs $\alpha x$... –  example Apr 3 '12 at 14:01
@example: $e^g$ can never be zero, so $e^g f$ has the same zeros as $f$. –  Hans Lundmark Apr 3 '12 at 17:15

Let me summarize the comments: I'm assuming that we talk about meromorphic functions.

1. 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)

2. 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.

3. 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).

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"...it follows that $f$ is a meromorphic function." Not surprising, since you assumed that to start. Perhaps you want a stronger conclusion: "...it follows that $f$ is a rational function." –  GEdgar Apr 5 '12 at 14:41
Indeed. Thanks for catching this unfortunate typo. –  t.b. Apr 5 '12 at 14:42