# Rational Functions are Determined By Locations and Multiplicities of Zeros and Poles (why?)

There is a theorem that says rational functions in the extended complex plane are exactly the meromorphic functions.

After this, my textbook draws the corollary: "...as a consequence, a rational function is determined up to a multiplicative constant by prescribing the locations and multiplicities of its zeros and poles."

I don't see how it follows. Plus I think I have an example. Take a rational function with no zeros and two poles of multiplicity one, at 0 and 1.

At least two such functions exist:
$f(x)=\frac{1}{x} + \frac{1}{x-1}$
$g(x)=\frac{1}{x} + \frac{2}{x-1}$

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$f$ has a zero at $1/2$. $g$ does not. – Chris Eagle Aug 11 '12 at 13:52
If you always reduce your rational functions to the form $p(x)/q(x)$, where $p$ and $q$ are polynomials, then the statement you are questioning starts to sound rather straightforward, no? – J. M. Aug 11 '12 at 13:54
Oh yes. I guess it was my false counter example that was confusing me. I think the book meant to say that meromorphic functions are determined by zeros and poles. – Mark Aug 11 '12 at 14:02
This question as been answered by comments. – Mark Aug 11 '12 at 14:02

The precise statement is the other way round: a function meromorphic on the whole Riemann sphere $S$ is rational.
This is completely false on $\mathbb C$: the function $e^z$ is meromorphic (even holomorphic) on $\mathbb C$, but certainly not rational on $\mathbb C$, nor anywhere else.

Now if $f,g\in Rat(S)$ are rational on the Riemann sphere $S$ and have the same zeros and poles (counted with multiplicities), then the quotient $\phi=f/g$ having neither zeros nor poles is holomorphic on the whole of $S$ and thus constant (say because $\phi$ is bounded on $S$ by compacity, hence bounded on $\mathbb C$ and Liouville's theorem applies).

"Yes,Georges, but I gave you a counterexample"
Your rational functions don't have the same zeros: $$f(\frac {1}{2})=0\neq g(\frac {1}{2})=-2$$

A very optional remark
That meromorphic functions on $S$ are rational is the simplest example of a very profound result in algebraic geometry: Serre's GAGA principle.

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Ah, I see that Chris had already noticed that f and g don't have the same zeros. I have just upvoted his comment. – Georges Elencwajg Aug 11 '12 at 14:39