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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: " 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
Mark: This is usually handled in one of the following ways. 1. Post a comment inviting the commenter who answered the question to post it as an answer. 2. Write an answer summarizing the comments, and mark it "community wiki" using the checkbox at the bottom of the answer form. 3. Write a detailed answer of your own. In all cases, you can then accept the answer. – Nate Eldredge Aug 11 '12 at 14:10
up vote 5 down vote accepted

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

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