Let's say that all I know about a function $f$ is that it is meromorphic on $\mathbb{C}_\infty$, and that $\{z_i\}$ is the sets of zeroes of $f$, each one of order $n_i$ and $\{\lambda_j\}$ its set of poles, each one of order $m_i$. Does $f$ gets completely determined, modulo a constant scalar, with only this information?

I think that I can show this as it follows: if $f$ is meromorphic on the Riemann Sphere, then it is a rational function. If $\alpha :\mathbb{C}_\infty\rightarrow\mathbb{C}$ is a holomorphic function, then it is constant. If $\alpha$ is not identically zero, then $\alpha f$ is a meromorphic function over $\mathbb{C}_\infty$ with the same set of zeroes and poles with the same orders.

It could possibly be a trivial fact but I'm attending some Riemann Surfaces lectures without having a formal graduate course in Complex Analisys in one variable. Please criticize this!


A hint: Given such an $f$, the set of zeros $z_i\in{\mathbb C}$ and their orders $n_i$, as well as the set of poles $\lambda_j\in{\mathbb C}$ and their orders $m_j$, define the rational function $$g(z):={\prod_j (z-\lambda_j)^{m_j}\over\prod_i (z-z_i)^{n_i}}\ f(z)\qquad(z\in{\mathbb C}) .$$

  • $\begingroup$ I can show that $g(z)$ has no zeroes and no poles over $\mathbb{C}$. I guess I have to show that "g" is constant in the Riemann Sphere. I believe it is the same proof in Miranda's "Algebraic Curves and Riemann Surfaces" for Theorem 2.1 (every meromorphic function over $\mathbb{C}_\infty$ is rational and vice-versa). I believe that what I stated is true: a meromorphic function is completely determined in the Riemann Sphere by its zeroes, poles and its orders modulo complex scalar. Thanks! $\endgroup$ – Marra Apr 20 '13 at 22:51

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