4
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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 $$g(z):={\prod_j (z-\lambda_j)^{m_j}\over\prod_i (z-z_i)^{n_i}}\ f(z)\qquad(z\in{\mathbb C}) .$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.