1-form on Riemann Surface Good evening, I can not prove the following result: 
Let $\omega $ be a meromorphic 1-form on $ \mathbb {C} _ {\infty} = \mathbb {C} \cup \infty $ such that $ \omega_{|\mathbb{C}} = f (z) dz $. Show that f is ratio of polynomial functions. 
Any suggestions on how to develop the demonstration? 
thanks.

As suggest in the comment, on another cover $\mathbb C$ (which corresponds to $\mathbb C\setminus\{0\} \cup \{\infty\}$ with coordinates $\tilde z$), write 
$$\omega = \tilde f(\tilde z)$$
On the intersection of two coordinate $\mathbb C\setminus \{0\}$, we have $z = 1/\tilde z$, then 
$$d\tilde z = d(1/ z) = -\frac{1}{ z^2} d z . $$
Write $f(z) = \sum_{n=k}^\infty a_n z^n$ and $\tilde f(\tilde z) = \sum_{n=m}^\infty b_n \tilde z^n$ for some $k, m\in \mathbb Z$, then 
$$ \sum_{n=k}^\infty a_n z^n dz = f(z) dz = \tilde f(\tilde z) d\tilde z = \sum_{n=m}^\infty b_n \tilde z^n d\tilde z = \sum_{n=m}^\infty b_n z^{-n} \bigg(-\frac{1}{z^2}\bigg)dz$$
$$ \Rightarrow \sum_{n=k}^\infty a_n z^n  = \sum_{n=m}^\infty (-b_n) z^{-n-2}$$
Then compare coefficients to see that $a_n = b_n = 0$ for large enough $n$. 
