¡Hi! Can anyone give me a hint for solve the following problem?
"Let $q(z)$ a rational function, and the degree of the denominator exceeds that of the numerator by $l$ and $l \geq 2$. Show
$$ \mathop{\sum}\limits_{c \in \mathbb{C}} Res_{c} q =0$$
I've tried used Growth lemma for rational functions and Residue theorem but i'm stuck.
The residue theorem is the way to go. Just integrate f(z) over $|z|=R$ and show $$\left| \int_{z=|R|}{f(z)dz}\right |<=\int_{|z|=R}{|f(z)||dz|}\to 0\text{ as }R\to\infty$$ This is clear, since $|f(z)|<\frac{K}{R^2}$ so large $R$ and some constant $K$, and $|dz|=Rd\theta$
Now $f(z)$ is meromorphic with finitely many poles, so for $R$ sufficiently large, all the poles are inside the contour, and the residue theorem gives the desired result.
EDIT In answer to the OP's comment:
$f(z)=\frac{p(z)}{q(z)}$ where $\deg q \ge \deg p+2$ Say $p(z)=az^k+O(z^{k-1}), q(z)=bz^n+O(z^{n-1})$ where $n-k\ge 2$. For $|z|$ sufficiently large, $$|p(z)|\le 2|a||z|^k, |q(z)|>=\frac{|b||z|^n}{2} \implies |f(z)|\le \frac{4|a|}{|b||z|^{n-k}}<=\frac{4|a|}{|b||z|^2}$$