Let $p(z),q(z) \in \mathbb{C}[z]$ two polynomials with coefficients in $\mathbb{C}$ s.t. $deg(p) = m$, $deg(q) = n$ and $n \ge m +2$. I need to show that
$$ \lim_{R \to \infty} \int_{|z| = R} \frac{p(z)}{q(z)}dz = 0$$
Here's my reasoning:
W.l.o.g. we can assume $p(z),q(z)$ to be monic and with no common zeros. Let $R$ be sufficiently large s.t. all the zeros of $q(z)$ are within $B(0,R)$ (the ball centered in $0$ with radius $R$).
By the Residue Theorem:
$$\int_{|z| = R} \frac{p(z)}{q(z)}dz = 2\pi i \sum_{r_j} Res_{r_j}\bigg( \frac{p(z)}{q(z)}\bigg) = -2\pi i \cdot Res_\infty \bigg( \frac{p(z)}{q(z)}\bigg)$$
where $r_j$ are the zeros of $q(z)$, which are also the poles of the ratio. So it is enough to show that $Res_\infty \bigg( \frac{p(z)}{q(z)}\bigg)=0$.
By definition: $Res_\infty \bigg( \frac{p(z)}{q(z)}\bigg) \triangleq Res_0\bigg(- \frac{g(z)}{z^2}\bigg) $, where, letting $z_i$ be the zeros of $p(z)$,
$$ g(z) := \frac{p(z^{-1})}{g(z^{-1})} = \frac{ (z^{-1} - z_1) \dotsb (z^{-1} - z_m)}{(z^{-1} - r_1) \dotsb (z^{-1} - r_n)} = \\ z^{n-m} \frac{ (1 - z\cdot z_1) \dotsb (1 - z\cdot z_m)}{(1 - z\cdot r_1) \dotsb (1 - z\cdot r_n)} = z^{n-m} \cdot h(z)$$
And conclude by noticing that $-g(z) z^{-2} = -z^{n-m-2} h(z)$ is the product of two holomorphic (in the origin) functions (as $n \ge m +2 $ and $h(z)$ has no singularities in a sufficiently small ball around the origin), thus its residue in $0$ is $0$.
I feel a bit sloppy in this reasoning as this way there is not a real need of a limit as $R \to \infty$, we can actually take $R$ sufficiently large, so it's like I'm proving something stronger. Moreover I think I'm making things harder than they are. Is there an easier way to show the same?