Showing $\lim_{R\to\infty}\int_T \frac{p(z)}{q(z)}dz=0$ if $T$ is a circle with radius R and $deg(q)\ge deg(p)+2$

Question

Showing $\lim_{R\to\infty}\int_T \frac{p(z)}{q(z)}dz=0$ if $T$ is a circle with radius R and $deg(q)\ge deg(p)+2$.

I got stuck in this stage, if I take a parameterization $\phi(t)$ I get: $$\|\lim_{R\to\infty}\int_T \frac{p(z)}{q(z)}dz\|\le\lim_{R\to\infty}\int_{0}^{2\pi} \|\frac{p(\phi(t))}{q(\phi(t))}\|\|\phi'(t)\|dt$$ now I'm trying to show that $\|\frac{p(\phi(t))}{q(\phi(t))}\|\alpha \frac{1}{R^2}$ which make sense, but, I failed to show that.

Answer 1

Here is a hint, from which you ought to be able to easily finish the problem. Begin by proving the following important lemma:

Lemma. If $f$ is a monic polynomial of degree $n$, then there exists an $R > 0$ such that

$$|z| \ge R \implies \frac 1 2 \le \frac{|f(z)|}{|z|^n} \le 2.$$

This is just a quantitative version of what you're aiming for.

To prove this, write $f(z) = z^n + a_{n - 1} z^{n - 1} + ... + a_0$ and see what happens when you divide by $z^n$.