Suppose $P(z)$ and $Q(z)$ are polynomials satisfying $deg(Q) \geq deg(P) +2$ and $\gamma_R$ is an arc parameterized by $z=Re^{i\theta}$ for $\theta \in [\alpha,\beta]$. Show that $ \lim_{R \to \infty}\oint_{\gamma_R} \frac{P(z)}{Q(z)} = 0$.
Right, so we have that since $P(z)$ and $Q(z)$ are polynomials, that $P(z) = a_0 +a_1z+....+a_nz^n$ and that $Q(z) = b_0 + b_1z + .....+b_mz^m$. From here we see that $\frac{P(z)}{Q(z)} = \frac{a_0 +a_1z+.....+a_nz^n}{b_0 + b_1z + .....+b_mz^m} = \frac{z^n(\frac{a_0}{z^n}+.....+an)}{z^m(\frac{b_0}{z^m}.....+bm)}=z^{n-m}\frac{\frac{a_0}{z^n}+.....+an}{\frac{b_0}{z^m}.....+bm}$. Since $\gamma_R$ is an arc parameterized by $z=Re^{i\theta}$ we have that $|\frac{P(z)}{Q(z)}| = |z^{n-m}| |\frac{\frac{a_0}{z^n}+.....+an}{\frac{b_0}{z^m}.....+bm}| \leq R^{n-m}|\frac{\frac{a_0}{z^n}+.....+an}{\frac{b_0}{z^m}.....+bm}|$.
From here I am unsure on how to bound the second term. I need to bound it so that I can use the $M-L$ estimate, if I can bound it I know where to go with it.
Let $P(z) = a_n z^n + \cdots a_1 z + a_0$ and $Q(z) = b_m z^m + \cdots + b_1 z + b_0$ with $m \geq n + 2$ (and $a_n, b_m \neq 0$). Let $R > 0$.
$$\bigg|\int_{\gamma_R} \frac{P(z)}{Q(z)} dz \bigg| \leq \ell(\gamma_R) \max_{z \in \gamma_R} \frac{|a_n z^n + \cdots a_1 z + a_0|}{|b_m z^m + \cdots + b_1 z + b_0|}$$
We now use the regular triangle inequality in the numerator and the opposite triangle inequality in the denominator. We also assume that $R$ is so large that we can omit the absolute value sign in the denominator after applying the opposite triangle inequality. We get
$$\bigg|\int_{\gamma_R} \frac{P(z)}{Q(z)} dz \bigg| \leq R (\beta-\alpha)\frac{a_n R^n + \cdots a_1 R + a_0}{b_m R^m - \cdots - b_1 R - b_0}$$
Now, it's easy to see that this expression goes to $0$ as $R \rightarrow \infty$ since we have polynomial in $R$ of degree $n+1$ in the numerator and of degree $m$ in the denominator, and by assumption $m > n+1$.