Does the contour integral of a rational fraction function in the complex plane vanish in large radius limit?

Question

Let $f(z)=\frac{z^m+az^{m-1}+\cdots+b}{z^n+cz^{n-1}+\cdots+d}$ be a rational fraction function of complex variable $z$, where the integers $n-m\geqslant 2$. Is the following integration limit $$\lim_{r\rightarrow+ \infty }\int_{\Gamma(r,\theta)}dz f(z)=0$$ correct ? How to prove it?

Note that $\Gamma(r,\theta)$ represents a circular arc whose centre is $0$, radius is $r$, and arc angle is $\theta\in(0,2\pi]$, and $\theta$ is fixed in the $r\rightarrow+ \infty$ limit.

Answer 1

$\lvert f(z)\rvert=\frac{1}{\lvert z\rvert^{n-m}}\frac{\lvert 1+az^{-1}+\cdots+bz^{-m}\rvert}{\lvert 1+bz^{-1}+\cdots+dz^{-n}\rvert}\leq \frac{1}{\lvert z\rvert^{n-m}}\frac{1+\lvert az^{-1}+\cdots+bz^{-m}\rvert}{1-\lvert bz^{-1}+\cdots+dz^{-n}\rvert}$.

When $\lvert z\rvert$ is large enough, we have $\lvert az^{-1}+\cdots+bz^{-m}\rvert<\frac{1}{10}, \lvert bz^{-1}+\cdots+dz^{-n}\rvert<\frac{1}{10}$,

thus $\lvert f(z)\rvert\leq\frac{1}{\lvert z\rvert^{n-m}}\frac{1+\frac{1}{10}}{1-\frac{1}{10}}\leq \frac{2}{\lvert z\rvert^{n-m}}\leq \frac{2}{\lvert z\rvert^{2}}$$\Rightarrow \lvert \int_{\Gamma(r, \theta)}f(z)dz\rvert \leq \int_{\Gamma(r, \theta)}\lvert f(z)\rvert\lvert dz\rvert\leq \int_{\Gamma(r, \theta)}\frac{2ds}{r^2}=\frac{4\pi}{r}\rightarrow 0 (r \rightarrow \infty)$

Answer 2

use the coordinate $z=re^{i \theta}$, and the integral become: $$ lim_{r\rightarrow +\infty}\frac{1}{r^{n-m-1}}\int _{0}^{2\pi}d\theta g(\theta, r)$$ where $g(\theta, r)$ is order 1 and thus the integral is finite, thus $\frac{1}{r^{n-m-1}}$ times the integral is 0 as r goes to infinite.