Integral of rational function in the complex plane

Question

Let $P$, and $Q$ be complex polynomials such that $\deg Q \ge \deg(P) + 2$

Prove that there exists $r > 0$ such that if $\gamma$ is a closed curve outside $\{z : |z| \le r\}$, then

$$\int \frac{P}{Q} \, dz=0,$$ the integral being over $\gamma$.

Don't know how to solve this.

Thanks in advance!

Answer 1

The solution is from Bak/Newman: use the ML inequality to show that $$\int_\gamma \frac{P}{Q} << 2\pi R\frac{A}{R^2},$$ where A is some constant and $R^2$ comes from the fact that $deg(Q) \geq deg(P) + 2$. Now, take a limit with $R\rightarrow \infty$. You get that your integral is zero: by the residue theorem, it is constant starting with R such that the curve contains all the poles. But, it tends to zero with $R\rightarrow\infty$, so it is zero.

For more details, see Bak/Newman Complex Analysis, section 11.1.

Answer 2

Hints:

1. The integral is independent of $r$ once you're out past the roots of $Q$

2. The integral is $O(1/r)$