Residue Calculus - Showing that the quotient of polynomials have integral $0$ in a simple closed contour in a special case. I'm having difficulty understanding the solution to the following problem.
In the solution below, I can't understand why since $b_m\neq 0$, the quotient of these polynomials is represented by a series of the form $d_0+d_1 z+d_2 z^2 +\cdots$. In fact, since the degree of the denominator is greater, shouldn't the quotient include negative powers of $z$? I would greatly appreciate if anyone could explain this to me.
Problem:

Solution:


What I'd do: We have $|P(z)/Q(z)|$ on the order of $1/|z|^2$ as $|z| \to \infty.$ By Cauchy, $$\int_C \frac{P(z)}{Q(z)}\,dz = \int_{|z|=R} \frac{P(z)}{Q(z)}\,dz$$ for large $R.$ Use the "M-L" estimate on the latter to see its absolute value is no more than a constant times $(1/R^2)\cdot2\pi R.$ The latter $\to 0$ as $R\to \infty.$ Conclusion: $\int_C \frac{P(z)}{Q(z)}\,dz = 0.$
