# Residue at infinity and quotient of complex polynomials

A textbook problem: Say I have $$f(z) = P(z)/Q(z)$$, where $$P(z) =a_0+a_1z+...+a_nz^n$$ and $$Q(z)=b_0+b_1z+...+b_mz^m$$ are both polynomials such that $$m\geq n+2$$, and that $$a_n\neq 0$$, $$b_m\neq0$$. I want to show that $$\int_C f(z) dz=0$$ if all zeros of $$Q$$ are inside a simple closed contour $$C$$.

A solution is provided this way: We of course firstly compute

$$\frac{1}{z^2}\frac{P(1/z)}{Q(1/z)}=\frac{a_0z^{m-2}+a_1z^{m-3}+...+a_nz^{m-n-2}}{b_0z^m+b_1z^{m-1}+...+b_m}$$

The solution then says that "As $$m-n-2\geq0$$ so that the numerator is a polynomial, and since $$b_m\neq 0$$, this $$\frac{1}{z^2}\frac{P(1/z)}{Q(1/z)}$$ is represented by a series of the form $$d_0+d_1z+d_2z^2+...$$." From here the result is clear by the theorem using residue at infinity, since this $$d_i$$ series has $$z^{-1}$$ coefficient $$0$$.

I have two questions:

1. How and why did it say that $$\frac{1}{z^2}\frac{P(1/z)}{Q(1/z)}$$ is represented by a series of that form?

2. I try concluding in another way instead: From the above computation, $$\frac{1}{z^2}\frac{P(1/z)}{Q(1/z)}$$ is clearly analytic at $$0$$ because $$b_m\neq 0$$ and the numerator is indeed a polynomial as above. Thus, the residue of $$\frac{1}{z^2}\frac{P(1/z)}{Q(1/z)}$$ at $$0$$, which is just the integral of it over any simple closed contour small enough only enclosing $$0$$ but not the zeros of the denominator (since they must be finitely many), is zero by Cauchy-Goursat. Is this argument valid? It may or may not be longer once I know the answer to question 1 above.

Thank you so much for the guidance!

Some ideas:

1. The function $$\;\cfrac1{z^2}\cfrac{P\left(\cfrac1z\right)}{Q\left(\cfrac1z\right)}\;$$ is analytic at $$\;z=0\;$$ and thus it has a Taylor series centered at $$\;z=0\;$$ , as said there.

2. Google "Residue at infinity", there you'll be able to see that if $$\;f\;$$ is a meromorphic function on and inside a closed, simple curve $$\;C\;$$ with positive orientation in the complex plane, and its singularities inside $$\;C\;$$ are all simple poles, then

$$\frac1{2\pi i}\oint_C f(z)\,dz=\text{Res}_{z=0}\,\left(\frac1{z^2}f\left(\frac1z\right)\right)$$

The above is very handy and saves lots of work when trying to calculate the residues at many finite, simple poles to evaluate an integral. Instead of that, only one resiude!

• Ouch! Number 1 is what I was missing. I can't believe it hahaha thanks a lot! Dec 19, 2020 at 2:53
• So basically, the first sentence of my number 2 above answers my number 1? I understand the use of residue at infinity, but what about the second sentence of my number 2, starting from "Thus, the residue of ... " where I deduce the residue at infinity value to be zero by Cauchy-Goursat? Is that part valid? Dec 19, 2020 at 3:06
• That very part, where you mention C-G theorem is, certainly,. true, though that in fact follows from the residue theorem directly, once we understand that function is analytic at $\;z=0\;$ and thus its Laurent series around this point is, in fact, a Taylor series and thu with coefficient $\;a_{-1}=0\;$ ...What is lacking, I believe, is to connect this result about the residue at infintiy with the value of the original integral...it is at least not clearly and firmly said anywhere. Dec 19, 2020 at 8:17
• I see. It's clear now to me. Thanks a lot. I just shared another complex analysis question math.stackexchange.com/questions/3954747/…, and hopefully you can check on that if you have time. Dec 19, 2020 at 9:48