# Does integration over a circle of the inverse of a polynomial of degree two or higher gives $0$?

Let $$p(z)=a_nz^n+...a_1z+a_0$$, $$a_n\neq 0$$. It is clear that if $$p(z)$$ is of degree one ($$n=1$$), then, by Cauchy's Integral Theorem, $$\int_{C_R}\frac{1}{p(z)}dz\neq 0$$, where $$C_R$$ represents a circle of sufficiently large radius $$R$$ (as it contains the pole $$z=-a_0$$.)

However, my text says that

$$\lim_{R\to\infty}\int_{C_R}\frac{1}{p(z)}dz=0$$

for all $$p(z)$$ of degree two or higher $$(n\geq 2)$$. How does one show this? In the previous part of the exercise, we are asked to show that for sufficiently large $$R$$, where $$|z|=R$$, it implies $$|p(z)|\geq R^n|a_n|/2$$. Maybe this helps in some way?

Note: Please show this without the residue theorem if possible.

You know that there exists an $$K$$ such that, for all $$z \in \mathbb C$$ such that $$| z | > k \implies |p(z)| \geq \frac{|a_n||z|^n}{2}.$$

So if $$R > k$$, then

$$\left| \oint_{C_R}\frac{dz}{ p(z) } \right| \leq l( C_R ) \times \sup_{z \in C_R} \left| \frac{1}{p(z)} \right| \leq 2\pi R\times \frac 2 {|a_n | R^n} = \frac{4\pi }{|a_n|} R^{1 - n}$$ and if $$n \geq 2$$, then $$\lim_{R \to \infty} R^{1- n} = 0.$$

Using the inequality you've already proven, you have $$\frac{1}{|p(z)|}\leq\frac{2}{|a_n|}\cdot\frac{1}{R^n}$$for $$z\in C_R$$. Then, by the triangle inequality, we have for sufficiently large $$R$$:

$$0\leq\left|\int_{C_R}\frac{1}{p(z)}\,dz\right|\leq\left(\int_{C_R}\,1\,|dz|\right)\cdot\left(\frac{2}{|a_n|R^n}\right)=(2\pi R)\cdot\left(\frac{2}{|a_n|R^n}\right).$$

Since $$n\geq2$$, the right-hand side goes to $$0$$ as $$R\to\infty$$. By the squeeze theorem, this implies that $$\int_{C_R}\frac{1}{p(z)}\,dz=0.$$

Just use the trivial inequality: an integral of an absolutely value of a function $$f$$ over a curve is not bigger than the length of the path times the maximum of $$|f|$$ along that path. So for a fixed $$R$$ the integral of $$|\frac{1}{p(z)}|$$ is not bigger than $$2\pi R\times\frac{2}{R^n |a_n|}$$, and since $$n>1$$ this expression goes to zero when $$R\to\infty$$. So from the triangle inequality we get:

$$|\int_{C_R}\frac{1}{p(z)}dz|\leq\int_{C_R}|\frac{1}{p(z)}|dz\leq 2\pi R\times\frac{2}{R^n |a_n|}\to 0$$