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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.

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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 $$\lim_{R \to \infty} R^{1- n} = 0.$$

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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 $$\int_{C_R}\frac{1}{p(z)}\,dz=0.$$

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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:

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

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