The complex integral of the reciprocal of polynomial is constant on sufficiently large circles

Question

Let $P$ be a polynomial of degree $n ≥ 2$. Suppose all the roots of $P$ lie in the disk $D_{r}(0)$. Let $R > r$ and $$I(R)=\int_{|Z|=R}{1\over P(z)}dz$$ Prove that the integral is constant.

I know that $\lim_{R\to \infty}I(R)=0$. How can I proceed?

Answer 1

Using residue theorem:

If $P$ is a polynomial of degree $n\ge 2$, then $P(z)$ is analytic, so $\frac{1}{P(z)}$ is analytic on $\mathbb{C}\setminus \{a_1,a_2,\cdots, a_k\}$, where $a_1,\dots,a_k$ are roots of $P(z)$. Then by residue theorem, $$ I(R)=2\pi i \sum \operatorname{Res}(f;a_k), $$ which is a constant.

Using partial fraction of a polynomial:

Every polynomial in $\mathbb{C}$ is factored as $$ P(z)=(z-a_1)^{r_1}(z-a_2)^{r_2}\cdots(z-a_k)^{r_k}. $$ Then $1/P(z)$ can be decomposed as $$ \frac{1}{P(z)}=\sum_{i=1}^k\sum_{j=1}^{r_i}\frac{1}{(z-a_i)^j}. $$ Then apply Cauchy's integral formula.