A question about complex integration of $\frac{1}{p(z)}$

Question

Let $p(z)$ be a polynomial of degree $n\ge 2$. Is it true that, there is a $R>0$ such that $$\int\limits_{|z|=R}{\frac{1}{p(z)}dz}=0?$$

My attempt is: there is a $R>0$ such that $|p(z)|\ge \frac{1}{2}|a_n||z|^n$, for $|z|\ge R$, where $p(z)=a_0+a_1z+\dotsm+a_nz^n$. Thus $\frac{1}{p(z)}$ is defined and continuous on $\Bbb C\setminus B(0,R)=\{z:|z|\ge R\}$. Also $\frac{1}{p(z)}$ is bounded by $\frac{2}{|a_n|R^n}$ on $|z|=R$. So I have $$\left|\int\limits_{|z|=R}{\frac{1}{p(z)}dz}\right|\le \frac{4\pi}{|a_n|R^{n-1}}.$$ and this tends to zero as $R\to\infty$.

But how to find a $R$ so that the integral vanishes ? Is it possible? Need help.

Answer 1

Let $r>0$ be so large that all roots of $p(z)$ lie within the disc $|z| < r$. Then for every $R \geqslant r$ we have

$$\int \limits_{|z| = R} \frac{1}{p(z)} \, \mathrm{d} z = \int \limits_{|z| = r} \frac{1}{p(z)} \, \mathrm{d} z.$$

But the left side tends to $0$ as $R \to \infty$, so

$$\int \limits_{|z| = r} \frac{1}{p(z)} \, \mathrm{d} z = 0$$

itself.

Answer 2

I got it wrong at first because i didn't notice that the degree of $p$ is greater than $1$. What you can easily observe is that $1/p$ has a holomorphic antiderivative in the annulus $D(0,R,\infty)$ where $D(0,R)$ contains all the roos of $p$.

Thus the integral equals zero.

To notice the difference, if $p$ had degree $1$ and assume $p(z)=z$ then $\int 1/z\neq 0$ because $\log z$ is not holomorphic in $D(0,R)$ because you cannot have a holomorphic angle. (The last are just informational).

Answer 3

Let $R$ be greater than the absolute values of all the roots of $p$. Then $$ \int_{|z|=R}\frac{dz}{p(z)} = -\int_{|w|=1/R}\frac{dw}{w^2p(1/w)}, $$ using the change of variables $w=1/z$. But the singularity of $\frac{1}{w^2p(1/w)}$ at $w=0$ is removable: $$ \lim_{w\to 0} w^2p(1/w) = \lim_{z\to \infty} \frac{p(z)}{z^2},$$ which is finite since $p$ has degree at least 2. Thus, the integrand in the right-hand integral extends to a holomorphic function of $w$ in the disk $|w|\leq 1/R$, so the integral is 0.