Contour Integral proof

**skyking** · February 3rd 2011, 05:22 PM

The problem:

Let $P(z)$ be a polynomial of degree greater than 1, with all of it's zeroes inside the circle $C_{R}=\{z:|z|=R\}$ . show that $\int_{C_{R}} \frac{dz}{P(z)} = 0$ .

My idea for a proof:

Since every polynomial over $C$ can be writen as $P(z)=(z-a_1)(z-a_2)...(z-a_k)$ , where $a_1,...a_k$ are all the zeroes of $P(z)$ (not necessarily unique). Than we can write $\int_{C_{R}} \frac{dz}{P(z)} = \int_{C_{R}} \frac{b_1}{z-a_1} dz + ... + \int_{C_R}} \frac{b_k}{z-a_k} dz$ .

Now we can say that for every $a_i$ we can find a circle $C_{a_{i}}$ such that $C_{a_{i}}\subset C_{R}, a_i\notin C_{a_{i}}$
and therfore $\frac {1}{z-a_i}$ is analytic in $C_{a_i}}$ so we get $\int_{C_{a_{i}}} \frac{dz}{z-a_i} =0$ , but since $C_{a_{i}}\subset C_{R}$ we have $\int_{C_{R}} f = \int_{C_{a_i}} f = 0$ and that shows what we needed to show.

I would appreciate some thoughts on this proof as I am not 100% sure it is valid.

**xxp9** · February 3rd 2011, 06:16 PM

consider the coordinate change $z \mapsto \frac{R^2}{z}$ , we have
$\oint_{C_R} \frac{dz}{P(z)}= R^2 \oint_{C_R} \frac{dz}{z^2P(R^2/z)}$
If $P(z)=a_n\Pi_{k=1}^n(z-z_k)$
$P(R^2/z)=a_n\Pi_{k=1}^n(R^2/z-z_k)=\frac{a_n}{z^n}\Pi_{k=1}^n(R^2-z_k z)$

Obviously $R^2/z_k, k = 1,2, ..., n$ are all the roots of $P(R^2/z)$ , and $|R^2/z_k|>R$ for any k. That is all the roots of $P(R^2/z)$ are outside of the circle $C_R$ ,

So $R^2 \oint_{C_R} \frac{dz}{z^2P(R^2/z)}$
= $R^2 \oint_{C_R} \frac{z^{n-2}dz}{a_n\Pi_{k=1}^n(R^2-z_k z)}$

Since n>=2 the integrand is analytic inside the circle. DONE.

**skyking** · February 3rd 2011, 07:03 PM

thanks for the reply, I found a hole in my proof as well so your direction is appreciated.

**skyking** · February 3rd 2011, 07:19 PM

Thinking about this some more, what happens if the root is zero?

**roninpro** · February 3rd 2011, 10:03 PM

Here is an alternative:

Let $P(z)=a_nz^n+a_{n-1}z^{n-1}+\ldots+a_1z+z_0$ . Since all of the roots are inside the circle $C_R$ , if $R'>R$ , then

$\displaystyle \oint_{C_R} \frac{1}{P(z)}\text{ d}z=\oint_{C_{R'}} \frac{1}{P(z)}\text{ d}z$

by Cauchy's integral theorem. (In other words, expanding the circle does not change the value of the integral.) Note that when $|z|$ is sufficiently large, then $|a_n/2||z|^n\leq |P(z)|$ . Then, when $R$ is large,

$\displaystyle \left|\oint_{C_R} \frac{1}{P(z)}\text{ d}z\right|\leq \oint_{C_R} \left|\frac{1}{P(z)}\right| |\text{d}z|\leq \oint_{C_R} \frac{1}{|a_n/2||z|^n}|\text{d}z|=\frac{2\pi R}{|a_n/2|R^n}=\frac{2\pi}{|a_n/2|R^{n-1}}$

(Keep in mind that since $n\geq 2$ , we have $n-1\geq 1$ .) Taking $R\to \infty$ , the right hand side goes to zero, so the integral must be zero.

**skyking** · February 4th 2011, 01:48 PM

Originally Posted by roninpro

Here is an alternative:

if $R'>R$ , then

$\displaystyle \oint_{C_R} \frac{1}{P(z)}\text{ d}z=\oint_{C_{R'}} \frac{1}{P(z)}\text{ d}z$

by Cauchy's integral theorem. (In other words, expanding the circle does not change the value of the integral.) Note that when $|z|$ is sufficiently large, then $|a_n/2||z|^n\leq |P(z)|$

Thanks for the reply.

One question. Dosent $\frac{1}{P(z)}$ needs to be analytic inside the circle to use Cauchys integral theorem. since the roots of $P(z)$ are inside the circle it clearly isn't. Am I missing something here?

**roninpro** · February 4th 2011, 05:39 PM

Originally Posted by skyking

Thanks for the reply.

One question. Dosent $\frac{1}{P(z)}$ needs to be analytic inside the circle to use Cauchys integral theorem. since the roots of $P(z)$ are inside the circle it clearly isn't. Am I missing something here?

Another way to state my claim is that the value of the contour integral is zero in the region of the expansion (so that expanding the circle does not affect the original integral value). Since $P(z)$ has no roots in that region, $1/P(z)$ is analytic there, so it is okay to use Cauchy's integral theorem.

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