Prove that $\oint_{|z|=r} {dz \over P(z)} = 0$

Question

I got stuck on this problem, hope anyone can give me some hints to go on solving this:

P is a polynomial with degree greater than 1 and all the roots of $P$ in complex plane are in the disk B: $|z| = r$. Prove that: $$\oint_{|z| = r} {{dz}\over{P(z)}} = 0$$ Here, the direction of the integral is the positive direction(actually, it can take whatever direction, because the value of the integral is 0).

What I tried so far: Applying D'Alembert-Gauss theorem, we can write $P(z) = (z-z_1)^{p_1}(z-z_2)^{p_2}...(z-z_n)^{p_n}$, here $z_i$ are complex numbers which different from each other. We can choose for each $i = 1,...,n$ a $r_i > 0$ small enough such that $B(p_i,r_i)$ are disjoint with each others and all belong to $B$. So use Cauchy theorem for compact Jordan region generated by $B$ and $B(p_i, r_i)$, it's easy to see that: $$\oint_{|z| = r} {{dz}\over{P(z)}} = \sum_1^{n}{\oint_{|z-z_i|=r_i} {{dz}\over{P(z)}}} = \sum_1^{n}{\oint_{|z-z_i|=r_i} {{\prod_{j \neq i}{1 \over {(z-z_j)^{p_j}}}}\over{(z-z_i)^{p_i}}}}$$. Then I tried to apply Cauchy theorem for each ${\oint_{|z-z_i|=r_i} {{\prod_{j \neq i}{1 \over {(z-z_j)^{p_j}}}}\over{(z-z_i)^{p_i}}}}$: $$f^{(k)}(z) = {k! \over {2 \pi i}} \oint_{\partial{B}}{{f(t) \over (t-z)^{k+1}}dt}$$, here $\partial{B}$ is notion for the boundary of the disk $B$

But I got stuck when trying to calculate the $(p_i - 1)$-th derivative for ${\prod_{j \neq i}{1 \over {(z-z_j)^{p_j}}}}$. I expect that each expression should be equal to 0, but I can't prove it. Anyone has any ideas to move on? If there's any point unclear, please don't hesitate to ask me. Thanks!

Answer 1

Using the ML inequality:

$$\left|\oint_{|z|=R}\frac{dz}{p(z)}\right|\le2\pi R\cdot\max_{|z|=R}\frac1{|p(z)|}\le2\pi R\frac1{R^n}\xrightarrow[R\to\infty]{}0$$

since $\;n\ge 2\; $ .

Why? Because of the maximum modulus principle:

$$p(z)=\sum_{k=0}^na_kz^k=z^n\sum_{k=0}^na_kz^{k-n}\stackrel{\forall\,|z|=R}\implies\left|p(z)\right|\ge|z|^n\left(\left|a_n\right|-\left|\frac{a_{n-1}}z\right|-\ldots-\left|\frac{a_0}{z^n}\right|\right)\ge |a_n|R^n$$

the last equality being true for $\;R\;$ big enough since the expression within the parentheses tends to $\;|a_n|\;$ .

Answer 2

Let $a_1,\ldots,a_r$ be the roots of $P(z)$, with multiplicities $m_1,\ldots,m_r$.

Consider the partial fraction decomposition of $\frac{1}{P(z)}$,

$$\frac{1}{P(z)}=\sum_{i=1}^r\sum_{j=1}^{m_i}\frac{c_{i_j}}{(z-a_i)^j}.$$

Then since $\text{deg}(P) > 1$,

$$0=\lim_{z\rightarrow\infty}\frac{z}{P(z)}=\lim_{z\rightarrow\infty}z\sum_{i=1}^r\sum_{j=1}^{m_i}\frac{c_{i_j}}{(z-a_i)^j}=\sum_{i=1}^rc_{i_1}.$$

Using this fact, and then applying Cauchy's integral formula (for derivatives) to $\int_{|z|=R}\frac{dz}{P(z)}$, we have,

$$\int_{|z|=R}\frac{dz}{P(z)}=\int_{|z|=R}\sum_{i=1}^r\sum_{j=1}^{m_i}\frac{c_{i_j}}{(z-a_i)^j}dz=\int_{|z|=R}\sum_{i=1}^r\frac{c_{i_1}}{z-a_i}dz=2\pi i\sum_{i=1}^rc_{i_1}=0.$$

QED

Answer 3

Another solution can be based on the change of variable $z = \frac{R^{2}}{w}$.

If $$ p(z) =cz^{m}\prod_{k=1}^{n-m}(z-a_{k}) $$ where $n \ge 2$, $0 \le m \le n$ and $c \neq0$ is a constant, then $$ p\left(\dfrac{R^{2}}{w}\right) = cR^{2m}w^{-n}\underbrace{\prod_{k=1}^{n-m}(R^{2}-a_{k}w)}_{= q(w)} = cR^{2m}\dfrac{q(w)}{w^{n}}. $$ Futhermore, if $0 < |a_{k}|<R$ then $\left|\frac{R^{2}}{a_{k}}\right| > R.$ We are now prepared to change the variable $z$ to $w = \frac{R^{2}}{z}$. The circle $|z| = R$ will be transformed to the circle $|w| = R$ but with a reversed orientation. Thus $$ \oint_{|z|=R}\dfrac{1}{p(z)}\, dz = -\oint_{|w|=R}\dfrac{w^{n}}{cR^{2m}q(w)}\cdot\dfrac{-R^{2}}{w^{2}}\, dw = \dfrac{R^{2-2m}}{c}\oint_{|w|=R}\dfrac{w^{n-2}}{q(w)} \, dw. $$ Since $n \ge 2$ and $q(w)$ has no zeros inside the circle $|w| = R$ the last integral will be $0$.