Show that the complex closed line integral $\oint\frac{\mathrm{d}z}{p(z)}$ is $0$ ($p$ is polynomial) Let $p$ be a polynomial of degree $n\geq2$ and has $n$ different roots $z_1,\dots,z_n$.
Prove that $\oint\frac{\mathrm{d}z}{p(z)}=0$ where the closed path is large enough so that all roots are in the interior of the path.
I need this as a lemma to a question that was asked in complex analysis. I tried to think of something clever to do using the Argument Principle but couldn't find a function whose logarithmic derivative would be proportional to $\frac{1}{p}$. I thought something along the lines of $p \cdot p^{(1)} \cdot p^{(2)} \cdot \dots \cdot p^{(n)}$ would work but had a bunch of things "left over" in the enumerator that ruined it. 
That's where I'm at... 
 A: Change coordinates: map
$$z\mapsto {1\over z}=\xi,\quad -{d\xi\over \xi^2} = dz$$
Then you have the integral around a large circle, $C_R$ transformed into
$$\oint_{C_{1/R}}{d\xi\over \xi^2p({1\over\xi})}$$
By the Cauchy estimate we have
$$\left|\oint_{C_{1/r}}{d\xi\over \xi^2p({1\over\xi})}\right|\le {2\pi\over R}\cdot R^2\cdot\sup_{\xi\in C_{1/R}}\left|p\left({1\over\xi}\right)\right|\le {C\over R}$$
because $\text{deg}(p)\ge 2$.
The way to see this is zero immediately is that, since $z=\infty$ is not a simple pole (since $\text{deg } p\ge 2$) there is no residue at infinity.
A: List the roots of $p$ as $z_1,\dots,z_n$, using the fundamental theorem of algebra. Let $r=\max |z_n|$. If you now integrate around any circle of radius $R>r$ centered at 0, the integral will be
$$2 \pi i \sum_{k=1}^n \text{Res} \left ( \frac{1}{p},z_k \right )$$
which does not depend on $R$. Now the triangle inequality gives us that $|p(z)| \geq c_1 R^n - c_2 R^{n-1}$ if $|z|=R$, so by an ML estimate we have that the integral is at most $\frac{2 \pi R}{c_1 R^n - c_2 R^{n-1}} \to 0$ as $R \to \infty$. But the integral does not depend on $R$, so it must be zero.
A: A direct estimate works as well. Note that the fact that the roots of $p$ are distinct is irrelevant, the degree of $p$ is what matters.
Let $n \ge 2$ be the degree of $p$. Choose $L$ so that for some $K>0$ we have $|p(z)| \ge K |z|^n$ for all $|z| \ge L$. Note that $z \mapsto { 1 \over p(z)}$ is analytic on $B(0,L)^c$. 
Let $\gamma_R(t) = R e^{i 2 \pi t}$
If $R,R' \ge L$, then $\gamma_R,\gamma_{R'}$ are homotopic on $B(0,L)^c$.
Hence we have $\int_{\gamma_R} {dz \over p(z)} = \int_{\gamma_{R'}} {dz \over p(z)}$.
If $R,R' \ge L$, we have 
$|\int_{\gamma_R'} {dz \over p(z)} | = |\int_{\gamma_R} {dz \over p(z)} | = |\int_0^1 {\gamma_R'(t) \over p(\gamma_R(t)) } dt |  \le \int_0^1 |{\gamma_R'(t) \over p(\gamma_R(t)) }| dt \le 2 \pi {R \over K R^n}$.
Since $R\ge L$ was arbitrary, we see that $\int_{\gamma_R'} {dz \over p(z)} = 0$.
