Complex integration and simple curve Let $\gamma$ be a simple curve which contains $n$ different points $z_{1}, z_{2},...,z_{n}$. Prove that: $$\int_{\gamma} \frac{1}{(z-z_{1})(z-z_{2})...(z-z_{n})} = 0 $$
I do not know how to do this exercise, I tried using partial fractions, but it is impossible to keep working them. If there is a theoretical argument really appreciate it.
 A: Assuming $n \ge 2$ (otherwise the claim is false), show that the integral is independent of $\gamma$ (as long as $\gamma$ is a simple closed curve enclosing all the points). Then take $\gamma$ as a circle of radius $R$ and compute the limit as $R \to \infty$ (this is the step where you need $n \ge 2$).
A: An identity coming from Lagrange interpolation:
$$\sum_{k=1}^n \frac{ \prod_{j\ne k} (z-z_k)}{ \prod_{j\ne k} (z_j-z_k)} =1 $$
Indeed, the LHS is a polynomial of degree $\le n-1$ that takes the value $1$ at $n$ distinct point $z_k$, $1\le k\le n$. 
Dividing both sides by $\prod_{k=1}^n (z-z_k)$ we get 
$$\frac{1}{\prod_{k=1}^n (z-z_k)} = \sum_{k=1}^n \frac{ 1}{ \prod_{j\ne k} (z_j-z_k)}\cdot \frac{1}{z-z_k}$$
This identity is equivalent to $\frac{1}{\prod_{k=1}^n (z-z_k)}$ having a simple pole at $z_k$ with residue 
$\frac{ 1}{ \prod_{j\ne k} (z_j-z_k)}$; we can also obtain the residue as $\frac{1}{(\prod_{l=1}^n (z-z_l))'_{z=z_k}}$. Back to the problem. 
Cauchy's formula tell us that 
$$\int_{\gamma} \frac{ 1}{ \prod_{j\ne k} (z_j-z_k)} = 2 \pi i \cdot (\sum \text{residues at poles inside } \ \gamma ) $$ or 
$$2\pi i \cdot \sum_{k=1}^n \frac{ 1}{ \prod_{j\ne k} (z_j-z_k)}$$. We need to show that 
this sum is zero (for $n>1$ ). Indeed, look again at the identity $\sum_{k=1}^n \frac{ \prod_{j\ne k} (z-z_k)}{ \prod_{j\ne k} (z_j-z_k)} =1 $. The coefficient of $z^{n-1}$ on RHS is $0$ ( since $n>1$) , and on LHS is $\sum_{k=1}^n \frac{ 1}{ \prod_{j\ne k} (z_j-z_k)}$, which is therefore also $0$.
A: Here's an expansion of the suggestion from @mrf.  First, fix $z_0\in\mathbb C$ and $M>0$ and note that for 
$$\left|z\right|\geq (M+1)\left|z_0\right|$$ 
we have
$$\left|z-z_0\right| \geq \left|z\right|-\left|z_0\right| \geq (M+1)\left|z_0\right| - \left|z_0\right| = M\left|z_0\right|,$$
by the reverse triangle inequality.  Applying this to each $z-z_k$ in the product we get
$$\left|\frac{1}{(z-z_1)(z-z_2)\cdots(z-z_n)}\right|\leq\frac{1}{M^n \left|z_1\right| 
\left|z_2\right|\cdots\left|z_n\right|}.$$
As a result, for large $R$, we have
$$\int_{C_R}\left|\frac{1}{(z-z_1)(z-z_2)\cdots(z-z_n)}\right|dz 
\leq
\frac{2\pi R}{(R-1)^n \left|z_1\right| 
\left|z_2\right|\cdots\left|z_n\right|} \rightarrow 0$$
as $R\rightarrow\infty$, provided $n>1$.  Of course, the curve $C_R$ is meant to be a positively oriented circle of radius capital $R$ and, since these are all homotopic, the value of the integral is independent of $R$ - so long as $R$ is large enough so that $C_R$ contains all the $z_k$s.  Thus,
$$\int_{C_R}\frac{1}{(z-z_1)(z-z_2)\cdots(z-z_n)}dz = 0.$$
Furthermore, the result holds for any curve $\gamma$ that is homotopic to $C_R$.

The same argument can be used to show that
$$\int_{C_R}\left|\frac{1}{(z-z_1)(z-z_2)\cdots(z-z_n)} - \frac{1}{z^n}\right|dz = 
\int_{C_R}\left|\frac{z^n-(z-z_1)(z-z_2)\cdots(z-z_n)}{z^n(z-z_1)(z-z_2)\cdots(z-z_n)}\right|dz \rightarrow 0$$ 
as $R\rightarrow\infty$. As a result, the integral must have the same value as 
$$\int_{C_R} \frac{1}{z^n}dz,$$
which is easy to compute directly.
