My question is related to this one, but it is not a duplicate, because I am not allowed to use Residues or Cauchy's Integral Formula to solve it. The only tools I have at my disposal are Cauchy's Theorem and Cauchy's Theorem for Multiply Connected Domains.
I need to solve the following:
Given a polynomial $p(z) = (z-z_{1})(z-z_{2})\cdots (z-z_{n})$, where $z_{k} \neq z_{j} $for $k \neq j$, let $L$ be a simple closed rectifiable curve that does not pass through any of the roots of $p(z)$. How many distinct values of $\int_{L}\frac{dz}{p(z)}$ can one obtain, at most, by changing $L$? (In other words, the integral $\int_{L}\frac{dz}{p(z)}$ is a function $J(L)$ of $L$. How many values can it take on, at most?)
Using partial fraction decomposition, we know that what we would want is to have $\displaystyle \int_{L}\frac{dz}{p(z)} = \int_{L}\frac{1}{(z-z_{1})(z-z_{2})\cdots(z-z_{n})}dz = \int_{L} \frac{A_{1}}{(z-z_{1})}+\frac{A_{2}}{(z-z_{2})}+\cdots + \frac{A_{n}}{(z-z_{n})}dz$.
So, $A_{1}(z-z_{2})\cdots(z-z_{n})+A_{2}(z-z_{1})(z-z_{3})\cdots(z-z_{n})+\cdots + A_{n}(z-z_{1})(z-z_{2})\cdots(z-z_{n-1}) = 1$,
and each $\displaystyle A_{k} = \frac{1}{(z_{k}-z_{1})(z_{k}-z_{2})\cdots (z_{k}-z_{k-1})(z_{k}-z_{k+1})(z_{k}-z_{n})}$.
But, I don't know where to go from there.
The other question mentioned that $\sum_{i=1}^{n}A_{i} = 0$, which makes absolutely no sense to me. If this is true, I would welcome an explanation why.
I get that as $L$ gets larger to encompass all of the $z_{i}$ (even if the $z_{i}$ are not in order from smallest to largest, we can, WLOG, order them that way), the integral will become zero, by Cauchy's Theorem. But, this is as far as I have been able to reason.
Feel free to be extremely detailed in your answer. I am completely clueless.