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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.

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  • Unless I have misunderstood, this will simply be $|\wp({ z_1,\cdots, z_n})|-1=2^n-1$ (the $-1$ being because if all roots lie in the interior, or none, we get the same value for $J(L)$, namely $0$, as you pointed out) – ocg Mar 08 '16 at 04:10
  • @JulienGodawatta, can you please explain in more detail where this came from? Specifically the $2^{n}$ part? Again, remember, I am extremely, completely clueless. And how to show this formally, please? Perhaps as an answer, with details, and not just as a comment. –  Mar 08 '16 at 04:19
  • Say $p$ has $n$ distinct roots, then $1/p$ has $n$ poles. Best case scenario, the integral around no roots, around each root, and around every pair of roots, every triplet $...$ around all $n$ roots, all yield different values. The number of these is precisely the power set of ${z_1,\cdots,z_n}$, which has size $2^n$. Noting that both the cases where the interior of $L$ contains all roots, and where it contains none, both give $J(L)=0$, we get $2^n-1$ possible values. – ocg Mar 08 '16 at 04:29
  • Two exceptions: degree of $p$ is $0$ or $1$, where we have respectively $1$ and $2$ possible values. This is because the interior of $L$ containing all roots $\Rightarrow J(L)=0$ is only true when $\text{deg},p\geqslant 2$ (and more generally, the sum of residues of the rational function $q/p$ is only $0$ when $\text{deg},p> \text{deg},q+1$) – ocg Mar 08 '16 at 04:34
  • @JulienGodawatta, is that the most rigorous way to write it? –  Mar 08 '16 at 04:44
  • Well, a less "wordy" way of writing that out be using residues (which are precisely the values of $J(L)$). Without this or the integral formula I cannot see how else you will answer this differently. Cauchy's theorem applies when $f$ is holomorphic inside $L$, which $1/p$ will not be unless it is constant, so the theorem is not relevant here. – ocg Mar 08 '16 at 04:52
  • More rigorous, well, I suppose you could write this as "if for each $S\subseteq {z_1,\cdots, z_n}$, taking $L$ such that its interior contains $S$ but not ${z_1,\cdots,z_n}\setminus S$, $J(L)$ takes different values, then we have at most $|\wp({z_1\cdots z_n})|$ etc.". As long as your statements are clear it is rigorous enough, in my opinion. – ocg Mar 08 '16 at 04:53
  • @JulienGodawatta, actually, last night it made sense to me that if the region encompassed all the roots, the integral would be zero. Tonight it doesn't. And can you please explain the $-1$ part again? –  Mar 09 '16 at 01:05
  • Sure, take $C_R$, a large circle around the origin (containing all roots). Consider $\left|\int_{C_R} \frac{\mathrm{d}z}{p(z)}\right|$. By parametrising $z=Re^{i\vartheta}$, you will see that the integrand will be of order $O(R^{1-n})$ i.e. will look like $1/(R^{n-1}+\text{stuff})$. This means that $\int_{C_R}1/p$ can be made arbitrarily small by taking $R$ arbitrarily large, i.e. it must be zero (note that this relies on the fact that the value of the integral does not change as $R$ grows, because we do not hit any other poles) – ocg Mar 09 '16 at 01:36

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