Number of distinct values of $ \oint_\gamma \frac{dz}{(z-a_1)(z-a_2)...(z-a_n)}$ for closed $ \gamma $

I've been asked to find the number of distinct values that $ \oint_\gamma \frac{dz}{(z-a_1)(z-a_2)...(z-a_n)}$ can take for simple closed curves $ \gamma $ not passing through any of the $ a_i $.

My thoughts so far:

I know I should be thinking about partial fractions and using Cauchy's Integral Formula. WLOG set $ |a_1| < |a_2| < ... < |a_n| $. Let $ I(\gamma) $ denote the interior of the closed curve $ \gamma $. Now there are $ n $ possible scenarios that could yield different values for the integral: $ a_1 \in I(\gamma) $, $ a_1 $ and $ a_2 \in I(\gamma) $, ... , $ a_1, a_2, ... , a_n \in I(\gamma) $. From the integral formula, each root $ a_i $ of $ p(z) = (z-a_1)...(z-a_n) $ will make a non-zero contribution iff $ a_i \in I(\gamma) $. Writing as partial fractions, we have:

$ \frac{1}{p(z)} = \frac{A_1}{z-a_1} + \frac{A_2}{z - a_2} + ... + \frac{A_n}{z - a_n} $, which leads to

$ A_1 (z-a_2)...(z-a_n) + A_2(z-a_1)...(z -a_n) + ... + A_n(z-a_1)...(z-a_{n-1}) = 1$ (*)

And so we see that the possible values of the integral are $ 2\pi i \left( A_1 \right) $, or $ 2\pi i \left( A_1 + A_2 \right) $, or ... $ 2 \pi i \left( A_1 + ... + A_n \right) $.

But the relationship (*) gives restrictions on the values of the $ A_i $. For example, $ A_1 + ... + A_n = 0 $ (by considering the coefficient of $ z^{n+1} $).

I'm unsure about making the final leap to the answer - how should I count the possible values of the $n $ sums? I would greatly appreciate any advice, either in the form of pointing out mistakes in my reasoning so far or hints on where to go next. I'd rather not have a full answer (or a hint that makes my task trivial). Thanks.

Solution 1:

That $\sum_i A_i=0$ is the only condition the $A_i$'s generically have to fulfill. Thus, the value of the integral can be $$A_1, A_2, \dots, A_n,$$ $$A_1 + A_2, A_1 + A_3, \dots, A_{n-1}+A_n,$$ $$\dots$$ $$A_{1}+A_2 + \dots + A_{n-1} =-A_{n}, -A_{n-1}, \dots , -A_1.$$

Of the first line there are $n={n\choose 1}$ distinct values. The second line yields ${n\choose 2}$ distinct values. In total, we have $$\sum_{k=0}^{n-1} {n \choose k} = 2^n -1$$ distinct values of the integral; $k=0$ counts the value zero which you get when you enclose nothing or everything. Note that for $n=1$ the formula does not work. As there are obviously two results (0 or $A_1$) possible.

Solution 2:

(this is a little correction of user10287's answer - we need to care about the orientation of the curve! so we get twice as many possibilities. It's the residue at infinity being $0$ that cuts the number back by $2$)

The curve divides the Riemann sphere to two parts. The integral is given by the sum of residues which are on the positive side of the curve. The residue at $\infty$ is $0$ (unless $n=1$!). So we get $2^n-1$ possibilities (or $2$ in the case $n=1$) ($2^n$ for the choice of the subset of $a_1,\dots,a_n$ which is on the positive side of the curve, and $-1$ for the case when nothing is on the positive side, giving the same answer as if everything is at on the positive side. It wouldn't fit as a comment, so I post it as an answer).

It is now a question whether all the remaining $2^n-1$ possibilities are different. The value of $A_1$ is $1/((a_2-a_1)(a_3-a_1)\dots(a_n-a_1))$ (multiply your equation by $z-a_1$ and then substitute $z=a_1$) (and similarly for other $A_i$'s). The question is whether there is any non-trivial identity $A_{i_1}+\dots+A_{i_k}=A_{j_1}+\dots+A_{j_l}$ besides $A_1+\dots +A_n=0$. It certainly depends on $a_i$'s - and for the moment I don't see why there is no such identity :)