# 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.

• I've had another thought: as soon as you fix $A_n$, the other possible values of the integral are all fixed also. So the question (I think) widdles down to "how many possible values of $A_n$ are there?". I think. – Mathmo Apr 30 '11 at 19:19
• It is true that when $\gamma$ goes around all the pole that the answer is zero. You can see that this is the case because you can imagine the complex plane to be a sphere (Riemann sphere) and then you can take the complement of $I(\gamma)$ (which has no poles inside) as the interior... – Fabian Apr 30 '11 at 19:20
• @Fabian: I don't understand what you're saying. What about $n=1$ and $a_1 = 0$? – t.b. Apr 30 '11 at 19:22
• I think the residue theorem is useful here. – Jose27 Apr 30 '11 at 19:25
• I guess I missed something, but why don't you consider the case when the interior of the loop $I(\gamma)$ contains $a_2$ and not $a_1$, just to name one? – Raskolnikov Apr 30 '11 at 19:27

## 2 Answers

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.

• Isn't this off by $1$? If $n=2$, the contour can enclose both, either, or neither, which is $4$ cases; both and neither give zero, so that's $3$ different values; your formula gives $2$. – Gerry Myerson May 1 '11 at 4:30
• @Gerry Myerson: you are right. I forgot that zero is also a number. I corrected the text. Thank you very much! – user10287 May 1 '11 at 8:15
• @user10287, can you please explain why the $A_{i}$ have to add up to zero? – ALannister Mar 8 '16 at 3:17

(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 :)

• The question asks for the number of values the integral can take - I interpret that to mean, the maximum. I'm confident you can choose the $a_i$ so there are no other relations. – Gerry Myerson May 1 '11 at 12:47
• @user8268, I'm currently working on this problem, but I am not allowed to use residues or Cauchy's Integral formula to solve it. Can you explain your answer without using residues? And feel free to give a more detailed solution than what the OP originally asked for. – ALannister Mar 8 '16 at 3:17