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.

  • $\begingroup$ 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. $\endgroup$ – Mathmo Apr 30 '11 at 19:19
  • $\begingroup$ 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... $\endgroup$ – Fabian Apr 30 '11 at 19:20
  • $\begingroup$ @Fabian: I don't understand what you're saying. What about $n=1$ and $a_1 = 0$? $\endgroup$ – t.b. Apr 30 '11 at 19:22
  • $\begingroup$ I think the residue theorem is useful here. $\endgroup$ – Jose27 Apr 30 '11 at 19:25
  • $\begingroup$ 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? $\endgroup$ – Raskolnikov Apr 30 '11 at 19:27

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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 $2$. $\endgroup$ – Gerry Myerson May 1 '11 at 4:30
  • $\begingroup$ @Gerry Myerson: you are right. I forgot that zero is also a number. I corrected the text. Thank you very much! $\endgroup$ – user10287 May 1 '11 at 8:15
  • $\begingroup$ @user10287, can you please explain why the $A_{i}$ have to add up to zero? $\endgroup$ – 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 :)

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson May 1 '11 at 12:47
  • $\begingroup$ @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. $\endgroup$ – ALannister Mar 8 '16 at 3:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.