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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.

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    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.2011-04-30
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    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...2011-04-30
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    @Fabian: I don't understand what you're saying. What about $n=1$ and $a_1 = 0$?2011-04-30
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    I think the residue theorem is useful here.2011-04-30
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    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?2011-04-30
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    @Theo Buehler: I don't know if I know what I'm saying... ;-)2011-04-30
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    @Raskolnikov: I think you're right - thanks for pointing that out. I didn't really think that bit through properly. So I need to make it a bit more general, but that's not too difficult.2011-04-30
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    @Theo Buehler: the case $n=1$ is special as the function does not fall off fast enough that the contour integral would be 0 if you evaluate it far away. So maybe I should have said it like that and not refer to some Riemann sphere...2011-04-30
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    @Mathmo, can you please explain more why $A_{1}+A_{2}+\cdots+A_{n} = 0$?2016-03-08

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