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One of the results of the Cauchy-Goursat theorem is that for any simple closed countour $C$ that contains the point $z_0$: $$\oint_C{\frac{dz}{(z-z_0)^n}} = \begin{cases}2\pi i & n=1 \\ 0 & n \text{ an integer} \ne 1 \end{cases} $$ I can see why this is true mathematically, but I don't understand the intuition behind it. For negative integers $n$, this makes sense because the function becomes a polynomial and the line integral is $0$, but why is it also true for other positive integers not equal to $1$? What makes a simple pole special such that the contour integral about it is $2\pi i$ but the contour integral about higher order poles is $0$? It seems to me that the graph of the pole would look similar regardless of it's order. Again, I know why this result is true mathematically; I am just confused about its physical interpretation.

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  • $\begingroup$ what makes you think that intuitively, integrating polynomial over line is 0? $\endgroup$ – user27126 Dec 11 '13 at 7:42
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    $\begingroup$ One difference is that all but $1/z$ has well defined antiderivative everywhere except the origin (Although $(In z)' = 1/z$, $Inz$ is defined only up to a branch) $\endgroup$ – user99914 Dec 11 '13 at 7:43
  • $\begingroup$ @Sanchez I can grasp the idea that the closed integral of a holomorphic function is $0$ because I can see from the graph how it is possible for the values to cancel out over a closed loop. But it's beyond me why it would be different for just a simple pole, I can't see why that is true by looking at a graph. $\endgroup$ – hesson Dec 11 '13 at 7:47
  • $\begingroup$ You can also write down the anti-derivative like John said, as long as $n \neq 1$. So if you can see that for polynomials, you should be able to see that for $n \ge 2$ too. As for $n = 1$, interpretation of integrating $dz/z$ as winding number is probably helpful. $\endgroup$ – user27126 Dec 11 '13 at 7:56
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A way you can see this is the following. Let's consider the integral : $$\int_{C} \frac{1}{z^{2}} dz$$ where $C$ is the curve given by the contour $|z|=1$, which is the circle of $radius \ R=\ 1$ centered at $(0,0)$. In this case $z_0=0$, $n=2$ and the things are really simple. Now let's try parametrizing the curve $C$: $$ z=e^{it}$$ $$dz=i(e^{it})dt$$ $$0<t<2\pi$$ In this case we use the formula : $$ \int_{C}\ f(z) \ dz = \int_{a}^{b}\ f(z(t))\ z'(t) \ dt $$ Plugging in the integral, the integral becomes : $$\int_{0}^{2\pi}\ \frac{i(e^{it})}{e^{2it}} dt = \int_{0}^{2\pi}\ i(e^{-it}) dt = \left. -e^{-it} \right|_{0}^{2\pi}=0 $$ This works for higher degree exponential too and also shifted from the origin $z_0=a$ if the multiple pole is contained within the curve $C$.

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  • $\begingroup$ You can use \left. \right|_0^{2 \pi} to make the vertical line the appropriate length. $\endgroup$ – André 3000 Mar 14 '14 at 18:31
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It should at least be clear that the integrals are 0 for $n \ne 1$ because $(z-z_{0})^{-n}$ has an obvious anti-derivative for $n \ne 1$.

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