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If $f(z)$ is analytic in a simply connected domain $D$, its integral over any closed contour is $0$. I don't quite understand how the idea of contracting the contour that encloses a pole follows from that. Specifically, why are we able to say the following integral has the same value over any closed contour that encloses $z_0$, a pole of order $m$,

$$\oint_C \frac{f(z)}{z-z_0}dz$$

where $C$ is the circle of radius $\rho$ centered at $z_0$. In other words, what allows us to be able to contract an arbitrary contour (that encloses $z_0$) to the circle $C$?

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    $\begingroup$ Your first sentence is incorrect. It would be true in a simply connected domain, or in a domain that contains every point inside the contour. $\endgroup$ – Robert Israel Oct 30 '14 at 6:40
  • $\begingroup$ Yes, I forgot that. Editing my statement right now. $\endgroup$ – sodiumnitrate Oct 30 '14 at 13:35
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The easiest way is to draw a picture. I shamelessly use a picture from an entirely different subject from Paul's Online Math Notes

enter image description here

Suppose the pole is in the center, and you were first integrating around the outer circle, and wanted instead to be integrating around the inner circle. If you were to perform the integral around the region indicated by $D_1$ in the picture, you get $0$. So $$\int_{C_1} + \int_{C_5} + \int_{C_2} + \int_{C_6} = 0.$$ Similarly, from the bottom, $$\int_{C_4} + \int_{-C_6} + \int_{C_3} + \int_{-C_5} = 0.$$ Adding these together, we see that $$\int_{C_1} + \int_{C_2} + \int_{C_3} + \int_{C_4} = 0,$$ or equivalently, $$\int_{C_1} + \int_{C_4} = \int_{-C_2} + \int_{-C_3},$$ which says exactly that integrating around the outer circle gives the same as integrating around the inner circle. Notice the negatives correct the orientation in the picture.

This works essentially as is for general contours as well. I just stole the first picture I could find instead of making my own.

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