If(adsbygoogle = window.adsbygoogle || []).push({}); Cis a simple closed contour such thatwlies interior toC, andn> 1, then

[tex]\int_{C} \frac{dz}{(z-w)^n} = 0.[/tex] I'm confused because the function [itex]f(z) = (z-w)^{-n}[/itex] has a pole atw, so it isn't holomorphic, but the integral is still zero. The Cauchy-Goursat Theorem says that iffis holomorphic over a simply connected domainDand ifCis a simple closed contour inDthen the contour integral offoverCis zero. Does this mean there exist functions which are not holomorphic but behave like holomorphic functions under contour integration? (Silly question, I know)

I also want to know how to prove that this integral is zero. Here is my attempt.

LetCbe as above. Since the (Jordan) interior ofCis open, there exists a small disk of radiusr> 0 aroundwsuch that its boundary, which I will parametrize by [itex]\gamma(t) = w + re^{it}[/itex] for [itex]t\in [0, 2\pi][/itex], is contained entirely interior toC. Then

[tex]\int_{C} \frac{dz}{(z-w)^n} = \int_0^{2\pi} \frac{1}{(\gamma(t) - w)^n}\gamma'(t) dt = \int_0^{2\pi} \frac{r i e^{it}}{r^n e^{int}} dt = \int_0^{2\pi} r^{1-n} i e^{i(1-n)t} dt = \frac{r^{1-n}}{1-n}(e^{2\pi (1-n)i} - 1) = 0.[/tex](I'm using the complex analysis notes at http://math.fullerton.edu/mathews/c2003/CauchyGoursatMod.html [Broken])

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# Contour integral of dz/(z-w)^n

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