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])

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**