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If C is a simple closed contour such that w lies interior to C, and n > 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 at w, so it isn't holomorphic, but the integral is still zero. The Cauchy-Goursat Theorem says that if f is holomorphic over a simply connected domain D and if C is a simple closed contour in D then the contour integral of f over C is 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.
Let C be as above. Since the (Jordan) interior of C is open, there exists a small disk of radius r > 0 around w such 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 to C. 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 )
[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 at w, so it isn't holomorphic, but the integral is still zero. The Cauchy-Goursat Theorem says that if f is holomorphic over a simply connected domain D and if C is a simple closed contour in D then the contour integral of f over C is 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.
Let C be as above. Since the (Jordan) interior of C is open, there exists a small disk of radius r > 0 around w such 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 to C. 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 )
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