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Here's the problem: Let $G$ be the punctured unit disk (i.e missing the point $0$). Let $f:G \mapsto \mathbb{C}$ be analytic. Suppose $\gamma$ is a closed curve in $G$ homologous to $0$ (that is the winding number of $\gamma$ about any point outside of $G$ is $0$). Then what is the value of $\int_{\gamma} f$?

I said that here Cauchy's Theorem apply since $f$ is analytic and $\gamma$ is homologous to $0$ in $G$ and since $0$ is not in $G$ so the winding number of $\gamma$ about $0$ is $0$. Is this true?

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Why has this question been downvoted? –  user786 Jul 24 '11 at 4:04
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If like you say $\gamma$ does not enclose $0$ and is otherwise in $G$, then there is a neighborhood of $\gamma$ that is simply connected and hence Cauchy's Theorem applies; the value is 0. I don't know why it was downvoted. –  anon Jul 24 '11 at 4:12
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Thx for the answer anon. –  user786 Jul 24 '11 at 4:14

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up vote 4 down vote accepted

Submitting my comment as an answer so the question can be done with. Since $\gamma$ does not enclose $0$ and is otherwise contained in the unit disk, there is a neighborhood of $\gamma$ that is simply connected and therefore Cauchy's Theorem applies, $\oint_\gamma f dz=0$.

EDIT: GEdgar has pointed out that if the path can cross itself then it's still technically possible for $\gamma$ to enclose $0$. Regardless, the curve will still be decomposable into a finite number of closed curves of two types - those that are contained in a simply connected neighborhood (for which CT applies), and those that encircle the origin in either clockwise or counterclockwise fashion (the former and latter are equal in number, hence all of these curves will cancel each other out). Alternatively, $\gamma$ is necessarily at least homeomoprhic to a closed curve contained in a simply connected neighborhood (without passing over $0$). I think that nails it down now.

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I assumed $\gamma$ was allowed to cross itself, for example. It is still true that the integral is zero, but not that it must be enclosed in a simply connected neighborhood. –  GEdgar Jul 24 '11 at 13:12
    
@GEdgar: Oh, if it crosses itself then it's still possible for it to enclose the value $0$ - took me a second to be able to visualize it. I'll amend my post with my augmented reasoning. (I was wondering why I got the downvote.) +1 for the help. –  anon Jul 24 '11 at 13:51

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