Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 0 down vote favorite

Would it be right to conclude that the closed integral $\int_P z^{1\over 3}dz=0$ for $P$ being the circle $|z-z_0|=|z_0|$ by Cauchy's theorem despite the fact the the integrand is not holomorphic at one point, namely $z=0$, on the path $P$?

share|improve this question
No, to apply Cauchy's theorem you must have a holomorphic function. Perhaps you can get this result by taking a limit after applying Cauchy's theorem... – GEdgar Feb 18 '12 at 22:17
@GEdgar: Would you mind saying a little bit more on what you mean by taking a limit after applying Cauchy's theorem? – Frank Feb 18 '12 at 22:20
You want an integral around a circle passing through $0$. You know an integral around a slightly smaller circle (not inclosing $0$) does have value $0$ by Cauchy's theorem. Try to use some limit argument now, to show the integral on the large circle is the limit of the integral on smaller circles. – GEdgar Feb 18 '12 at 22:24
@GEdgar: Thanks, but doesn't that require some sort of continuity or something of the sort? For example if our bad point is a singularity we can't really take the limit? I might just be confused... – Frank Feb 18 '12 at 22:32
Continuity, yes. And a fairly simple application of it. But (in answer to your question) this is not simply an immediate application of Cauchy's theorem. – GEdgar Feb 19 '12 at 3:26
show 1 more comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.