Talk:Cauchy's integral formula

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I think that you need also to demand the the subset U is also connectable. i.e. U is a domain in the complex plain. MathKnight 22:49, 21 Feb 2004 (UTC)

That's tacitly already there, in the comments about the disk. Michael Hardy 20:05, 27 Feb 2004 (UTC)

[edit] Statement of the theorem

It expresses the fact that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disk.

I think it needs to be said that the function is holomorphic on a region slightly larger than the disk.

No, being holomorphic inside of the disk and continuous up to the boundary of the disk is enough. You see, you can always integrate not on the boundary of the disk, but on the boundary of a slightly smaller disk inside of it, and then have that smaller disk tend to the original disk. The Cauchy integral formula will hold on the boundary of the small disk (as the function will be holomorphic around the small disk) but then, when you take the limit as the small disk approaches the original disk, the Cauchy integral formula will continue to hold on the big disk. Oleg Alexandrov (talk) 18:46, 24 November 2005 (UTC)
Oh, I missed that. You're right, of course. Brian Tvedt 03:33, 26 November 2005 (UTC)

the disk D … is completely contained in U

I think completely contained is intended to mean that the closure of D is contained in U, which would take care of my objection. Not sure how standard this terminology is, though.

Another thing: why the focus on disks? Isn't the theorem usually stated in terms of regions bounded by a simple closed curve?

Brian Tvedt 13:24, 24 November 2005 (UTC)

I guess the disk is the simplest and the most important case. If you start with a region bounded by a simple closed curve, you can always shrink it to a disk without changing the value of the integral. Simple closed curves however have the disadvantage that you must use Jordan's theorem which says that the concept of inside is well-defined, which you don't need to worry about if you deal with a disk. Oleg Alexandrov (talk) 18:46, 24 November 2005 (UTC)
I think we should follow most books on the subject, which typically do state the theorem for a simple closed curve, though it suffices to prove it for a disk, with some kind of hand-waving around the Jordan curve issue, Actually I just noticed that the article states that the theorem holds for a "closed rectifiable curve" in the region but doesn't mention that the region needs to be simply connected. Brian Tvedt
Yes, the region must be simply-connected, or alternatively, the curve together with its interior must be in the region (the latter is a bit more general). Oleg Alexandrov (talk) 04:20, 26 November 2005 (UTC)

[edit] Wrong equation?

\oint_C {z^2 \over z^2+2z+2}\,dz = \oint_{C_1} {\left({z^2 \over z-z_2}\right) \over z-z_1}\,dz + \oint_{C_2} {\left({z^2 \over z-z_1}\right) \over z-z_2}\,dz

When I go back again, I get \oint_C {2z^2 \over z^2+2z+2}\,dz. Is the equation wrong (one z too much, or a 0.5 missing)? --Abdull 17:33, 7 June 2006 (UTC)

[edit] moduli and variable substitution

  1. The article speaks of variable substitution. May Integration by substitution be meant here?
  2. Also, ...[...] Clearly the poles become evident, their moduli are less than 2 and thus lie inside the contour [...]. What is meant with moduli?

Thanks, --Abdull 17:37, 7 June 2006 (UTC)

1. Yes. 2. The modulus of a complex number z = a + bi is |z|=\sqrt{a^2+b^2}, and one checks that –1–i and –1+i both have modulus √2, which is less than 2, as claimed. -lethe talk + 22:38, 7 June 2006 (UTC)
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