Talk:Cauchy's integral formula

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)

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)

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)

Remember that the first contour is broken up into two distinct contours, one enclosing circle for each singularity. The integrands only add if the contours of integration are the same. So you can't simply add the integrands to get $$\oint_C {2z^2 \over z^2+2z+2}\,dz$$. Rpchase 06:11, 18 December 2006 (UTC)

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)

Seeking comments on major edit of this article
I'm proposing a major update of this article. You can read what the new version would look like at my user page. The following is a summary of the changes.


 * I reworked the organization of the article a bit, adding some new subsections.
 * I've re-written the statement of the theorem so that it is now stated for a simple, closed contour C and not just for a disk.
 * I added a subsection on Cauchy's integral formula as an integral representation for holomorphic functions.
 * I replaced the proof sketch with a complete proof of the theorem.
 * I added an additional example of the use of Cauchy's integral formula, and expanded the example that was there to explain why the contour integral in it can be replaced with two contour integrals.
 * I included an example of the use of Cauchy's differentiation formula.
 * I've added several illustrations.
 * I've added several new references.
 * I removed the statement that for holomorphic functions "differentiation is equivalent to integration," which is a notion so vague as to be not even wrong.

All comments are welcome, including negative ones. In particular, I want to know if you think I'm headed in the right direction concerning this entry; obviously, I think that I am, but there would not be any point in making changes that other users do not find are for the better. I suspect that the re-written article is too long, and I would like to know what all interested readers think of the length. I think the my draft proposal is an improvement on the current article, but you certainly can disabuse me of that idea if that's how you feel. Frank 1729 (talk) 19:41, 19 January 2008 (UTC)


 * I really dislike the way you spread formulas out vertically. This also makes the extra trivial steps you include extra annoying. Example:

\begin{align} \oint_{C_1} {g(z)} &= \oint_{C_1} {f_1(z) \over z-z_1}\, dz\\ &=2\pi i\cdot f(z_1)\\ &=2\pi i\cdot \frac{z_1^2}{z_1-z_2}\\ &=2\pi i\cdot \frac{(-1+i)^2}{-1+i-(-1-i)}\\ &=2\pi i\cdot \frac{-2i}{2i}\\ &=2\pi i (-1)\\ &=-2\pi i. \end{align} $$
 * that I would write like
 * $$\oint_{C_1} {g(z)}\, dz = \oint_{C_1} {f_1(z) \over z-z_1}\, dz = 2\pi i f(z_1) = 2\pi i \frac{z_1^2}{z_1-z_2} = 2\pi i \frac{(-1+i)^2}{-1+i-(-1-i)} = -2\pi i.$$
 * You missed a "dz" in the first term and I dislike your use of "\cdot" for normal multiplication.
 * I really like your effort at expanding and improving this article though. I will take another look after you fix my above concern. --MarSch (talk) 11:41, 20 January 2008 (UTC)


 * Your comments concerning vertical spacing are well taken. As you know, many authors (or editors) prefer to align equations and inequalities vertically, and others like more of a horizontal style; yet others mix it up a bit. By breaking up a chain of equations vertically I was hoping that the mathematical text would fit monitors of different sizes without the reader having to scroll horizontally. For instance, your suggested edit of my work actually runs off the right side of my monitor here at home, not too badly, though; but your line really runs off the screen on my ultra mobile pc; I suspect it will look fine on my monitor at work. However, my vertical lines read well on all the screens, large and small, that I've tested it on. I think, though, that I can still improve my style and I will try to follow your suggestion of horizontal of spacing while at the same time keeping in mind how the article will look in different monitors.


 * I missed one dz and I have corrected that. I often use cdots for my own benefit while I'm editing and remove them afterward. I will probably remove many of them now. Thanks for your input by the way; I have found it very useful. I hope you give my draft another look. Frank 1729 (talk) 14:57, 20 January 2008 (UTC)


 * seems rather long-winded. in particular, the more general version hinges on on fact that path integrals are invariant under homotopy of paths. filling in trivial gaps in the proof from the current version, while merely quoting this result, adds nothing. Mct mht (talk) 23:47, 21 January 2008 (UTC)


 * Mct mht, thanks for your comments. Writing the sketch of a proof is difficult because in skipping steps, even ones that you may think are trivial, can lead to mistakes. For instance, the inequality


 * $$\left | \frac{1}{2 \pi i} \oint_C { {f(z) \over z-a} \,dz} - f(a) \right |$$



\leq \frac{1}{2 \pi i} \oint_C \frac{ |f(z) - f(a)| } {z-a} \,dz \rightarrow 0. $$


 * in the current version contains a mathematical error that renders it meaningless. I believe you were the one who wrote that line, last August, so at the very least you should fix your error as soon as possible. There are those who think that the Wikipedia should present only proof sketches for the more technical theorems; I respect that view. But you should also respect the view of those like myself who think that complete and correct proofs are not necessarily long-winded; they are just correct. Frank 1729 (talk) 02:51, 22 January 2008 (UTC)


 * well, you caught the missing "| &middot; |"... let me state again, if i may: adding trivialities while completely glossing over the essential point of the generalization is pointless. a more sensible and honest expansion of article would seem to include a word or two about why path integrals are invariant under homotopy of paths. anyway, just a suggestion. Mct mht (talk) 03:27, 22 January 2008 (UTC)


 * Your point about adding a little explanation about why integrals are invariant under homotopy is a good one. I'm not sure it will be easy to do but I will give it a try. I don't quite see eye to eye with you concerning what you call trivialities; the proof I wrote is really a fairly standard one and appears, with minor variations, in many textbooks. However, I won't add it to the article because unless I get a lot of support for it, and I don't think I will. Instead, I think I might re-write the proof sketch slightly to connect a couple of ideas together. You or, another use, could revert it back later on if you'd like. Thanks for you input. It is appreciated. Frank 1729 (talk) 03:47, 22 January 2008 (UTC)


 * ok, Frank, have at it. hope to see you around. Mct mht (talk) 04:25, 24 January 2008 (UTC)

"Bach's" differentiation formula?
For some reason, this article claimed that the Cauchy differentiation formula (i.e. derivative version of Cauchy integral formula) is sometimes called "Bach's differentiation formula". I expect this is almost certainly vandalism. It was introduced on 9th Aug 2009 from an IP user. I've certainly never heard this name and a quick Google turns up only Wikipedia pages and a forum post that I expect got the name from Wikipedia.

I've reverted it to "Cauchy". Tcnuk (talk) 12:00, 16 April 2010 (UTC)

Example is unclear
It is never stated what exactly 'f' is, what is its relation to 'g', and then it is imemdiately used to calculate something. It took me quite some time to figure out what is going on. It would be very benefitial if someone who understands what is going on would fix it and maybe add a little bit of "reasoning" about what is going on. 94.112.136.34 (talk) 15:54, 3 June 2014 (UTC)