Talk:Cauchy-Riemann equations

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[edit] Holomorphic implies analytic

If f(z)=u(x,y)+ i v(x,y)satisfies cauchy reimann equation alongwith condition that u,v and all its 1st order patial derivatives are continuous, then the function is analytic ...........plz prove that --anon

See Holomorphic functions are analytic. Oleg Alexandrov (talk) 21:30, 9 November 2005 (UTC)

[edit] "Not necessarily sufficient" ??!!

If a condition is "not necessarily sufficient" then it is not sufficient, surely?

I don't think the use of the more complex term is helpful. I'll take it out, unless someone else wants to, or wants to make a case for it.

David Young

I'm working on the survey article "complex analysis". I'm puzzled by the "necessary but not sufficient" language in this article. I'd like to rewrite this, because I don't think it's true. But I don't want to start another controversy inadvertently.

  • If the Cauchy-Riemann equations are satisfied in a neighborhood, f(z) has a first derivative there. It doesn't matter whether the first partial derivatives are continuous functions or not ... we just have to be able to integrate them to get u and v.
  • By the Cauchy-Goursat theorem, if f(z) has a first derivative in a neighborhood, it's analytic there. No requirement of continuity has to be imposed on the partials to do this proof, either.
  • So the Cauchy-Riemann equations are in fact sufficient to prove analyticity, indirectly.

Am I missing something here? DavidCBryant 15:48, 7 December 2006 (UTC)

I've been thinking about this some more, and I think the current language (necessary but not sufficient) may have been written by somebody who only studied one textbook. Some authors develop the theory of functions by assuming that the partial derivatives in C-R are continuous, and eventually get around to Cauchy-Goursat. Other authors start off with complex integration, and prove Cauchy-Goursat first. Just a thought. DavidCBryant 17:05, 7 December 2006 (UTC)
I'm pretty sure the article is correct as it stands, as some sort of continuity assumption is required as well as the Cauchy-Riemann equations. For example: let f(x+iy) = 0 if one or both of x, y is zero, and f(x+iy) = 1 if neither of x, y is zero. Then, for this f, all the partial derivatives exist at the origin, and all are zero, so that the C-R equations hold. All this is for 'differentiability at a point'. of course, and you are probably right if we are considering C-R equations in a region. Madmath789 17:21, 7 December 2006 (UTC)
Oops! Please forgive me for getting some facts mixed up. It's just been too long since I thought about some of this basic stuff. Proceeding directly from the C-R equations and trying to prove that holomorphic ==> analytic is the wrong way to go. Anyway, I probably shouldn't have even asked this question here in the first place, if I'd been thinking more carefully the other day. DavidCBryant 20:04, 8 December 2006 (UTC)

I don't know enough to contribute to this article, but I did notice that "necessary but not sufficient" and "if and only if" are both used to describe the same thing; one way or another something is wrong there. 192.5.109.49 (talk) 01:11, 15 May 2008 (UTC)

The C-R conditions are necessary, but not sufficient. However, if a function is smooth enough (C^1 I think), the C-R conditions become sufficient. That is to say, the C-R conditions are necessary, and with a small extra assumption they become sufficient. Oleg Alexandrov (talk) 04:58, 15 May 2008 (UTC)

[edit] History

According to ru:Условия Коши — Римана, these equations were first published by D'Alambert in 1752. In 1777 Euler connected these equations to the analyticity of complex functions. Cauchy used these equations to construct the theory of functions in 1814. Riemann's dissertation on the theory of functions was published in 1851. Unfortunately, the Russian article does not have any references. Can anyone look into this?(Igny 15:09, 7 July 2006 (UTC))


I have edited the article, adding this information. Also I deleted the following nonsense

The relation has this interpretation: x and y must be constant with respect to \bar z. This expresses the concept that an analytic function is "truly" a function of a single complex variable, rather than of a real vector.

It simply contradicts with x=.5(z+\bar z), y=.5(z-\bar z).

[edit] Why the Cauchy-Riemann Equations are Satisfied

An complex analytical function is continuous at a point Zo if and only if it's limit exist and is the same independent of what direction the limit takes. Here Cauchy is taking advantage of the fact that \lim_{\Delta x \rightarrow 0, \Delta y = o} is equivalent to \lim_{\Delta x = 0, \Delta y \rightarrow 0}.

—The preceding unsigned comment was added by 72.71.218.71 (talk • contribs) .

is that supposed to make any sense? --MarSch 14:32, 8 December 2006 (UTC)

[edit] More general form

There is a higher dimensional analogue of the CauchyRiemann equations involving a complex structure J^{\mu}_{\nu} See e.g. http://en.wikipedia.org/wiki/Pseudoholomorphic_curve It would be nice if this could be added here.


I think the polar form section would be better represented with the "r" terms. It suggests divide by zero problems in a way that could be confusing. The constants should simply be moved o the other side. You can see this form in Sarason's "Notes on Complex Function Theory." Comments? Are there any reasons that the current form is better? 24.7.83.80 (talk) 09:25, 13 February 2008 (UTC) Alex Kaiser ( UC Berkeley )

[edit] Rewrite attempt

The current version of Alternative formulation is not incorrect. I would just like to point out a certain slight abuse of notation, f in f(x,y) is not the same as in f(z,\bar z). I tried to make it clear that when you calculate df/dz, you mean d/dz f(x(z,\bar z),y(z,\bar z)), not d/dz f(z,\bar z). (Igny 04:11, 7 February 2007 (UTC))

There is a slight abuse of notation indeed. But the calculation is done correctly, using the chain rule, as far as I see. In your edit you spelled out fully the Jacobian matrix in the chain rule for 2 variables, but I don't think that it adds rigor, and it looked harder to understand. Oleg Alexandrov (talk) 04:51, 7 February 2007 (UTC)
I agree the calculation is correct in principle, and in general the current version is ok by me. I consider my concern about the notation noted (no puns intended). By the way, the Jacobian is calculated in the current version as well, just not named such. (Igny 05:16, 7 February 2007 (UTC))
You are right that in the current version one calculates the jacobian also (you have to, it is a variable change we are talking about). But the way it is now done is easier to follow than in your proof I think.
So the question is then whether it is worth making the proof more formal (and more complicated) for the sake of fixing a slight abuse of notation which I think is used quite a lot in such cases. I don't have a good answer. Oleg Alexandrov (talk) 16:23, 7 February 2007 (UTC)

Does anyone have a print reference for this alternative formulation, because to me it look very dodgey. In particular, it appears to assume that \frac{\partial \bar z}{\partial z}=0 which is just plain wrong no matter how you look at it. I think this section needs to go until someone can provide a print reference. IN any case, I'm putting a citation needed flag on that section. ObsessiveMathsFreak (talk) 18:51, 7 December 2007 (UTC)

This formulation is the standard one in some physics applications (CFT and strings). See lectures by Paul Ginsparg or Polchinski Vol I chapter II. —Preceding unsigned comment added by 83.57.132.73 (talk) 20:17, 24 February 2008 (UTC)

[edit] Analyticity

Is there a difference between analytic and holomorphic in this article? It sounds confusing to me.

There is no difference, a holomorphic function of a complex variable is the same as an analytic function, although the latter term is more general, being used for functions of real variables too. Oleg Alexandrov (talk) 15:33, 8 August 2007 (UTC)