Talk:Residue (complex analysis)

message: Problems at infinity, as always. Would it not be worth while explaining (briefly) that the residue is properly defined for differential forms and not for functions? 80.58.23.107 18:09, 25 Nov 2004 (UTC)

message: In the expression that follows the line "The integral now collapses to a much simpler form. Recall," a minus sign is missing (after the integral, where it says "a is an element of....").
: Thanks. Next time, you can fix it yourself! :) Dysprosia 09:05, 11 Apr 2005 (UTC)
thread_title:  Minus sign 

message: It would be helpful if this page had the general definition of the residue on it. (I would add it but I cannot rememeber it, come looking :P)
:I have added such a definition.  It depends on the book you are reading though, whether the residue is defined in terms of Laurent series or more generally.Rgrizza (talk) 18:48, 25 March 2008 (UTC)
replies: text: Actually, the sentence is not wrong: after all, it is a contour integral of a meromorphic function (the function f/2πi). Rather, the objetion is that a contour integral is just a way of computing the residue, and not really the definition of it. The residue of f at z0 shoud be better definied as the coefficient c that makes f(z)-c/(z-z0) exact (that is, a derivative of another meromorphic function) locally at z0. In other words, f(z) writes as a derivate g'(z), plus a non-derivate part c/(z-z0) (whence the term "residue").--pma 06:53, 9 May 2010 (UTC)
thread_title:  General Definition 

message: The calculation of the series method example is wrong. The function 1/z also has a non-trivial series in (z-1), which has to be multiplied with the series as it stands there. Luckily the first term of it is 1 and the value of the residue won't change by it, but had it been a non-singular function at z=-1, this wouldn't have been the case. It will complicate the example though, maybe make it less pedagogic... David  12:06, 7 June 2007 (UTC)
replies: text: I corrected it. David  10:26, 10 August 2007 (UTC)
thread_title:  Example series method 

message: The motivational example claims to evaluate an integral "...without standard integral theorems that are available..." If we have no such theorems at hand then how are we able to "Recall that...
:\oint_C {1 \over z^n} \,dz=0,\quad n \in \mathbb{Z},\mbox{ for }n \ne 1."? 
Where C is a simple closed plane curve bounding an open region containing 0. For example, we could explicitly calculate the integral when C is the unit circle, but how do we know that the integral is path independent (provided the path doesn't pass through 0)?  Δεκλαν Δαφισ   (talk)  19:36, 2 April 2009 (UTC)
thread_title:  Motivational Example 

message: In the section Examples with subsections titled Residue of a monomial and Application of monomial residues: not even one residue is calculated.
This is simply bad writing.
In the next section, titled Calculating residues, we are finally given the usual definition of a residue:
"According to the residue theorem, we have:
: \operatorname{Res}(f,c) = {1 \over 2\pi i} \oint_\gamma f(z)\,dz
"where γ traces out a circle around c in a counterclockwise manner. ..."
This should be the definition of a residue, and it should appear much, much earlier in the article. 2601:200:C000:1A0:A446:C53:9C70:A477 (talk) 16:32, 12 December 2022 (UTC)
replies: text: I will move examples sections down to reflect this. EditingPencil (talk) 10:54, 20 November 2023 (UTC)
thread_title:  Crucial information delayed much too long