# Talk:Residue (complex analysis)

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)

## Minus sign

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)

That is not true: If C is parameterized by exp(i*t) for 0<t<2Pi, then Int[z^-a,z,C]=Int[i*exp(-a*i*t)*exp(i*t),t,0,2Pi]=Int[i*exp((1-a)i*t),t,0,2Pi]. If a != 1, then this is (exp((1-a)i*2Pi)-exp((1-a)i*0))/(1-a), but exp(2Pi)=exp(0), so this expression is 0; however, if a=1, then this is Int[i,t,0,2Pi]=(2Pi-0)i=i*2Pi. I fixed the markup. Julyo 6:21, 16 Apr 2004 (UTC)

## Vote for new external link

Here is my site with residue calculus example problems. Someone please put this link in the external links section if you think it's helpful and relevant. Tbsmith

http://www.exampleproblems.com/wiki/index.php/Complex_Variables

## General Definition

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)

It seems to me that the introductory defining sentence: 'the residue is a complex number equal to the contour integral of a meromorphic function along a path enclosing one of its singularities' is wrong, because the factor i/2Pi is omitted. Maybe add a "is proportional to..."? --93.220.115.185 (talk) 09:40, 8 May 2010 (UTC)

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)

## Example series method

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)

I corrected it. David 10:26, 10 August 2007 (UTC)

## Motivational Example

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...

${\displaystyle \oint _{C}{1 \over z^{n}}\,dz=0,\quad n\in \mathbb {Z} ,{\mbox{ for }}n\neq 1.}$"?

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)