Residue Function

*To*: mathgroup at smc.vnet.net*Subject*: [mg76060] Residue Function*From*: "Dana DeLouis" <dana.del at gmail.com>*Date*: Tue, 15 May 2007 05:04:28 -0400 (EDT)

Hello. I am studying the subject of Residues, and came across the following example in an article. Basically, it says the Residue of the following equation is 2. However, both Mathematica 5.2 & 6.0 (windows) gives it as -2. equ = Exp[2/z]; Residue[equ, {z, Infinity}] -2 >From what little I know, the residue comes from the ^-1 term in the series... Normal[Series[equ, {z, Infinity, 3}]] 1 + 4/(3*z^3) + 2/z^2 + 2/z Coefficient[%, z^(-1)] 2 The above shows 2, as in the article. I don't see where Mathematica arrives at -2. Anyone familiar with this topic to comment? Thanks in Advance. Dana DeLouis

**Follow-Ups**:**Re: Residue Function***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>