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Residue Function

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  • Subject: [mg76060] Residue Function
  • From: "Dana DeLouis" <dana.del at>
  • 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

equ = Exp[2/z]; 

Residue[equ, {z, Infinity}]


>From what little I know, the residue comes from the ^-1 term in the

Normal[Series[equ, {z, Infinity, 3}]]

1 + 4/(3*z^3) + 2/z^2 + 2/z

Coefficient[%, z^(-1)]


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

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