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Re: Re: Residue Function

  • To: mathgroup at
  • Subject: [mg76324] Re: [mg76271] Re: Residue Function
  • From: Carl Woll <carlw at>
  • Date: Sat, 19 May 2007 04:52:28 -0400 (EDT)
  • References: <>

Dana DeLouis wrote:

>Thank you everyone for your help.  Here's what I have so far.
>The original question was to find the Residue of Exp[2/z].
>The article showed using Mathematic returning 2.  However, it returns
>unevaluated for me.
>Residue[Exp[2/z], {z, 0}]
>Series doesn't help here either.
>Series[Exp[2/z], {z, 0, 5}]
>In fact, most of the examples in books I am studying are not evaluated using
>the function "Residue" in Mathematica.
>It appears to not work well with Exp, and Trig Functions examples.
>I believe you have to use Infinity in the above example, but I am not sure
>yet as to the proper way to do it.
One possibility is to use NResidue:


In[3]:= NResidue[Exp[2/z], {z, 0}, Radius -> 1] // Chop
Out[3]= 2.

As the help for Residue mentions, Residue will usually only work 
Mathematica is able to find a power series at that point. Since Exp[2/z] 
has a Laurent expansion but not a Taylor series expansion at z=0, Series 
doesn't work.

Since Exp[2/z] does have a "Taylor" series expansion at the point 
z=Infinity, another possibility is to find this series and pick off the 

In[4]:= s = Series[Exp[2/z], {z, Infinity, 3}] // Normal
Out[4]= 1+2/z+2/z^2+4/(3 z^3)

In[5]:= Residue[s, {z, 0}]
Out[5]= 2

Carl Woll
Wolfram Research

>I have received feedback that the Residue function did in fact work in older
>versions of Mathematica.  However, beginning in version 5, some
>functionality was removed as it was causing errors elsewhere.  At least I
>know now why most of my examples are not working.
>I was also given this version, but it's a little over my head at this point.
>     {t, 0, 2*Pi}, Assumptions -> r > 0]
>Valeri also posted an interesting equation that returned -2 using version
>However, in Mathematica 6.0, it returns 0.    Hmmm! 
>I received an email stating that this "appears to be a bug" with version 6.0
>at this time.
>Another major concern mentioned was that this returned 0 also.  
>FullSimplify[-(Integrate[Exp[2/z], {z, 1, I}] + 
>          Integrate[Exp[2/z], {z, I, -1}] + Integrate[Exp[2/z], 
>            {z, -1, -I}] + Integrate[Exp[2/z], {z, -I, 1}])/(2*Pi*I)]
>But this returned -2.
>Chop[-(NIntegrate[Exp[2/z], {z, 1, I}] + NIntegrate[Exp[2/z], 
>            {z, I, -1}] + NIntegrate[Exp[2/z], {z, -1, -I}] + 
>          NIntegrate[Exp[2/z], {z, -I, 1}])/(2*Pi*I)]
>It appears to be an Integrate bug somewhere with 6.0, but I'm not an expert
>here to comment on it.
>Thank you everyone for your time.  Interesting subject "Residues."

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