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Re: Is this possible? Residue computation leads to complexInfinity?


kiki wrote:
> In[15]:=
> \!\(FullSimplify[Residue[\(\[ImaginaryI]\ z\)\/\(1 - \((2 + 4\ a)\)\ z\^2 + 
> z\
> \^4\), \@\(a + 1\) - \@a], \ a > 0]\)
> 
> Out[15]=
> \!\(Residue[ComplexInfinity, \(-\@a\) + \@\(1 + a\)]\) 
> 
> 
The syntax is not correct. Try

In[1]:=
FullSimplify[Residue[(I*z)/(1 - (2 + 4*a)*z^2 + z^4),
    {z, Sqrt[a + 1] - Sqrt[a]}], a > 0]

Out[1]=
(I*(-1 + Sqrt[1 + 1/a]))/(8*(-1 - a + Sqrt[a + a^2]))

Regards,
/J.M.


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