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

• To: mathgroup at smc.vnet.net
• Subject: [mg60723] Re: Is this possible? Residue computation leads to complexInfinity?
• From: Peter Pein <petsie at dordos.net>
• Date: Mon, 26 Sep 2005 01:36:03 -0400 (EDT)
• References: <dh5hl9\$dlc\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

kiki schrieb:
> In[15]:=
> \!\(FullSimplify[Residue[\(\[ImaginaryI]\ z\)\/\(1 - \((2 + 4\ a)\)\ z\^2 +
> z\
> \^4\), \@\(a + 1\) - \@a], \ a > 0]\)
>
> Out[15]=
> \!\(Residue[ComplexInfinity, \(-\@a\) + \@\(1 + a\)]\)
>
>
kiki,

have a look at the syntax (can be found - guess where -in the
documentation):

"Residue[expr, {x, x0}] finds the residue of expr at the point x=x0"

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

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

and

Union[Assuming[a > 0, FullSimplify[
(Residue[(I*z)/(1 - (2 + 4*a)*z^2 + z^4), #1] & ) /@
({z, z /. #1} & ) /@ Solve[1 - (2 + 4*a)*z^2 + z^4 == 0, z]]]]

returns the residues at the poles of your function:
{-(I/(8*Sqrt[a*(1 + a)])), I/(8*Sqrt[a*(1 + a)])}

Peter

--
Peter Pein, Berlin
GnuPG Key ID: 0xA34C5A82
http://people.freenet.de/Peter_Berlin/

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