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/