Re: Poles and Complex Factoring

*To*: mathgroup at smc.vnet.net*Subject*: [mg52176] Re: [mg6011] Poles and Complex Factoring*From*: godfreyau2003 at hotmail.com (Godfrey)*Date*: Sun, 14 Nov 2004 04:30:12 -0500 (EST)*References*: <5dp3kk$6nq@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 11 Feb 1997 01:29:40 -0500, peter wrote: >Dear All, > >I know how to calculate the residue of a fuction using Mathematica, but how can I >use Mathematica to calculate the order of a complex pole? > >It would also be nice for Mathematica to tell me if a particular singularity is an >essential singularity, removable singularity or a pole...but this is not >necessary; just icing on the cake. > >Also, is there a way to factor polynomials with imaginary roots? >Something like: > > Factor[ x^2 + 2x + 10 ] = (x - 1 + 4.5 I)(x - 1 - 4.5 I) > >Much thanks in advance! > >Peter > >-- >Birthdays are good for you: A federal funded project has recently determined >that people with the most number of birthdays will live the longest..... >-=><=-+-+-=><=-+-+-=><=-+-+-=><=-+-+-=><=-+-+-=><=-+-+-=><=-+-+-=><=-+-+-=><=- > I BOYCOTT ANY COMPANY THAT USES MASS ADVERTISING ON THE INTERNET Let x^2 + 2x + 10 = (x-a)(x-b) Obviously, a and b are the roots of the equation x^2 + 2x + 10 = 0 ------ (*) By quadratic forumula, we can find that the roots of (*)=2+sqrt(10) or 2-sqrt(10) Therefore, x^2 + 2x + 10 = {x-[2+sqrt(10)]}{x-[2-sqrt(10)]} ie. x^2 + 2x + 10 = [x-2-sqrt(10)][x-2+sqrt(10)] Factorization of other polynomials with imaginary roots can be done in the same way.