       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)
>
>
>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.

```

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