Re: Searching zeros of complex function

From: Gerry Thomas (gfthomas_at_sympatico.ca)
Date: 05/14/04 

Date: Fri, 14 May 2004 14:13:03 -0400

"Steven G. Kargl" <kargl@troutmask.apl.washington.edu> wrote in message
news:c82tii$ahq$1@nntp6.u.washington.edu...
> In article <f73pc.34245$dr1.940468@news20.bellglobal.com>,
> "Gerry Thomas" <gfthomas@sympatico.ca> writes:
> >
> > "Miguel Ángel Solano Vérez" <solanom@unican.es> wrote in message
> > news:c81se1$bbe$1@ccserver13.unican.es...
> >> I have a square complex matrix. A cuadrant of this matrix has terms of
> >> the type exp(j*beta*d). being beta the unknown. I'm looking for a
rutine
> >> (fortran) which gives the zeros of the determinant of this matrix, or
> >> simple a rutine which gives zeros of a complex function. I have been
> >> looking in the IMSL package but i do not find anything.
> >
> > As det(A) is a constant it doesn't have any roots. det(A-zI)=0 is a
> > polynomial of the order of A, n say, and its n roots are the
eigenvalues of
> > A. IMSL has routines to determine the eigenvalues of general complex A.
> >
>
> | 1 exp(j*beta*d) |
> | | = det(A) = -exp(j*beta*d)
> | 1 0 |
>
> How is det(A) a constant when the OP stated beta is an unknown?

The problem posed was how to find the roots of det(A) for arbitrary b and
was not how to find b. In the current post to the OP I gave him a hint and
in a followup post I gave him the answer to his homework assignment. As for
your question, specify b and you'll know det(A), a constant. Keep in mind
that this burlesque is not MATLAB/Maple or Mathematica. It's interesting to
note that on sci.math.num-analysis his post was pretty well ignored as it's
customary for respondents there to spot homework and at most to provide a
hint. Here it degenerates to vain pillory by delicate and fragile
narcissistic egos who provide a veritable anthropological feast of their
untempered ids.

The show's over, or is it Mission Accomplished, :-). 

-- 
E&OE
Ciao,
Gerry T.

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