Re: FindRoot with complex roots
- To: mathgroup at smc.vnet.net
- Subject: [mg9752] Re: [mg9694] FindRoot with complex roots
- From: "Sherman.Reed" <sherman.reed at worldnet.att.net>
- Date: Tue, 25 Nov 1997 00:06:55 -0500
- Sender: owner-wri-mathgroup at wolfram.com
Paul
I changed your equations as follows. I believe they were mistakes:
r*Cos[x]+zs*Sin[x]-zwCos[x]Sin[x]/Sqrt[n^2-Sin^2[x]]=0
^^^^^^^^^^^^^^^ ^^^^^^^^
zw*Cos[x]*Sin[x] (Sin[x])^2
and it worked just fine on the first crack.
hope his helped.
Sherman C. Reed
sherman.reed at worldnet.att.net
----------
> From: hines <hines at drea.dnd.ca>
To: mathgroup at smc.vnet.net
> To: mathgroup at smc.vnet.net
> Subject: [mg9752] [mg9694] FindRoot with complex roots
> Date: Friday, November 21, 1997 12:31 AM
>
> Dear Newsgroup
>
> I've been having trouble with the FindRoot function when the root(s) is
> complex. One such function is:
>
> r*Cos[x]+zs*Sin[x]-zwCos[x]Sin[x]/Sqrt[n^2-Sin^2[x]]=0
>
> where r,zs,zw are real and zs<0, and n complex. Typical values are
> r=100,zs=-10, and zw=20, n=1.0067+0.01I. When I use the substitution
> Sin[x]=y, Cos[x]=Sqrt[1-y^2] FindRoot works fine. When I try and solve
> it in trig expressions, it gives me errors like "Function... is not a
> length 1 list of numbers at x= .789...-.00123I" The FindRoot function
> handles the trig forms just fine when I set n = a purely real number
> and the root(s) are real.
>
> Paul Hines