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Re: FindRoot with complex roots

  • To: mathgroup at smc.vnet.net
  • Subject: [mg9727] Re: FindRoot with complex roots
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Tue, 25 Nov 1997 00:06:24 -0500
  • Organization: University of Western Australia
  • Sender: owner-wri-mathgroup at wolfram.com

Paul Hines wrote:

> 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

There are 3 problems with the syntax of this expression:

[1] No multiplication sign (or space) between zw and Cos[x]
[2] = instead of ==
[3] Sin^2[x] instead of Sin[x]^2
 
> 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.  

For these values I get

In[1]:= r*Cos[x] + zs*Sin[x] - (zw*Cos[x]*Sin[x])/Sqrt[n^2 - Sin[x]^2]
== 0 /. 
  {r -> 100, zs -> -10, zw -> 20, n -> 1.0067 + 0.01 I}

Out[1]=
            20 Sin[x] Cos[x]
-(------------------------------------) + 100 Cos[x] - 10 Sin[x] == 0
                                    2
  Sqrt[1.01334 + 0.020134 I - Sin[x] ]
 
and then FindRoot works fine

In[2]:= FindRoot[-((20*Sin[x]*Cos[x])/ Sqrt[1.01334489 + 0.020134*I - 
         Sin[x]^2]) + 100*Cos[x] - 10*Sin[x] == 0, {x, 0.1}]

Out[2]= {x -> 1.29469 + 0.0215355 I}

Cheers,
	Paul 

____________________________________________________________________ 
Paul Abbott                                   Phone: +61-8-9380-2734
Department of Physics                           Fax: +61-8-9380-1014
The University of Western Australia            Nedlands WA  6907       
mailto:paul at physics.uwa.edu.au  AUSTRALIA                            
http://www.pd.uwa.edu.au/~paul

            God IS a weakly left-handed dice player
____________________________________________________________________


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