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MathGroup Archive 2006

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Re: Re: Trigonometric form of complex numbers

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64199] Re: [mg64181] Re: Trigonometric form of complex numbers
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sun, 5 Feb 2006 04:44:53 -0500 (EST)
  • Reply-to: hanlonr at cox.net
  • Sender: owner-wri-mathgroup at wolfram.com

p[x_]:=Abs[x]*(Cos[Arg[x]]+I*Sin[Arg[x]]);

f[y_,n_Integer]:=Module[{x},
      p/@(x/.Solve[x^n==y,x])];

r=f[z,3]

{Abs[z]^(1/3)*(Cos[Arg[z]/3] + I*Sin[Arg[z]/3]), 
  Abs[z]^(1/3)*(Cos[Arg[(-(-1)^(1/3))*z^(1/3)]] + 
    I*Sin[Arg[(-(-1)^(1/3))*z^(1/3)]]), 
  Abs[z]^(1/3)*(Cos[Arg[(-1)^(2/3)*z^(1/3)]] + 
    I*Sin[Arg[(-1)^(2/3)*z^(1/3)]])}

FullSimplify[Times@@r]

z

However, you can get the sign inverted

r=f[5+I*6,2]

{61^(1/4)*(Cos[Arg[-Sqrt[5 + 6*I]]] + 
    I*Sin[Arg[-Sqrt[5 + 6*I]]]), 
  61^(1/4)*(Cos[(1/2)*ArcTan[6/5]] + 
    I*Sin[(1/2)*ArcTan[6/5]])}

FullSimplify[Times@@r]

-5 - 6*I


Bob Hanlon

> 
> From: ivan.svaljek at gmail.com
To: mathgroup at smc.vnet.net
> Subject: [mg64199] [mg64181] Re: Trigonometric form of complex numbers
> 
> I'm looking for a solution that would produce something like this:
> r (cos f + i sin f).
> 
> And would return n roots of a complex number in such form.
> 
> Mathematica itself does nothing to return roots of complex numbers. Am
> I wrong ?
> 
> 


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