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Roots of complex polynomials

  1. Aug 6, 2005 #1
    Hi all

    Jut had a question. How do I go about finding the general formula for roots of the complex poly [tex]{z}^{n}-a[/tex] where a is another complex number.

    Do I just go [tex]{z}^{n}=a[/tex]??? :S so complicated this things!!!

    Thanks in advance!
     
  2. jcsd
  3. Aug 6, 2005 #2

    lurflurf

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    It is easier if z is written in geometric form
    let a=u+i v where u and v are real
    z=r(cos(t)+i sin(t)) where 0<=r
    then
    z^n-a becomes
    (r^n)(cos(n t)+i sin(n t))-a
    so r should be chosen by
    r^n=|a|=sqrt(a^2+b^2)
    cos(n t)+i sin(n t)=a/|a|
    so
    cos(n t)=u/|a|
    sin(n t)=v/|a|
    tan(n t)=v/u
    t=(1/n)Arctan(v/u)
    this is one solution but n can be found using the usual methods of solving trigonometric equations.
    in fact
    t=(1/n)Arctan(v/u)+2pi k/n
    k=0,1,2,...,n-2,n-1,
    thus
    z=r(cos(t)+i sin(t))
    =|a|^(1/n)(cos((1/n)Arctan(v/u)+2pi k/n)+i sin((1/n)Arctan(v/u)+2pi k/n
    ))
    k=0,1,2,...,n-2,n-1
     
    Last edited: Aug 6, 2005
  4. Aug 7, 2005 #3

    AKG

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    It's actually easiest to understand in polar co-ordinates. [itex]a = |a|\exp (i\theta )[/itex]. So simply take |z| to be |a|1/n. Take angles [itex]\theta _k = \frac{\theta + 2k\pi}{n}[/itex] for values k = 0, 1, ..., n-1. Then your roots are the complex numbers [itex]|z|\exp (i\theta _k)[/itex] for the n values for k.
     
    Last edited: Aug 7, 2005
  5. Aug 7, 2005 #4

    HallsofIvy

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    Since ei&theta;= cos(&theta;)+ i sin(&theta;), AKG and lurflurf are saying exactly the same thing!
     
  6. Aug 7, 2005 #5

    AKG

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    We're saying very different things. lurflurf is saying that it's easiest to see it in geometric form, I'm saying it's easiest to see it in polar form.
     
  7. Aug 7, 2005 #6

    HallsofIvy

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    Writing a+ bi as r(cos(θ)+ i sin(θ)) is writing the complex number in polar form.
     
  8. Aug 8, 2005 #7

    AKG

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    Oops, my mistake. Then whatever form my post was in is easier to understand. What would it be called? Exponential form? Euler form?
     
  9. Aug 8, 2005 #8

    HallsofIvy

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    "Exponential" form would do. Engineering texts talk about "Cis &theta;" which is shorthand for "cos(&theta;)+ i sin(&theta;)" but, yes, exponential form is a lot simpler (though more sophisticated) than "cis".
     
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