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Complext roots

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the roots for

    [tex](z^N-a^N) = 0[/tex]

    when [itex]a^N<0[/itex] is allowed.


    2. Relevant equations

    No idea.

    3. The attempt at a solution

    I know that the roots are

    [tex]z_k=ae^{i(2\pi k/N)}, k = 0, 1, ... , N-1.[/tex]

    How do I find that solution? I'm reading a signals book, and it assumed that this is obvious. It should also be obvious that this is not a homework problem, so if possible please just tell me what I'm missing without hints. I'm studying and in a rush. Thanks!!
     
  2. jcsd
  3. Oct 29, 2007 #2

    Dick

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    exp(i*2*pi*k/N)^N=exp(2i*pi*k)=1 for k an integer. Take z_k to the Nth power. exp(2i*pi*k)=cos(2*pi*k)+i*sin(2*pi*k).
     
  4. Oct 30, 2007 #3
    couldn't you use the relationship
    [tex]z - \frac{1}{z} = 2{\bf{i}}\sin \theta[/tex] and De Movire's Theorem?
     
  5. Oct 30, 2007 #4

    HallsofIvy

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    There are two things you are "missing". The first is that you're formula is not correct- the N "Nth roots of a" are given by [itex]z_k= (a^{1/N}e^{i(2\pi k)/N}[/itex] )([itex]a^{1/N}[/itex] here is the "principal" real Nth root of the positive real number a, [itex]\sqrt[N]{a}[/itex]) and the fact that you don't seem to have tried to use that formula!
    [itex]z_1= \sqrt[N]{a}[/itex] with k= 0.
    [itex]z_2= \sqrt[N]{a}e^{2\pi i/N}[/itex] with k= 1.
    [itex]z_3= \sqrt[N]{a}e^{4\pi i/N}[/itex] with k= 2.
    etc.
     
  6. Nov 10, 2007 #5
    Thanks for your help, Halls. I posted this the night before a test in a cram.

    But now I'm coming back to it and trying to understand, and I still don't get it! I tried using the formula, but the z_k's you have up there make no sense! I just don't see what the complex exponential is doing! And when I plug in sample values for N, that exponential evaluates to crazy things.

    I think I'm having a mental block and missing something obvious. Also, was my formula really wrong? Here is exactly what the book says:

    [​IMG]
     
  7. Nov 10, 2007 #6
    [tex]z^n = a^n[/tex] let [tex]\zeta[/tex] be the [tex]n[/tex]-th root of unity and [tex]w = \sqrt[n]{a} = \exp (\log (a)/n)[/tex] then the roots are [tex]w,w\zeta,...,w\zeta^{n-1}[/tex]. (The square root here and log are the complex-valued functions).
     
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