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Solving for the roots of unity of a complex number

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Find both square roots of the following number:

    -15-8i




    2. Relevant equations

    De Moivre's thm: rn(cos(n[itex]\sigma[/itex]) + i sin(n[itex]\sigma[/itex])


    3. The attempt at a solution

    So to use De Moivre's I have to find the modulus and the argument.

    actually in this question r = 1, what I'm left to have to try and figure out is how to obtain -15 - 8i
    from (cos(n[itex]\sigma[/itex]) + i sin(n[itex]\sigma[/itex])

    All I could think of possibly doing is setting Cos[itex]\sigma[/itex] = -15 and
    iSin[itex]\sigma[/itex] = -8 but even doing that doesn't really provide me with much information. Unless that's what I'm suppose to work with.
     
  2. jcsd
  3. Apr 21, 2013 #2

    fzero

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    ##r\neq 1##, go back and compute ##|-15-8i|##. Your method should work once you use the correct value of ##r##.
     
  4. Apr 21, 2013 #3


    Ok. Well before I had said the 1, the modulus I got was 17. Using that relation, then what I would end up doing is finding this relationship:

    z2 = -15-8i but now I converted -15-8i into polar form:

    z2 = 17(cos[itex]\sigma[/itex] + iSin[itex]\sigma[/itex])

    now all that is left is for me to find the roots of unity of this equation.

    Actually in the conversion to polar form, how would I write the relations:

    i.e: Cos[itex]\sigma[/itex] = -15/17 ?

    would it be as [itex]\sigma[/itex] = Cos-1(-15/17)?
     
  5. Apr 21, 2013 #4

    fzero

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    You want to solve

    $$z^2 = 17 ( \cos 2\sigma + i \sin 2\sigma).$$

    A root of unity is something slightly different. Namely, a complex number of the form

    $$w_k = \cos \frac{2\pi k}{n} + i \sin \frac{2\pi k}{n},~~k=0,1,\ldots, n-1$$

    satisfies ##(w_k)^n = 1##. The ##w_k## are the nth roots of unity. What you want to do is take the square root of a complex number. The two concepts are related.

    For the principal root you can indeed solve ##\cos 2\sigma = -15/17##. There is another root obtained by adding a multiple of ##i\pi## to ##\sigma##.
     
    Last edited: Apr 21, 2013
  6. Apr 21, 2013 #5


    I can? This is where my rusty trig skills come to bite me:

    Cos 2[itex]\sigma[/itex] = -15/17

    How is this going to be anywhere near a clean solution of the form (pi/3) or something of the sort?

    So what I'm doing is the same "process" as finding the roots of unity, but instead it is with respect to another value that is not 1?
     
  7. Apr 21, 2013 #6

    fzero

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    I would do the calculation with a few significant figures. At the end, you should be able to look at the answer and figure out the exact result in whole numbers.

    Yes. A root of unity is a solution to ##z^n =1##. Instead of ##1##, you have ##-15 - 8i##. The DeMoivre theorem is used in both cases.
     
  8. Apr 21, 2013 #7

    SteamKing

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    If you are trying to find a square root using DeMoivre's theorem, n = 1/2 instead of n = 2.
     
  9. Apr 21, 2013 #8

    fzero

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    From the statement of De Moivre's theorem, the OP set the problem up as "find ##z## if ##z^2 = -15-8i##." You can indeed set the problem up your way, but it could add an additional form of confusion to change the OP's notation.
     
  10. Apr 21, 2013 #9


    Yeah I know the way he's talking about, but I found it more confusing when trying to find the rest of the trig relations. Thanks for the help.
     
  11. Apr 21, 2013 #10

    SammyS

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    The complex number, -15-8i, is in the third quadrant of the complex plane.

    cos-1(-15/17) is in the second quadrant.

    You can express [itex]\sigma[/itex] in several ways.

    -cos-1(-15/17) is in the third quadrant. Add multiples of 2π to that.

    Alternatively: ##\ \pi+\cos^{-1}(15/17)\ ## is in the third quadrant. Add multiples of ##\ 2\pi\ ## to that.
     
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