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Homework Help: Finding Complex Roots

  1. Aug 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Find all complex solutions of z^6 + z^3 + 1

    (z^3 + 1)/(z^3 - 1) = i


    2. Relevant equations



    3. The attempt at a solution
    I am going crazy with trial and error with these, there must be some systematic method or tricks that I am oblivious of. For the second question I factored it to:

    [(z+1)(z^2 - z + 1)]/[(z-1)(z^2 + z +1)] = i

    obviously we have conditions z cannot equal 1. But also z cannot = -1/2 +- sqrt(3)i What happens here?
     
  2. jcsd
  3. Aug 5, 2013 #2
    Substitute ##u = z^3## and solve for ##u##.
     
  4. Aug 5, 2013 #3

    lurflurf

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  5. Aug 5, 2013 #4
    Wicked alright let u = z^3

    We have u^2 + u + 1 = 0

    then u = -1/2 +- [(√3)/2]i

    Now given that we took u = z^3, can I just cube the results to get the answers for z??
     
  6. Aug 5, 2013 #5
    You know that

    [tex]z^3 = -\frac{1}{2} \pm i \sqrt{3}[/tex]

    Then what is ##z##?
     
  7. Aug 5, 2013 #6
    OMG how silly of me yes you'd take the cube root of the positive and negative answers each so,

    z = (-1 + i3^1/6)/(2^1/3)

    z = (-1 - i3^1/6)/(2^1/3)
     
    Last edited: Aug 5, 2013
  8. Aug 5, 2013 #7
    However we would have 6 roots to this polynomial would we not, by the fundamental theorem of algebra?
     
  9. Aug 5, 2013 #8
    I have no idea what you just did.
     
  10. Aug 5, 2013 #9
    Oh no, do you get something different when you take the cube root?
     
  11. Aug 5, 2013 #10
    For any nonzero complex number, there are in fact three cube roots! Did you cover in class how to find these?
     
  12. Aug 5, 2013 #11
    No my book never mentioned anything about this, but it makes sense that we have a positive and a negative version of the root. That would give me four if I included the conjugates to the one I wrote, what would the 3rd set be??
     
  13. Aug 5, 2013 #12
    Did you learn that you can write any complex number ##z## as

    [tex]z = r(\cos(\theta) + i \sin(\theta))[/tex]

    Did you learn how to find ##z^n## using this representation?
     
  14. Aug 5, 2013 #13
    That kind of looks like the set up to DeMoivre theorem?

    I know if it is r(cosθ + isinθ)^n = r^n(cos nθ + isin nθ)

    But how does this help us?
     
  15. Aug 5, 2013 #14
    Right, so we need to find ##z = r(\cos(\theta) + i\sin(\theta))## such that

    [tex]r^3 (\cos(3\theta) + i\sin(3\theta) ) = \frac{1}{2} \pm i\sqrt{3}[/tex]

    So, try to write the RHS in the same ##r(\cos(\theta) + i\sin(\theta))## form, and see if you can find ##r## and ##\theta##.
     
  16. Aug 5, 2013 #15
    Ok we have cosine of an angle corresponding to 1/2 and the sin corresponding to (sqrt3)/2 so the
    θ in question should be ∏/3 (60 degrees).

    The r = (1/2)^2 + (sqrt3/4)^2
    r = 1/4 + 3/4
    r = 1
     
  17. Aug 5, 2013 #16
    OK, so you need to find ##r## and ##\theta## such that

    [tex]r^3(\cos(3\theta) + i \sin(3\theta)) = \cos(\pi/3) + i\sin(\pi/3)[/tex]

    This comes down to

    [tex]r^3 = 1,~\cos(3\theta ) = \cos(\pi/3),~\sin(3\theta) = \sin(\pi/3)[/tex]

    Can you find ##r## and ##\theta## that satisfy this?
     
  18. Aug 5, 2013 #17
    Sure so r = 1 and θ=1pi/9
     
  19. Aug 5, 2013 #18
    Your ##r## is correct, your ##\theta## is not. There are multiple ##\theta## that solve this by the way.
     
  20. Aug 5, 2013 #19
    cos(pi/3) = cos(3θ)
    cant I take cos inverse of both sides yielding

    pi/3 = 3θ?

    then θ=1/9 pi of course times some scalar k?
     
  21. Aug 5, 2013 #20
    No, you can't just take the cos inverse of both sides. That way you'll lose solutions. However, ##\pi/9## is correct, but there are two other solutions.

    When I ask you to solve ##\cos(x) = \cos(y)##. How would you do it? Try to draw a graph if you don't see it.
     
  22. Aug 5, 2013 #21
    Cosine is an even function so throw in -pi/9 as well.

    As for the third how about 17∏/9?
     
  23. Aug 5, 2013 #22
    Remember, it needs to satisfy ##\sin(3\theta ) = \sin(\pi/3)## as well.
     
  24. Aug 5, 2013 #23
    Ah yes so forget the negative ninth pi, was 17pi/9 correct? And 19pi/9?
     
  25. Aug 5, 2013 #24
    Stop guessing. If you know ##\cos(x) = \cos(y)## and ##\sin(x) = \sin(y)##, then what do you know about ##x## and ##y##??

    Well, certainly ##x=y## is a possibility. But so is ##x=y + 2\pi##. Indeed, adding ##2\pi## doesn't change sine and cosine. Adding ##4\pi## doesn't change it either. Subtracting ##2\pi## doesn't change it either.

    So we see that if ##\cos(x) = \cos(y)## and ##\sin(x) = \sin(y)##, then ##x = y + 2k\pi## for some ##k\in \mathbb{Z}##.

    Now, apply this to your problem.
     
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