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De Moivre's theorem

  1. Oct 27, 2011 #1
    Hi guys, just a simple question.

    I need to find all solutions of z^6 = 1.

    I need to change that to |z|^6(cos 6θ + isin6θ) = 1 and the prof wrote → 1(1+0i). How did the prof do this? The rest of the question I understand. It's just that one step still stumps me. Thanks
     
  2. jcsd
  3. Oct 27, 2011 #2

    jambaugh

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    I'm not clear on your question. Are you asking why 1 = 1+0i? or 1 = 1*1 = 1*(1+0i)?

    Note: A more compact notation is: [itex] r\cdot(\cos(\phi) + i\sin(\phi)) = r e^{i\phi}[/itex]. (This is the analytic extension of the exponential function to the complex plane, the real and imaginary components of [itex]e^{i\phi}[/itex] are the cosine and sine functions respectively.)

    In this notation: [itex] 1 = 1\cdot e^{0 i} = 1\cdot e^{(0 + 2\pi k)i}[/itex] for [itex] k = 0, 1, 2, 3, \ldots[/itex]

    Then all solutions can be obtained by taking the 6th root of all forms of 1:
    [itex] z = [ e^{0 + 2\pi k}]^{1/6} = e^{\frac{2 \pi}{6} k} = e^{\frac{\pi}{3} k} = cos(\frac{\pi}{3} k ) + i \sin(\frac{\pi}{3} k)[/itex]
    iterate through values of k until you begin repeating.
     
  4. Oct 27, 2011 #3
    For that question, here is the answer the prof wrote.
    Find all the solutions of Z^6 = 1.

    |z|^6(cos6θ + isin6θ)

    |z|^6 = 1
    |z| = 1
    Therefore:
    1(1 +0i) <-- don't get how the prof got to here.

    cos6θ = 1 sin6θ = 0

    6θ = 0+n(2π)
    = n(π/3)

    so 0, π/3, 2π/3, π, 4π/3, 5π/3

    z1 = (cos0 + isin0) = 1
    Blah Blah Blah..
     
  5. Oct 27, 2011 #4

    jambaugh

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    Again I'm not clear on your question.
    Are you saying you have difficulty understanding why [itex]1 = 1+0i?[/itex] or [itex]1 = 1\cdot(1) = 1\cdot(1+0i)[/itex]?

    If not let me fill in some gaps in the exposition.
    -----------------------------------------------------
    Find all solutions to [itex] Z^6 = 1[/itex].

    DeMoivre's theorem states that for [itex]Z = |Z|(\cos(\theta) + i\sin(\theta) )[/itex], we have [itex]Z^n = |Z|^n (\cos(n\theta) + i\sin(n\theta))[/itex].

    Apply this here by assuming [itex]Z = |Z|(\cos(\theta) + i\sin(\theta) )[/itex] and setting
    [itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1[/itex] which in polar form is [itex]1 = 1(1 + 0i )[/itex]

    First solve for the magnitude... (magnitude of l.h.s. must equal magnitude of r.h.s.):

    Since [itex] |Z|^6 = |1| = 1[/itex] you have [itex] |Z| = \sqrt[6]{1}=1[/itex].

    Now apply the known magnitude of Z...

    therefore [itex]Z^6 = 1( \cos(6\theta) + i\sin(6\theta)) = 1(1+ 0i)[/itex]
    and thus [itex] \cos(6\theta) = 1, \sin(6\theta) = 0[/itex]...


    the rest if solving for the thetas such that this is true.
     
  6. Oct 27, 2011 #5
    Sorry for being unclear, but I think it's the step in your explanation that turning the De Moivre theorem to polar form. How do you do that?

     
  7. Oct 28, 2011 #6

    jambaugh

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    De Moivre's theorem is already in polar form.

    Are you saying you don't understand that 1 = x + iy with x = 1 and y = 0?

    Can't you plot complex numbers in the complex plane? Plot 1 in the complex plane. Where is it? At what angle and how far from the origin?

    EDIT:
    [itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1[/itex]
    [itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1 + 0[/itex]
    [itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1+ 0 i[/itex]
    [itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) =1( 1+ 0 i )[/itex]
     
  8. Oct 28, 2011 #7
    I don't understand why Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1 equals to 1( 1+0i)

     
  9. Oct 28, 2011 #8

    jambaugh

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    Which = in this equation don't you understand?

    The first is just setting n=6 in De Moivre's theorem. Do you have De Moivre's theorem in front of you? Pull out your textbook and look at it.

    The 2nd "=" is the original equation you're trying to solve.

    The 3rd "=" is OBVIOUS. Each step I showed in the last post is an identity substitution. 1=1+ 0 because 0 is the additive identity.
    0i = 0 by a very basic algebraic lemma.
    (1+0i) = 1(1+0i) because 1 is the multiplicative identity.

    I'm going to bed now. If you can be more explicit in exactly what it is you do not understand, rather than just pointing at a step, I'll leave it to others to enlighten you.

    EDIT: Sorry about being so terse. I was sleepy and cranky. I've slept now and am neither. So please ignore/forgive my rudeness. Is there anything I can do to clarify this further?
     
    Last edited: Oct 28, 2011
  10. Nov 23, 2011 #9
    would u mind can u answer this for me... please EVALUATE: e^e^m(pi+i pi/6)?
     
  11. Dec 12, 2011 #10

    jambaugh

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    Tell me the value of m.

    [tex]z=e^{m\pi (1+i)/6} = e^{m\pi/6}[\cos(m\pi/6) + i \sin(m\pi /6)] = x+iy[/tex]
    [tex]e^z = e^{x+iy} =e^x\cdot e^{iy}= e^x(\cos(y)+i\sin(y))[/tex]
     
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