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Homework Help: Z^3-1=0 (express using de moivres theorum)

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Use De Moivre's theorem to obtain solutions to the equation (Z^3)-1=0

    2. Relevant equations

    z^n= rθ^n(cosnθ+isinnθ)

    3. The attempt at a solution

    I am having difficulty on figuring out how to meld the equation to De Moivres Theorem
     
  2. jcsd
  3. Oct 17, 2012 #2

    Dick

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    Write z=r(cos(θ)+isin(θ)). So you have z^3=1. What are the possibilities for r and θ? Start with r. By convention we pick r>=0. What must r be?
     
    Last edited: Oct 17, 2012
  4. Oct 17, 2012 #3
    r would be 1 and θ=0 right?
     
  5. Oct 17, 2012 #4

    Dick

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    Good start! That's one solution. Can you find two more? Try to think of other values of θ that might work. cos(3θ) has to be 1 and sin(3θ) has to be 0.
     
    Last edited: Oct 17, 2012
  6. Oct 17, 2012 #5
    ok so

    z^3= 1^3(cos(3θ)+isin(3θ))
     
  7. Oct 17, 2012 #6

    Dick

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    Not so special. cos(2π)=1 and sin(2π)=0. Suggest a value for θ using that.
     
  8. Oct 17, 2012 #7

    2pi!
     
  9. Oct 17, 2012 #8
    Sorry I keep taking too many hints, I get it after I post that I don't understand
     
  10. Oct 17, 2012 #9

    HallsofIvy

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    The number "1" has "polar form" r= 1 and [itex]\theta= 0[/itex] because the number 1= 1+ 0i is represented by the point (1, 0) which is at distance 1 from (0, 0) and angle 0 with the positive x-axis. Alternatively, x= 1+ 0i= 1(cos(0)+ i sin(0)).

    Now, cosine and sign are periodic with period [itex]2\pi[/itex]. So the x-axis, in addition to being "[itex]\theta= 0[/itex]" is also "[itex]\theta= 2\pi[/itex]" and "[itex]\theta= 4\pi[/itex]".

    (As well as "[itex]6\pi[/itex]" or "[itex]9\pi[/itex]" or even "[itex]-2\pi[/itex]", etc. but those are not important. Do you see why?)
     
  11. Oct 17, 2012 #10

    Because it is really the same position right? 2pi=4pi in terms of position on the circle?
     
  12. Oct 18, 2012 #11

    Dick

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    True. But we are mainly interested in values of θ between 0 and 2π. The others are really just 'copies'. If you put θ=2π you get the same z value as θ=0. I was hoping you'd say 2π/3. Do you see why that works and is more interesting?
     
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