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Complex numbers

  1. Dec 27, 2014 #1
    1. The problem statement, all variables and given/known data
    [tex] If arg(\frac{z-ω}{z-ω^2}) = 0, \ then\ prove \ that\ Re(z) = -1/2 [/tex]
    2. Relevant equations
    ω and ω^2 are non-real cube roots of unity.

    3. The attempt at a solution
    arg(z-ω) = arg(z-ω^2)
    So, z-ω = k(z-w^2)
    Beyond that, I'm not sure how to proceed. Using the rotation formula may also be required, but I'm not sure how to use it here.
    Furthermore, I can sort of intuitively visualise the answer, but I'm not sure how to prove it analytically.
     
    Last edited: Dec 27, 2014
  2. jcsd
  3. Dec 27, 2014 #2

    SammyS

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    If ω is one of the non-real cube roots of unity, then what is ω2, and how is it related to ω ?
     
  4. Dec 27, 2014 #3
    1 + ω + ω^2 = 0
    [tex] ω = e^\frac{i2\pi}{3}[/tex]
    How does that help?
     
  5. Dec 27, 2014 #4

    SammyS

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    I was thinking in terms of the complex conjugate.
     
  6. Dec 27, 2014 #5
    ω is the conjugate of ω^2.
     
  7. Dec 28, 2014 #6

    Dick

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    You have that (z-ω)=k(z-ω^2) where k is real and positive. Take the real part of both sides.
     
  8. Dec 28, 2014 #7

    haruspex

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    Your target is Re(z). How else can you write that?
     
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