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Complex numbers

  1. Jan 12, 2008 #1
    find all complex numbers satisfying z^3=-8
    i have no idea how to solve this can someone help please?
    thank u
  2. jcsd
  3. Jan 12, 2008 #2
    Think -8 as -8 + 0i... Then convert this into polar coordinates.
  4. Jan 12, 2008 #3
    in polar coordinates it will be:
  5. Jan 14, 2008 #4
    i think that if you write down z=rexp(ix) would be easier...
    you find three roots on the cricle with radius r=(-8)^(1/3).

  6. Jan 14, 2008 #5


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    Are you saying you have been asked to find cuberoots and have never heard of D'Moivre's theorem? Why that's awful! Almost as bad posting a homework problem under the mathematics thread!
  7. Jan 14, 2008 #6

    Shooting Star

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    He actually doesn't need D'Moivre's theorem, only the three cube roots of unity.

    Hi sara_87,

    Have you done the complex cube roots of one in class? You know, omega, omega^2 etc.? Show some work.
  8. Jan 14, 2008 #7


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    Or just basic remainder and factor theorem....f(z)=z^3+8=0. But the other ways are faster.
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