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Precal hw on complex numbers

  1. Nov 24, 2004 #1
    Z^6 + 64 = 0
    there are suppose to be 6 solutions :confused:
    need help on how to find the roots
     
    Last edited: Nov 24, 2004
  2. jcsd
  3. Nov 24, 2004 #2
    Are you familiar with de Moivre's theorem? The problem is pretty easy then. Alternatively, you can factor the equation [tex]Z^6 + 64=0[/tex] into
    [tex](Z^3 - 8i)(Z^3+8i)[/tex] from which you can factor more.
     
    Last edited by a moderator: Nov 24, 2004
  4. Nov 24, 2004 #3
    opps, sorry, i must 've made myself unclear, the goal is to find the 6 values of z, but thanks for heling on the roots! :smile:
     
  5. Nov 24, 2004 #4

    Gokul43201

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    Write [itex]z^6 = r^6(cos 6\theta + isin 6\theta) [/itex] and solve.
     
  6. Nov 25, 2004 #5
    Vsage's point was that, from there, you can factor further, thus getting the six linear roots (hint: think about sum and difference of cubes).
     
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