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Solving a complex numbered cubic equation

  1. Nov 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve the equation ## z^3 − z^2 + z − 1 = 0 ## first by inspection, and then by the
    method described above. where Z is a complex number. (Alan F. Beardon, Algebra and Geometry)

    The method described above is shown in the attachment.

    2. Relevant equations
    The method is shown in the attachment.

    3. The attempt at a solution

    Solve by inspection means to draw the graph of this equation and check where it intersects the x axis??
    Shown in the attachment. the solution is 1 if i am not wrong

    My algebraic solution is different. why? My attempt is also in the attachement.

    Can you please also explain why there are six solutions for ## z ## when #z^3# has two solutions because of quadratic equation. cube root of any number has only one solutions which is the number itself. e.g. ## \sqrt[3]{-1} = -1*-1*-1 or \sqrt[3]{1} = 1*1*1## so cube root only gives one solution then what do they mean by six solutions??


    danke...
     

    Attached Files:

  2. jcsd
  3. Nov 28, 2014 #2

    SteamKing

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    By inspection means just that: look at the terms of the equation and substitute some guesses which can be evaluated using mental arithmetic.

    For the equation in the OP, looking at how the signs of the terms alternate, guessing that z = 1 is a solution is a solid hunch, since the magnitudes of z, z2, and z3 are all 1.
     
  4. Nov 29, 2014 #3

    HallsofIvy

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    "By inspection" often means just trying simple numbers. What do you get if you put z= 1 into the equation?
     
  5. Nov 29, 2014 #4

    epenguin

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    Or you could look at the fist two terms, then look at the second two terms and notice these paddies have, er, a something in common. ;)
     
    Last edited: Nov 29, 2014
  6. Nov 29, 2014 #5

    haruspex

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    It does not say there are six solutions for z. It says there are six solutions for ##\zeta##, but pairs of these produce the same value for z.
     
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