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Complex number equation

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data

    Solve

    [tex]
    z^3 - 3z^2 + 6z - 4 = 0

    [/tex]

    3. The attempt at a solution

    I tried factoring a z and quadratic equation but went nowhere

    Input apreciated
     
  2. jcsd
  3. Sep 22, 2008 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Well there is a constant term in the equation, so z is clearly not a factor.

    The first thing to take note of is that this is a 3rd degree polynomial and so there must be 3 roots. Moreover, at least one of those roots must be real. So try to find a real root by inspection: plug in z=0, z=1, z=-1, z=2...etc. until you find a root z_0 and then factor out a (z-z_0) to obtain a quadratic equation you can then solve to find the other two roots.
     
  4. Sep 22, 2008 #3
    One thing that you could try out is assume [tex]a+ib[/tex] is a solution to the equation, plug it in and equate the real and imaginary parts individually equal to zero. That should give you two degree 3 equations which you can individually try to solve...
     
  5. Sep 22, 2008 #4
    Ok so 1 works for z0

    If I factor a z-1 how would the equation look?
     
  6. Sep 22, 2008 #5
    Use any polynomial dividing technique

    (Z^3- 3Z^2+ 6 Z-4)/(z-1)= z^2 -2z+4

    So your eqn is:

    (z^2 -2z+4) (z-1)=0

    The rest simple algebra!
     
  7. Sep 22, 2008 #6

    Defennder

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    Simply divide the f(z) expression by z-1 using polynomial long division. The result is how it would look (if you did it correctly).
     
  8. Sep 22, 2008 #7
    THanks for the help
     
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