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Homework Help: Solving a polynomial with complex properties!

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    If z = cosθ + isinθ, show that zn + z-n = 2cosnθ. Hence solve the equation 3z4 - z3 + 4z2 - z + 3 = 0 for complex z.
    (Hint: Reduce the given equation to an equation with the unknown cosθ)


    2. Relevant equations
    z = cosθ + isinθ
    zn + z-n = 2cos(nθ)
    3z4 - z3 + 4z2 - z + 3 = 0
    cos2θ = 2cos2θ - 1

    3. The attempt at a solution
    From de Moivre's Theorem
    zn = cos(nθ) + isin(nθ) AND z-n = cos(-nθ) + isin(-nθ) = cos(nθ) - isin(nθ)
    Hence zn + z-n
    = cos(nθ) + isin(nθ) + cos(nθ) - isin(nθ)
    = 2cos(nθ)

    3z4 - z3 + 4z2 - z + 3 = 0
    Divide both sides by z2
    3z2 - z + 4 - z-1 + 3z-2 = 0

    Rearranging to
    3z2 + 3z-2 - z - z-1 + 4 = 0
    3(z2 + z-2) - (z + z-1) + 4 = 0

    Using zn + z-n = 2cos(nθ)
    3(2cos2θ) - (2cosθ) + 4 = 0

    Substituting cos2θ = 2cos2θ - 1
    3(2[2cos2θ - 1]) - 2cosθ + 4 = 0

    Simplifies down to
    6cos2θ - cosθ - 1 = 0

    Solving for cosθ
    (2cosθ - 1)(3cosθ + 1) = 0
    Therefore cosθ = 1/2 OR cosθ = -1/3
    cosθ = 1/2, θ = π/3 or 5π/3
    cosθ = -1/3, θ = cos-1(-1/3)
    (Note: π = funny looking pi)

    Hence (using complex conjugate is also a root of polynomial)
    z1 = 1/2 + sqrt(3)/2 i
    z2 = 1/2 - sqrt(3)/2 i
    z3 = -1/3 + sin[cos-1(-1/3)] i
    z4 = -1/3 - sin[cos-1(-1/3)] i

    I'm pretty confident that I've done it right until the cosθ = -1/3 and getting sin[cos-1(-1/3)] in the roots z3 and z4. So my question is have I the whole question correctly? If not can you show me how it should be done. Thanks in advance!
     
  2. jcsd
  3. Mar 28, 2010 #2

    vela

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    You did it correctly, though you should simplify sin[arccos(-1/3)]. You can always check your answers by plugging them back into the original equation.
     
  4. Mar 29, 2010 #3
    Ok thankyou (:

    umm another question unrelated
    but now I have another concern

    for a function to be STRICTLY increasing, is the point 0 actually included? Because me and my friends all have different opinions. So the gradient at x = 0 is 0 so would the function still be considered as strictly increasing if it included x=0?
     
  5. Mar 30, 2010 #4

    vela

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    Strictly increasing means f'(x)>0 for all x. If you have f'(x)≥0, the function is monotonically increasing.
     
  6. Mar 30, 2010 #5

    rock.freak667

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    Just to add one point to your original post, when I did questions like these in Further Math, it was always best to show that z=0 was not a root before dividing by any power of z.
     
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