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Finding Roots of complex numbers (solve non-linear multi-variable system of equation)

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data
    1. z^6=(64,0)
    2. z^4=(3,4)


    2. Relevant equations
    These are expanded out into Real and Imaginary components (treat them seperate):
    1. REAL (EQ 1) - x^6-15x^4y^2+15x^2y^4-y^6=64
    IMAG (EQ 2) - 6x^5y-20x^3y^3+6xy^5=0
    From here, you basically solve these for all six roots.
    2. REAL (EQ 1) - x^4-6x^2y^2+y^4=3
    IMAG (EQ 2) - 4x^3y-4xy^3=4


    3. The attempt at a solution
    These must be done algebraically, not using Euler angle components (answer would be trivial)

    For #1, I have EQ 2 broken down into
    3z^2+3v^2=10zv where z=x^2 and v=y^2.
    I know the solution is z= 3z and z=3/z, which I can then plug into EQ 1, and all my answers will be given. My algebra is just lacking to get those 2 answers for z.

    #2 is a lot more difficult and there is no 'zero equation'. Professor gave us hint to make both sides =12 at the end, subtract them to get an equation with 0; this gives me 3 equations

    EQ 2 - xy(x^2-y^2)=1
    EQ 1 - x^4+y^4=3(1+2x^2y^2)
    EQ 3 (new formed, what I will likely end up using to solve x in terms of y) - x^4-3x^3y-6x^2y^2+3xy^3+y^4=0

    My algerbra must be lacking. Not looking for a given answer, but any hints that will help me solve these non-linear multi-variable system of equations
     
  2. jcsd
  3. Sep 10, 2012 #2
    Re: Finding Roots of complex numbers (solve non-linear multi-variable system of equat

    Is there some reason you're avoiding using the polar (complex exponential) form of complex numbers?
     
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