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Solving a polynomial with complex coefficients

  1. Nov 1, 2015 #1
    1. The problem statement, all variables and given/known data
    z^6+(2i-1)z^3-1-i=0

    2. Relevant equations


    3. The attempt at a solution
    I know that I must put k=z^3 and solve the quadratic. But I'm not able to simplify the quadratic. I get the square root of (-8i+1)

    What am I supposed to do ?
     
  2. jcsd
  3. Nov 1, 2015 #2

    SteamKing

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    The square root of (1-8i) is, wait for it, another complex number!

    If you know how to express (1-8i) in polar form using Euler's formula, then finding the square root should be a snap.

    https://en.wikipedia.org/wiki/Complex_number
     
  4. Nov 1, 2015 #3

    SammyS

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    Check that result from quadratic formula again.

    I get something much simpler under the square root.
     
  5. Nov 2, 2015 #4

    ogg

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    FWIW, given any complex number z, then assume √z = a+bi and so z = (a+bi)² which means that Real(z²) = a²-b² and Im(z²) = 2abi; two equations, two unknowns. simple to solve, i hope.
     
  6. Apr 28, 2016 #5
    ogg, I think you meant to write:
    Re##(z)=a^2-b^2## and Im##(z)=2ab##
     
  7. Apr 28, 2016 #6

    SammyS

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    Right. The imaginary part doesn't include the imaginary unit, i , by the usual convention.

    Hello, Ian Taylor. Welcome to PF !

    If you notice, this thread is 1/2 year old.
     
  8. Apr 29, 2016 #7
    Thanks SammyS. He had also written ##z^2## rather than ##z##. which contradicts his original definition.
     
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