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Math Help - complex numbers polynomial

  1. #1
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    complex numbers polynomial

    the question is "If z = 1+i is a zero of the polynomial z^3 + az^2 + bz + 10 - 6i, find the constants 'a' and 'b' given that they are real."

    now becuase 1+i is a solution i am going to assume 1-i is a solution becuase the polynomial has real co-efficients. so (z - 1 - i)(z - 1 + i) = z^2 - 2z + 2 as a quadratic factor

    im suspecting there could be a problem with the question becuase on my CAS calculator 1+i is certainly a solution but 1-i is NOT (substituting the answers from the back of the book a=6, b=-8)
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  2. #2
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    If the coefficients are ALL real, then complex roots come in conjugate pairs. But, if you introduce even one complex coefficient, all bets are off. You do know that (z-1-i) is a factor of the polynomial. Why not try either polynomial long division or synthetic division?
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  3. #3
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    If the coefficients are real, then complex roots always occur as complex conjugates. You are correct there. But the coefficients are not all real (you have -6i at the end).

    I would start by long dividing your cubic by the linear factor z - 1 - i, to get the quadratic factor.
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  4. #4
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    ok i'll try dividing by the given factor i didnt realise that there was a complex co-efficient i thought that only applied to the terms with a "z".
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  5. #5
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    It is a coefficient: of the z^{0} term!
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  6. #6
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    ok i divided the cubic and the remaining quadratic is z^2 + (1 + i + a)z + a + b + (a + 2)i + (b + 8 + (2a + b - 4)i)/z - 1 - i

    ive verified this is correct on the CAS calculator but i dont know how to get the answer, the "quadratic" has 4 terms
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  7. #7
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    But you know it's a factor, so the remainder must be 0...
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  8. #8
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    so the end part (b+8+(2a+b-4)i)/(z-1-i), MUST equal zero?

    -----> b+8+(2a+b-4)i = 0
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  9. #9
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    ignore previous post
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  10. #10
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    Correct, but what to do becomes more obvious if you write it as \displaystyle b + 8 + (2a + b - 4)i = 0 + 0i, because now you equate real and imaginary parts to get two equations to solve simultaneously for \displaystyle a,b.
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  11. #11
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    ok ive solved it and know what i did right/wrong, i just abandoned a couple of easy solving techniques along the way
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