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Thread: complex numbers

  1. #1
    Member Jones's Avatar
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    complex numbers

    Hello,

    I have the complex polynomial: $\displaystyle z^3+(3+i)z^2+(4-2i)z+(a-3i) = 0$

    I need to determine the parameter a, so that the equation gets a real negative root.
    I don't know any root of the polynomial, if i did it would have been easy.

    Polynomial division?
    Any ideas?
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  2. #2
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    Substitute $\displaystyle z=-1$. You will see what $\displaystyle a$ must equal.
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  3. #3
    Member Jones's Avatar
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    I do, care to elaborate on the science behind your methodology?
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  4. #4
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    Quote Originally Posted by Jones View Post
    I do, care to elaborate on the science behind your methodology?
    There is no methodology here.
    If $\displaystyle z=-1$ gives a zero result then $\displaystyle -1$ is negative root.
    $\displaystyle \left( { - 1} \right)^3 + \left( {3 + i} \right)\left( { - 1} \right)^2 + \left( {4 - 2i}\right)\left( i \right) + \left( {a - 3i} \right) = 0$

    Do you see what value of $\displaystyle a$ makes that true?

    There is no mystery there if you understand anything about this question.
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  5. #5
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    What Plato is saying is that this equation will have many different roots depending on what a is. And that you can make one root to be anything you want by choosing a correctly. In particular, to be sure that the equation has at least one real, negative, root, you can just choose z to be any real, negative, number you want.

    for example, if you chose to make that root -2 instead of -1, you would have the equation $\displaystyle (2)^3+ (3+ i)(-2)^2+ (4- 2i)(-2)+ (a- 3i)= 0$ and solve that for a.

    However, I see a difficulty with this. It is possible that the form (a- 3i) is supposed to indicate that a is a real number. If so, that should have been said, but if so then that makes the problem harder!

    Assuming that a must be real, we can separate the equation into real and imaginary parts:
    Real part: $\displaystyle z^3+ 3z^2+ 4z+ a= 0$
    Imaginary part: $\displaystyle z^2- 2z- 3= 0$.
    (I have used the fact that z is a real number in separating.)

    Now you can solve that second, quadratic, equation for z and see what values of a allow a negative z that satifies both equations.
    Last edited by HallsofIvy; Jul 27th 2010 at 04:08 AM.
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  6. #6
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    Quote Originally Posted by Plato View Post
    There is no methodology here.
    If $\displaystyle z=-1$ gives a zero result then $\displaystyle -1$ is negative root.
    $\displaystyle \left( { - 1} \right)^3 + \left( {3 + i} \right)\left( { - 1} \right)^2 + \left( {4 - 2i}\right)\left( i \right) + \left( {a - 3i} \right) = 0$
    Typographical error: That "$\displaystyle (4- 2i)(i)$" should be "$\displaystyle (4- 2i)(-1)$".

    Do you see what value of $\displaystyle a$ makes that true?

    There is no mystery there if you understand anything about this question.
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