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Thread: complex numbers.pt2

  1. #1
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    complex numbers.pt2

    Problem:

    Is it possible to determine the real constant a so that
    $\displaystyle z=2+i$
    is a root for the equation
    $\displaystyle z^3+az^2+13z-10=0$

    I started out like so that if p(b)=0, where p(z) is a polynomial of a degree n,
    then p(z)=(z-b)*f(z) where f(z) is a polynomial of a degree n-1

    meaning that

    $\displaystyle z^3+az^3+13z-10=(z-(2+1))(z^2+dz+e)$

    but apparently that was wrong..

    Sorry for all the questions, but I have an exam coming up this monday
    Last edited by Greenb; Jan 30th 2009 at 01:37 PM. Reason: Misstyping in equation
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  2. #2
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    Quote Originally Posted by Greenb View Post
    Problem:

    Is it possible to determine the real constant a so that
    $\displaystyle z=2+1$
    is a root for the equation
    $\displaystyle z^3+az^3+13z-10=0$

    I started out like so that if p(b)=0, where p(z) is a polynomial of a degree n,
    then p(z)=(z-b)*f(z) where f(z) is a polynomial of a degree n-1

    meaning that

    $\displaystyle z^3+az^3+13z-10=(z-(2+1))(z^2+dz+e)$

    but apparently that was wrong..

    Sorry for all the questions, but I have an exam coming up this monday
    Do you mean $\displaystyle z = 2 + {\color{red}i}$ is the root?

    Just substitute into the cubic, simplify and equate the real and imaginary parts of the result to zero. Then solve for $\displaystyle a$.
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  3. #3
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    This is what I got:

    $\displaystyle (z-(2+i))(z^2+dz+e)=$
    $\displaystyle =z^3+dz^2+ez-2z^2-2dz-2e-iz^2-idz-ie$

    somehow I think I need the values for d and e to proceed..

    How do I "equate the real and imaginary parts of the result to zero" seeing that I should leave the left side intact?
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  4. #4
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    Please check your OP.
    It must contain some mistakes.
    You want $\displaystyle 2+\color{red}i$ to be a root of $\displaystyle z^3+az^{{\color{red}2}}+13z-10$.
    If I have guessed correctly, there is a simple answer for $\displaystyle a$.
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  5. #5
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    Ah didn't see that I had typed it wrong Anyway, yes a=-6 but I really can't see how.
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  6. #6
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    Quote Originally Posted by Greenb View Post
    a=-6 but I really can't see how.
    Indeed, that is what I got.
    $\displaystyle \left( {2 + i} \right)^3 + a\left( {2 + i} \right)^2 + 13\left( {2 + i} \right) - 10 = \left( {3a + 18} \right) + i\left( {4a + 24} \right)$
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  7. #7
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    I thought way more complicated and was actually totally off the track hehe. Thanks for clearing it up for me, I now see that mr fantastic also meant this solution, I was trying to fit in his advice into my solution ^^
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