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Math Help - complex number equation

  1. #1
    Member Jones's Avatar
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    complex number equation

    Hi,

    I would like some help with this.

    The equation p(z)=z^4-2z^3+4z^2-2z+3 is given.

    Show that one of the roots are i.


    So i verified that i is indeed a root. And since complex roots are "mirrored", another root must be -i.

    The factor theorem.
    (z-i)(z+i) = z^2 +1

    Then i factored the equation using polynomial division and the result is
    (z^2+1)(z^2-2z+3)

    the second bracket is a simple quadratic.
    \frac{-2}{-2} \pm \sqrt{\bigg(\frac{-2}{2}\bigg)^2 -3}

    =  1 \pm \sqrt{1-3}

    This is where i get it wrong, isn't the square root expression supposed to evaluate to 2i ?

    The correct answer is 1\pm i\sqrt{2} however, the result im getting is 1\pm 2i

    Any ideas?
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  2. #2
    Junior Member
    Joined
    Feb 2010
    Posts
    29
    Quote Originally Posted by Jones View Post
    Hi,
    =  1 \pm \sqrt{1-3}

    This is where i get it wrong, isn't the square root expression supposed to evaluate to 2i ?

    The correct answer is 1\pm i\sqrt{2} however, the result im getting is 1\pm 2i

    Any ideas?
    No, you have it right.

    =  1 \pm \sqrt{1-3} = 1 \pm \sqrt{-2} = 1 \pm \sqrt{-1*2} = 1 \pm i\sqrt{2}

    Just remember that the square root doesn't go away just because you have a negative number under it. In the same way \sqrt{8} = \sqrt{4*2} = 2\sqrt{2}, right?
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