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Math Help - Complex numbers problem.

  1. #1
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    Complex numbers problem.

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  2. #2
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    Quote Originally Posted by .::MadMaX::. View Post
    Why not just use the Quadratic Formula?

    Z^2 + (5 - 2i)Z + 6(1 - i) = 0


    Z = \frac{-(5 - 2i) \pm \sqrt{(5 - 2i)^2 - 4(1)[6(1 - i)]}}{2(1)}

     = \frac{-5 + 2i \pm \sqrt{25 - 20i + 4i^2 - 24(1 - i)}}{2}

     = \frac{-5 + 2i \pm \sqrt{25 - 20i - 4 - 24 + 24i}}{2}

     = \frac{-5 + 2i \pm \sqrt{-3 + 4i}}{2}.


    Now to find \sqrt{-3 + 4i}, convert to polars and use DeMoivre's Theorem.
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    Quote Originally Posted by .::MadMaX::. View Post


    I really can't understand what you're trying to do: why not the good-old and well-known formula to find the roots of a quadratic equation?!!?

    We have z^2+(5-2i)z+6(1-i)=0\Longrightarrow a=1\,,\,b=5-2i\,,\,c=6(1-i)\Longrightarrow Discriminant=\Delta=b^2-4ac=21-20i-24(1-i)=-3+4i , so:

    z_{1,2}=\frac{-b\pm \sqrt{\Delta}}{2a}=\frac{-(5-2i)\pm \sqrt{-3+4i}}{2}=\frac{(-5+2i)\pm(1+2i)}{2}=\left\{\begin{array}{l}-2+2i\\-3\end{array}\right.

    Tonio
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    Quote Originally Posted by .::MadMaX::. View Post
    In both both methods you are really assuming that Z^2= 0. That's why you got the wrong answer!
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    Thank you all,but I just don't know how to handle this:



    EDIT: Anyone?
    Last edited by .::MadMaX::.; January 25th 2010 at 07:55 AM.
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