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Math Help - Complex numbers: solving equation

  1. #1
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    Complex numbers: solving equation

    Here's what I've done so far.

     \displaystyle \begin{align*} z^2 = -8 + 6i \\(x + yi)^2 = -8 + 6i \\ (x^2 - y^2) + (2xy)i = -8 + 6i \\ \\ x^2 - y^2 = -8 \textrm{ and } 2xy = 6 \end{align*}

    Using the second equation:

     \displaystyle \begin{align*} x = \frac{3}{y} \end{align*}

    Substituting into the first equation:

     \displaystyle \begin{align*} \frac{9}{y^2} - y^2 = -8 \\ 9 - y^4 = -8y^2 \\ -y^4 + 8y^2 + 9 = 0 \end{align*}

    And that is where I'm stuck, I thought of using a horner scheme but still didn't make it



    P.S: Could someone indicate how to correctly set the alignment of the last equation ?
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    MHF Contributor MarkFL's Avatar
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    Re: Complex numbers: solving equation

    An alternate approach would be to rewrite the right side:

    -8+6i=1+2\cdot3i+9i^2=(1+3i)^2
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    MHF Contributor MarkFL's Avatar
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    Re: Complex numbers: solving equation

    To continue your method, write your quadratic in y^2 as:

    y^4-8y^2-9=0

    \left(y^2-9 \right)\left(y^2+1 \right)=0

    Since y must be real, we are left with:

    (y+3)(y-3)=0
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    Member kalyanram's Avatar
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    Re: Complex numbers: solving equation

    Quote Originally Posted by theintervurt View Post
    -y^4 + 8y^2 + 9 = 0 \end{align*} [/tex]
    You can continue this line of thought thought. This is a quadratic in y^2. Solve for y^2 and get suitable values for x.
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    Re: Complex numbers: solving equation

    You're fine - you just need to know a useful little trick to continue. Some 3-term polynomials are actually quadratics in disguise.

    y^4 - 8y^2 - 9 = 0 is (y^2)^2 - 8(y^2)^1 - 9 = 0, so it's "really" u^2 - 8u - 9 = 0. Solving (I'll just factor it):

    (y^2 - 9)(y^2 + 1) = 0, so y^2 = 9 or y^2 = -1.

    When you start with z = x+iy, you're saying x and y are real numbers. Thus you can ignore the imaginary solutions for y, leaving only y = 3 and y = -3.

    This trick can be useful. Similar examples: q^{14} + 5q^7 + 6 = 0, 4w^{100} - 21w^{50} + 20 = 0, r^{8/3} + 5r^{4/3} - 5 = 0, and so forth.

    -----

    By the way, your approach is completely fine, but can also do it by putting everything into polar form (this approach works better with higher powers than quadratics):

    There's an ERROR somewhere in what follows, but the approach is correct:

    Let z=Re^{i\theta}, -8+6i = 10e^{i\alpha}, where \alpha = arctan(6/(-8)) = arctan(-.75). Then

    z^2 = -8+6i \implies R^2e^{i2\theta} = 10e^{i(\alpha + 2\pi k)} \implies R = \sqrt{10}, 2\theta = \alpha + 2k\pi, where k is any integer.

    Then \theta = \alpha/2 + \pi k, so z = \sqrt{10}e^{i\alpha/2} or z = \sqrt{10}e^{i ( \alpha /2+ \pi )}.

    Note that tan(\alpha) = -.75, so tan(\alpha /2) = \frac{tan(\alpha)}{1+\sqrt{1+tan^2(\alpha)}} = \frac{(-3/4)}{1+\sqrt{1+(-3/4)^2}}

    = \frac{(-3/4)}{1+\sqrt{25/16}} = \frac{(-3/4)}{1+(5/4)} = \frac{(-3/4)}{(9/4)} = -1/3.

    Thus z = \sqrt{10}e^{i arctan(-1/3)} or z = \sqrt{10}e^{i ( arctan(-1/3) + \pi )}.

    Thus z = \sqrt{10}e^{- i arctan(1/3)} or z = -\sqrt{10}e^{-i arctan(1/3)}.

    Thus z = \sqrt{10}(cos(-arctan(1/3)) + i sin( -arctan(1/3)) or z = -\sqrt{10}(cos(-arctan(1/3)) + i sin( -arctan(1/3)).

    Thus z = \sqrt{10}(cos(arctan(1/3)) - i sin(arctan(1/3)) or z = \sqrt{10}(-cos(arctan(1/3)) + i sin(arctan(1/3)).

    Triangle is 1, 3, \sqrt{10}, so cos(arctan(1/3)) = 3/\sqrt{10}, sin(arctan(1/3)) = 1/\sqrt{10}.

    Thus z = \sqrt{10}((3/\sqrt{10}) - i (1/\sqrt{10}) or z = \sqrt{10}( -(3/\sqrt{10})+ i (1/\sqrt{10}).

    Thus z = 3 - i or z = -3 + i.

    Ugg - I blundered in here somewhere. It should be 1 + 3i and -1 - 3i. But the approach is correct, and good to know.
    Last edited by johnsomeone; September 11th 2012 at 11:31 PM.
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    Re: Complex numbers: solving equation

    Thank you all for the answers.
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    Lightbulb Re: Complex numbers: solving equation

    z=\pm \sqrt{-8+6i}=\pm (1+3i)
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