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Math Help - Complex Roots/Multiplication

  1. #1
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    Complex Roots/Multiplication

    I'm having trouble multiplying out the conjugates to obtain a quadratic for the following. Both quadratics should multiply to the polynomial x^4 + 3 when finished...any help??

    <br />
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)\cdot \left(x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)<br />

    second pair:

    <br />
\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)\cdot \left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)<br />
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  2. #2
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    Remember that,
    (x-z_1)(x-z_2)=x^2-(z_1+z_2)x+z_1z_2
    Given,
    \left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)
    When expanded,
    \left( x - \frac{\sqrt{2}\sqrt[4]{3}}{2}+\frac{\sqrt{2}\sqrt[4]{3}i}{2} \right)
    Note, (and watch those signs ) that,
    z_1=\frac{\sqrt{2}\sqrt[4]{3}}{2}-\frac{\sqrt{2}\sqrt[4]{3}i}{2}
    In the conjugate,
    \left(x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)<br />
    When expanded,
    \left( x-\frac{\sqrt{2}\sqrt[4]{3}}{2}-\frac{\sqrt{2}\sqrt[4]{3}i}{2} \right)
    Note, (and watch those signs ) that,
    z_2=\frac{\sqrt{2}\sqrt[4]{3}}{2}+\frac{\sqrt{2}\sqrt[4]{3}i}{2}
    Thus,
    z_1+z_2=\sqrt{2}\sqrt[4]{3}
    And,
    z_1z_2=\left( \frac{\sqrt{2}\sqrt[4]{3}}{2} \right)^2+\left( \frac{\sqrt{2}\sqrt[4]{3}}{2} \right)^2=\sqrt{3}
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  3. #3
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    Hello, harold!

    Recall that: . (a + b)(a - b)\:=\:a^2 - b^2


    Both quadratics should multiply to the polynomial x^4 + 3 when finished.

    First pair: . \left( x- \sqrt[4]{3}\left[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}\right] \right)\cdot \left(x-\sqrt[4]{3}\left[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}\right]\right)<br />

    Second pair: . \left( x- \sqrt[4]{3}\left[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}\right] \right)\cdot \left( x-\sqrt[4]{3}\left[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}\right]\right)<br />

    Note that: \sqrt[4]{3}\cdot\sqrt{2}\,=\,\sqrt[4]{12} .and \left(\frac{\sqrt[4]{12}}{2}\right)^2 = \frac{\sqrt{12}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}

    First pair: . \left( x- \sqrt[4]{3}\left[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}\right] \right)\cdot \left(x-\sqrt[4]{3}\left[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}\right]\right)<br />

    . . = \;\left( x- \frac {\sqrt[4]{12}} {{2}} + i \frac {\sqrt[4]{12}} {{2}}\right)\cdot \left(x - \frac {\sqrt[4]{12}} {{2}} - i\frac {\sqrt[4]{12}} {{2}}\right)

    . . = \;\left(\left[ x- \frac {\sqrt[4]{12}} {{2}}\right] + i \frac {\sqrt[4]{12}} {{2}}\right)\cdot  \left(\left[x - \frac {\sqrt[4]{12}} {{2}}\right] - i\frac {\sqrt[4]{12}} {{2}}\right)\quad\Leftarrow\quad(a + b)(a - b)

    . . = \;\left(x- \frac {\sqrt[4]{12}} {{2}}\right)^2 - \left(i \frac {\sqrt[4]{12}} {{2}}\right)^2\quad\Leftarrow\quad a^2 - b^2

    . . = \;\left(x^2 - \sqrt[4]{12}x + \frac{\sqrt{3}}{2}\right) - \left(-\frac{\sqrt{3}}{2}\right)

    . . = \;x^2 - \sqrt[4]{12}x + \sqrt{3}


    Similarly, the second pair becomes: . x^2 + \sqrt[4]{12}x + \sqrt{3}


    We have: . \left(x^2 - \sqrt[4]{12}x + \sqrt{3}\right)\cdit\left(x^2 + \sqrt[4]{12}x + \sqrt{3}\right)

    . . =\;\left[(x^2 + \sqrt{3}) - \sqrt[4]{12}x\right]\cdot\left[(x^2+\sqrt{3}) + \sqrt[4]{12}x\right]\quad\Leftarrow (a - b)(a + b)

    . . = \;(x^2 + \sqrt{3})^2 - (\sqrt[4]{12}x)^2\quad\Leftarrow\quad a^2 - b^2

    . . = \;x^4 + 2\sqrt{3}x^2 + 3 - \sqrt{12}x^2

    . . = \;x^4 + 2\sqrt{3}x^2 + 3 - 2\sqrt{3}x^2

    . . = \;x^4 + 3 . . . ta-DAA!

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