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Thread: 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 $\displaystyle x^4 + 3$ when finished...any help??

    $\displaystyle
    \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)
    $

    second pair:

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

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


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

    First pair: .$\displaystyle \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)
    $

    Second pair: .$\displaystyle \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)
    $

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

    First pair: .$\displaystyle \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)
    $

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

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

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

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

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


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


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

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

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

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

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

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

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