Complex Roots/Multiplication

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• Jul 22nd 2006, 04:48 PM
harold
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??

$
\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:

$
\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)
$
• Jul 22nd 2006, 06:14 PM
ThePerfectHacker
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 :eek: ) 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)
$

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 :eek: ) 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,
$z_1z_2=\left( \frac{\sqrt{2}\sqrt[4]{3}}{2} \right)^2+\left( \frac{\sqrt{2}\sqrt[4]{3}}{2} \right)^2=\sqrt{3}$
• Jul 23rd 2006, 09:44 AM
Soroban
Hello, harold!

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

Quote:

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)
$

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)
$

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)
$

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

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

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

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

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

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!