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- Jul 13th 2006, 08:37 AMharoldDeMoivre's Theorem problem
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- Jul 13th 2006, 09:44 AMThePerfectHackerQuote:

Originally Posted by**harold**

$\displaystyle x^4+3=\sqrt[4]{3} (1-i)(1+i)(-1+i)(-1-i)$ - Jul 13th 2006, 11:58 AMThePerfectHackerQuote:

Originally Posted by**harold**

$\displaystyle x^4=-3$

In polar complex form,

$\displaystyle x^4=3(\cos \pi+i\sin \pi)$ - Jul 13th 2006, 06:11 PMThePerfectHackerQuote:

Originally Posted by**harold**

As you understand (I hope),

$\displaystyle x^4=3(\cos \pi +i\sin \pi)$

In order to use de Moiver's theorem you write,

$\displaystyle x^4=3[\cos (\pi +2\pi k)+i\sin (\pi +2\pi k)]$

Now you use de Moiver's theorem (actually it is not cuz his theorem if for integral exponents not rational).

Thus,

$\displaystyle x=3^{1/4}\left[\cos \left( \frac{\pi +2\pi k}{4} \right) +i\sin \left( \frac{\pi +2\pi k}{4} \right)\right]$

And each solution is for $\displaystyle k=0,1,2,3$ - Jul 15th 2006, 07:01 PMThePerfectHackerQuote:

Originally Posted by**harold**

(Basically you use the factor theorem $\displaystyle a$ is zero of $\displaystyle f(x)$ if and only if $\displaystyle (x-a)$ divides $\displaystyle f(x)$) - Jul 15th 2006, 08:01 PMThePerfectHacker
Ignore what I said before I made a mistake.

-------

This is the factorization

$\displaystyle

\left(

x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$$\displaystyle

\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)$$\displaystyle

\left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right) $$\displaystyle

\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)

$ - Jul 16th 2006, 06:07 AMThePerfectHackerQuote:

Originally Posted by**harold**

The first pair of conjugates are,

$\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)

$

The polynomial they produce is,

$\displaystyle x^2-\sqrt{2}\sqrt[4]{3}x+\sqrt{3}$

The other pairs of conjugates are,

$\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)

$

The polynomial they produce is,

$\displaystyle x^2+\sqrt{2}\sqrt[4]{3}x+\sqrt{3}$

So the factorization of,

$\displaystyle x^4+3$ in $\displaystyle \mathbb{R}[x]$ is,

$\displaystyle [x^2+\sqrt{2}\sqrt[4]{3}x+\sqrt{3}][x^2-\sqrt{2}\sqrt[4]{3}x+\sqrt{3}]$