# DeMoivre's Theorem problem

• Jul 13th 2006, 08:37 AM
harold
DeMoivre's Theorem problem
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• Jul 13th 2006, 09:44 AM
ThePerfectHacker
Quote:

Originally Posted by harold
Using DeMoivre's Theorem (or a possible simpler method if there is one) find all the roots of $\displaystyle x^4 + 3$.

Yikes!!

I presume you can to factor this, based on your other post on irreducible polynomials?

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

Originally Posted by harold
Hi TPH,

Yes--I think I have to use this for my other post. What did you substitute in for demoivre's theorem?

You have,
$\displaystyle x^4=-3$
In polar complex form,
$\displaystyle x^4=3(\cos \pi+i\sin \pi)$
• Jul 13th 2006, 06:11 PM
ThePerfectHacker
Quote:

Originally Posted by harold
Hi TPH
Could you run through how you got the factored version using the above? Also, in the other forum, I still don't see how to bring it all together to conclude irreducibility (as well as factor part b). Thanks in advance for any help.
harold

I am not going to do every single step, you are a big boy you can handle it yourself.

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 PM
ThePerfectHacker
Quote:

Originally Posted by harold
Isn't the above on the right of the = sign just a constant? How does it equal the left?

I found the four solutions using $\displaystyle k=0,1,2,3$ from DeMoivre's Thm...how do I form these into a product of linear factors? I'm also having trouble doing this for part b) for my irreducible polys thread.

Tell me your solutions, so I can see if you are doing it correctly.

(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 PM
ThePerfectHacker
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) $$\displaystyle \left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right) • Jul 16th 2006, 06:07 AM ThePerfectHacker Quote: Originally Posted by harold And so now I can use the above as a guide to answer all my irreducible poly questions in the following way: Since \displaystyle \left( x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$$\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)$
is written as a product of irreducibles in $\displaystyle \mathbb C[x]$ (since each is degree 1) and
$\displaystyle \left( x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$,$\displaystyle \left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)$ are not in $\displaystyle \mathbb R[x]$ but notice they are conjugates, so $\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) $$\displaystyle \left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)$$\displaystyle =x^4+3$ is in $\displaystyle \mathbb R[x]$ and is irreducible in $\displaystyle \mathbb R[x]$

That is the factorization in $\displaystyle \mathbb{C}[x]$, good. Now find its factorization in $\displaystyle \mathbb{R}[x]$. The trick is to look for the conjugates. Remember that, $\displaystyle (x-z_1)(x-z_2)=x^2-x(z_1+z_2)+z_1z_2$, where $\displaystyle z_1+z_2$ is no longer complex and $\displaystyle z_1z_2$ is no longer complex.

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