# Math Help - DeMoivre's Theorem problem

1. ## DeMoivre's Theorem problem

...

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

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

$x^4+3=\sqrt[4]{3} (1-i)(1+i)(-1+i)(-1-i)$

3. 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,
$x^4=-3$
In polar complex form,
$x^4=3(\cos \pi+i\sin \pi)$

4. 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),
$x^4=3(\cos \pi +i\sin \pi)$
In order to use de Moiver's theorem you write,
$x^4=3[\cos (\pi +2\pi k)+i\sin (\pi +2\pi k)]$
Now you use de Moiver's theorem (actually it is not cuz his theorem if for integral exponents not rational).
Thus,
$x=3^{1/4}\left[\cos \left( \frac{\pi +2\pi k}{4} \right) +i\sin \left( \frac{\pi +2\pi k}{4} \right)\right]$
And each solution is for $k=0,1,2,3$

5. 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 $k=0,1,2,3$ 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 $a$ is zero of $f(x)$ if and only if $(x-a)$ divides $f(x)$)

6. Ignore what I said before I made a mistake.
-------
This is the factorization
$
\left(
x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$
$
\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)$
$
\left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$
$
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)
$

7. 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 $
\left(
x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$
$
\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)$
$
\left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$
$
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)
$

is written as a product of irreducibles in $\mathbb C[x]$ (since each is degree 1) and
$
\left(
x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$
, $
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)
$
are not in $\mathbb R[x]$ but notice they are conjugates, so $
\left(
x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$
$
\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)$
$
\left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$
$
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)
$
$=x^4+3$ is in $\mathbb R[x]$ and is irreducible in $\mathbb R[x]$
That is the factorization in $\mathbb{C}[x]$, good. Now find its factorization in $\mathbb{R}[x]$. The trick is to look for the conjugates. Remember that, $(x-z_1)(x-z_2)=x^2-x(z_1+z_2)+z_1z_2$, where $z_1+z_2$ is no longer complex and $z_1z_2$ is no longer complex.

The first pair of conjugates are,
$
\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,
$x^2-\sqrt{2}\sqrt[4]{3}x+\sqrt{3}$
The other pairs of conjugates are,
$
\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,
$x^2+\sqrt{2}\sqrt[4]{3}x+\sqrt{3}$
So the factorization of,
$x^4+3$ in $\mathbb{R}[x]$ is,
$[x^2+\sqrt{2}\sqrt[4]{3}x+\sqrt{3}][x^2-\sqrt{2}\sqrt[4]{3}x+\sqrt{3}]$