Thread: Factorising polynomial over complex numbers

1. Factorising polynomial over complex numbers

The question:
Factorise the polynomial $\displaystyle z^4 + 2z^2 - 3$ over the complex numbers.

My attempt:
Let $\displaystyle u = z^2$

$\displaystyle u^2 + 2u -3 = 0$

$\displaystyle \frac{-2 \pm \sqrt{4 - 4(1)(-3)}}{2}$

u = 1, -3

(u - 1)(u + 3) = 0
$\displaystyle (z^2 - 1)(z^2 + 3) = 0$

What do I do from here? Do I take z^2 = 1 (for example) then apply De'Movres theorem to find the roots of unity? Or is there a better way? Thanks.

2. Originally Posted by Glitch
The question:
Factorise the polynomial $\displaystyle z^4 + 2z - 3$ over the complex numbers.

My attempt:
Let $\displaystyle u = z^2$

$\displaystyle u^2 + 2u -3 = 0$

$\displaystyle \frac{-2 \pm \sqrt{4 - 4(1)(-3)}}{2}$

u = 1, -3

(u - 1)(u + 3) = 0
$\displaystyle (z^2 - 1)(z^2 + 3) = 0$

What do I do from here? Do I take z^2 = 1 (for example) then apply De'Movres theorem to find the roots of unity? Or is there a better way? Thanks.
If u=z^2

2u != 2z

Suppose you got:

(z^2 - 1)(z^2 + 3)

(z^2 - 1)(z^2 + 3) =(z-1)(z+1)(z-i*sqrt(3))(z+i*sqrt(3))

3. Originally Posted by Also sprach Zarathustra
If u=z^2

2u != 2z
Sorry, forgot to add the square to the Q.

4. $\displaystyle \displaystyle u^2 + 2u - 3 = (u - 1)(u + 3)$

$\displaystyle \displaystyle = (z^2 - 1)(z^2 + 3)$

$\displaystyle \displaystyle = (z - 1)(z + 1)[z^2 - (i\sqrt{3})^2]$

$\displaystyle \displaystyle = (z - 1)(z + 1)(z - i\sqrt{3})(z + i\sqrt{3})$.

The solutions should now be obvious.

5. Because those are roots of purely real numbers, you don't really need to use DeMoivre's formula.

6. Originally Posted by HallsofIvy
Because those are roots of purely real numbers, you don't really need to use DeMoivre's formula.
Now that I think about it, that's pretty obvious. I hope my brain is functioning better tomorrow for my exam. >_<

Thanks again guys!