# Factorising polynomial over complex numbers

• March 19th 2011, 06:23 PM
Glitch
Factorising polynomial over complex numbers
The question:
Factorise the polynomial $z^4 + 2z^2 - 3$ over the complex numbers.

My attempt:
Let $u = z^2$

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

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

u = 1, -3

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

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.
• March 19th 2011, 06:34 PM
Also sprach Zarathustra
Quote:

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

My attempt:
Let $u = z^2$

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

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

u = 1, -3

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

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))
• March 19th 2011, 06:51 PM
Glitch
Quote:

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

2u != 2z

Sorry, forgot to add the square to the Q.
• March 19th 2011, 07:22 PM
Prove It
$\displaystyle u^2 + 2u - 3 = (u - 1)(u + 3)$

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

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

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

The solutions should now be obvious.
• March 20th 2011, 03:38 AM
HallsofIvy
Because those are roots of purely real numbers, you don't really need to use DeMoivre's formula.
• March 20th 2011, 05:02 AM
Glitch
Quote:

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!