# Solve in Complex Numbers

• Oct 27th 2007, 01:27 PM
oceanmd
Solve in Complex Numbers

Solve this equation in the complex number system:x^3-8=0
I have only a real number answer, which is 2. How to solve it in complex numbers?

Thank you
• Oct 27th 2007, 01:50 PM
red_dog
$x^3-8=0\Leftrightarrow(x-2)(x^2+2x+4)=0$
$x_1=2$
$x^2+2x+4=0\Rightarrow\triangle=-12\Rightarrow x_{2,3}=\displaystyle\frac{-2\pm 2i\sqrt{3}}{2}=-1\pm i\sqrt{3}$
• Oct 27th 2007, 02:09 PM
oceanmd
Thank you very much
• Oct 27th 2007, 02:09 PM
galactus
You could take the more 'complex' route and use:

$w=-2\left(cos(\frac{\pi}{3}+\frac{2k{\pi}}{3})+isin(\ frac{\pi}{3}+\frac{-2k{\pi}}{3})\right)$, k=0,1,2

Let k=0 and get $w=-1-\sqrt{3}i$

k=1: $w=2$

k=2: $w=-1+\sqrt{3}i$