Will you please help me solve this equation.

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

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- Oct 27th 2007, 12:27 PMoceanmdSolve in Complex Numbers
Will you please help me solve this equation.

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, 12:50 PMred_dog
$\displaystyle x^3-8=0\Leftrightarrow(x-2)(x^2+2x+4)=0$

$\displaystyle x_1=2$

$\displaystyle x^2+2x+4=0\Rightarrow\triangle=-12\Rightarrow x_{2,3}=\displaystyle\frac{-2\pm 2i\sqrt{3}}{2}=-1\pm i\sqrt{3}$ - Oct 27th 2007, 01:09 PMoceanmd
Thank you very much

- Oct 27th 2007, 01:09 PMgalactus
You could take the more 'complex' route and use:

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

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

k=1: $\displaystyle w=2$

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