# Solving Complex Numbers

1. Dec 29, 2008

### missmerisha

I hope, I've posted this question in the right section.

1. The problem statement, all variables and given/known data
Solve the fooling equation over C

z^3+ 8 = 0

3. The attempt at a solution

First Attempt
z^3 = -8
cube root (2 ^3) = cube root (8 i^2 )
z = 2i

Second Attempt
z^3 = -8
z ^3 = -2 ^3
so, z = -2

2. Dec 29, 2008

### Vagrant

Or you could convert -8 into polar form, then using De Moivre's theorem get all three cube roots.

3. Dec 29, 2008

### missmerisha

We're learning Polar Form next year and I have never heard of De Moivre's Theorem.

So, is my second attempt incorrect?

4. Dec 29, 2008

### gabbagabbahey

Your first attempt is incorrect; $(2i)^3=-8i\neq -8$

Your second attempt is not incorrect but it is incomplete: $z^3+8=0$ is a 3rd degree polynomial equation; so it must have three roots. You have correctly found one root $z=-2$, but you still need to find the other two.

One method is to divide your polynomial $z^3+8$ by $z+2$ (Since z=-2 is a root, you know (z+2) must be a factor of the polynomial) which will leave you with a quadratic that you can solve to find your other two roots.

5. Dec 29, 2008

### missmerisha

thanks
I've got it now.