Finding Complex Roots of Equation z^3+8=0

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I just bombed a quiz because it was 2 questions and this was one of them:

Find all three complex roots of the following equation (give answers in polar and rectangular form)

z^3+8=0

Looks easy enough,

z=2e^{-i\frac{\theta}{3}}

This is where I think I completely realized I wasn't sure what I was doing. My roommate suggested I look for the roots of unity which I know that:

r^n(cos(n\theta)+isin(n\theta))=1+i*0

so if I want to consider mine it should be:

8^{1/3}(cos(\frac{\theta}{3})+isin(\frac{\theta}{3})=-8

so then

\theta=\frac{k2\pi}{3}

is this the right track?
 
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I'm not sure what you're trying to do (eg, your second to last latex line doesn't look right). Just keep in mind that there will be a unique solution for r, and then you're left with finding the cube roots of -1 (not 1). If it helps, these are the sixth roots of 1 that aren't also cube roots of 1.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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