How do I find all complex numbers that are solutions of $z^3 +8i=0$? My workbook has answers in the standard form $a+bi$. I would assume that to solve this, I would expand the complex number $z$ into trigonometric form to deal with that exponent. This is what I have so far using De Moivre's Theorem:
$(r^3(\cos(3\theta)+i\sin(3\theta)) +8i=0$. 
However, I do not know where to go from here. Thank you!

$(r^3(\cos(3\theta)+i\sin(3\theta)) +8i=0$
So continue:
$(r^3(\cos(3\theta)+i\sin(3\theta)) = -8i= 8(0 + i(-1))$
$(r^3(\cos(3\theta)+i\sin(3\theta)) = 8((\cos \frac {3\pi}2+2k\pi) + i\sin(\frac{3\pi}2+2k\pi))$
So $r^3 = 8$ and $3\theta = \frac {3\pi}2 + 2k\pi$
So $r =2$ and $\theta = \frac \pi 2 + \frac 23k\pi$ 
So there are three solutions:
1)if $k =0$ then $2(\cos \frac \pi 2 + i\sin \frac \pi 2) = 2i$.
Check.   $(2i)^3 = 2^3i^2 = 8*i^2 *i = -8i$.  That's good.
2) if $k = 1$ then $2(\cos (\frac \pi 2+\frac {2\pi} 3) + i \sin (\frac \pi 2+\frac {2\pi} 3))= 2(\cos \frac {7\pi}6) + i\sin \frac {7\pi}6) = -\sqrt 3 - i$.
Check:  $(-\sqrt 3-i)^3 = -[(\sqrt 3)^3 + 3(\sqrt 3)^2i + 3(\sqrt 3)i^2 + i^3] = $
$-[3\sqrt 3 + 9i - 3\sqrt 3 - i] = 8i$.  It works.
and if $k = 2$ then 3) $2(\cos (\frac \pi 2+\frac {4\pi} 3) + i \sin (\frac \pi 2+\frac {4\pi} 3))= 2(\cos \frac {11\pi}6) + i\sin \frac {11\pi}6) = \sqrt 3 - i$.
Check:  $(\sqrt 3-i)^3 = (-\sqrt 3)^3 + 3(-\sqrt 3)^2i + 3(-\sqrt 3)i^2 + i^3] = $
$-3\sqrt 3 - 9i +3\sqrt 3 + i = -8i$.  It works.
