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!
Best Answer
$$z^3=-8i\iff \begin{cases} |z|^3=|-8i|=8\\ \arg{z^3}=3\arg z=-\dfrac{\pi}{2}+2k\pi \end{cases}$$ $\iff$ $$ \begin{cases} |z|=2\\ \arg{z}=-\dfrac{\pi}{6}+\dfrac{2\pi}{3}k, \quad k\in\{0,1,2\} \end{cases}$$ Can you take it from here?