Find all complex numbers z so that z^{3 }= -8i.
I know I have to do something with polar form, but I'm really lost.
If $\displaystyle z=re^{i\varphi}$, then $\displaystyle z^3=r^3e^{3i\varphi}$. Note that $\displaystyle -i=e^{i(3\pi/2)}$. So you need to find all $\displaystyle 0\le\varphi<2\pi$ such that $\displaystyle 3\varphi\equiv3\pi/2\pmod{2\pi}$. There are three such $\displaystyle \varphi$. And, of course, you need to find $\displaystyle r$.
This is posted in basic algebra forum. Reply #2 is correct but advanced.
Notation: $\displaystyle r\exp(i\theta)=r\cos(\theta)+i~r\sin(\theta)~.$ So $\displaystyle {\left[ {r\exp \left( {i\theta } \right)} \right]^n} = {r^n}\exp \left( {i\theta n} \right)$
Let $\displaystyle \rho = 2\exp \left( {\frac{{ - i\pi }}{6}} \right)$, so $\displaystyle \rho^3=-8i$. We have one root.
Let $\displaystyle \xi = \exp \left( {\frac{{2\pi i}}{3}} \right)$, then you need to show that $\displaystyle \rho\cdot\xi~\&~\rho\cdot\xi^2$ are the other two roots.