How to solve $z^3=\frac12-\frac12i$ After simplifying the equation: 
$(2i)^9z^3=(1+i)^{17}$
I got at the end that: $z^3=\frac12-\frac12i$ 
but I didn't know how to take it from here, and how to find $z_1$, $z_2$, $z_3$. 

Hint: write $$
(1-i) = \sqrt 2 e^{-i\pi / 4} = \sqrt 2(\cos(-\pi / 4) + i \sin(-\pi / 4))
$$
