How to solve $z^3=\frac12-\frac12i$

Question

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 $z_1$, $z_2$, $z_3$.

Answer 1

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