Solve: $z^6-(-1-i)^3=0$ > $$z^6-(-1-i)^3=0$$ $$z^6=(-1-i)^3$$ $$z^6=(1-3i-3+i)$$ $$z^6=(-2-2i)$$ $$-2-2i: \text{ }r=\sqrt{(-2)^2+(-2)^2}=\sqrt{8}\text{ , }\theta=arctan(1)-\pi=-\frac{3}{4}\pi$$ (Why for $x<0,y<0,arctan(\frac{y}{x})-\pi$ and for $x<0,y>0,arctan(\frac{y}{x})+\pi$?) $$z^6=\sqrt{8}cis(-\frac{3}{4}\pi)$$ $$z_{i}={8}^\frac{1}{12}cis(\frac{-\frac{3}{4}\pi+2\pi k}{6})\text{ ,} k=\\{0,1,2,...,5\\}$$ $$z_{0}={8}^\frac{1}{12}cis-\frac{\pi}{8}$$ $$z_{1}={8}^\frac{1}{12}cis-\frac{5\pi}{24}$$ $$z_{2}={8}^\frac{1}{12}cis-\frac{13\pi}{24}$$ $$z_{3}={8}^\frac{1}{12}cis-\frac{7\pi}{8}$$ $$z_{4}={8}^\frac{1}{12}cis-\frac{29\pi}{24}$$ $$z_{5}={8}^\frac{1}{12}cis-\frac{37\pi}{24}$$ Are the stages correct? is there a shorter way/approch to solve this?

The equation is equivalent to $z^6-w^3=0$ or $(\dfrac{z^2}{w})^3=1$, where $w = -1-i.$ This will be slightly easier to write down since it is very easy to describe the third roots of unity, which are $1, \omega, \omega^2$ with $\omega = \cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$.