Hi Gibsonj338 
Find all cubic roots of
\(z=-1+i \mbox{ distance of z to the origin }=\sqrt{1^2+1^2}=\sqrt2\\ \mbox{cubic toots so there are 3 of them so they are } \frac{2\pi}{3} radians\; apart.\\ z=\sqrt2(cis\frac{3\pi}{4})\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{4}\div3)),\quad\sqrt[6]{2}(cis((\frac{3\pi}{4}+2\pi)\div3)),\quad \sqrt[6]{2}(cis((\frac{3\pi}{4}+4\pi)\div3))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{4}\div3)),\quad\sqrt[6]{2}(cis(\frac{11\pi}{4}\div3)),\quad \sqrt[6]{2}(cis(\frac{19\pi}{4}\div3))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{12})),\quad\sqrt[6]{2}(cis(\frac{11\pi}{12})),\quad \sqrt[6]{2}(cis(\frac{19\pi}{12}))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{\pi}{4})),\quad\sqrt[6]{2}(cis(\frac{11\pi}{12})),\quad \sqrt[6]{2}(cis(\frac{19\pi}{12}))\\ or\ \)
\(\sqrt[3]{z}=\frac{1+i}{\sqrt[3]{2}},\quad\sqrt[6]{2}(cos(\frac{11\pi}{12})+i\;sin(\frac{11\pi}{12})),\quad\sqrt[6]{2}(cos(\frac{19\pi}{12})+i\;sin(\frac{19\pi}{12}))\\ \sqrt[3]{z}=\frac{1+i}{\sqrt[3]{2}},\quad\sqrt[6]{2}(-cos(\frac{\pi}{12})+i\;sin(\frac{\pi}{12})),\quad\sqrt[6]{2}(cos(\frac{5\pi}{12})-i\;sin(\frac{5\pi}{12})) \)
