list the 3 complex roots of -1 by rewriting -1 in trogonometric form and applying DeMoivre's Theorem. you may leave your answers in trigonometris form
$\displaystyle z^3=-1=cis(\pi+2k\pi)=cis(\pi(2k+1))\,,\,\,with\,\,\,ci s(x):=e^{xi}\,,\,\,x\in\mathbb{R}\,,\,\,k\in\mathb b{Z}$ . In order to get all the different elements though we restrict $\displaystyle k=0,1,2$ (why? Check the trigonometric circle).
So $\displaystyle z=(-1)^{1/3}=cis(\pi(2k+1))^{1/3}=cis\left(\pi\frac{2k+1}{3}\right)$ (here we used De Moivre's theorem) , so now just input $\displaystyle k=0,1,2$ and obtain the three complex roots of $\displaystyle -1$ ...as simple as that!
Tonio