Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 1 / 1C / a / Solution

De Moivre's theorem states that $(\cos \theta +i\sin \theta {)}^n=\cos (n\theta )+i\sin (n\theta )$ for integer $n$. For $z\ne 0$, write its polar form as $z=r{e}^{i\theta }$, where $r=|z|>0$. If $w=s{e}^{i\phi }$ and $w^n=z$, equality of moduli gives $s=r^{1/n}$ and equality of arguments gives $n\phi =\theta +2\pi k$. Thus the distinct roots of a complex number areChanging the chosen argument of $z$ only permutes these roots. For $z=0$ the only root is zero. Taking $r=8$ and $\theta =\pi$ gives the cube rootsCubing any of them returns $-8$; geometrically they are equally spaced on the circle of radius two.