Given the complex number $z = r ( \cos \theta ) + i \sin ( \theta ) )$, the $\boldsymbol n^{\bf th}$ roots of $z$ are given by:
${ \sqrt [ n ] { r } ( \operatorname { cos } ( \frac { \theta + 2 k \pi } { n } ) + i \operatorname { sin } ( \frac { \theta + 2 k \pi } { n } ) ) \quad \text { for } k = 0,1,2 }$