Find the cube roots of i in the form $a + b{\text{i}}$, where $a,{\text{ }}b \in \mathbb{R}$.

${\text{i}} = \cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}$     (A1)

${z_1} = {{\text{i}}^{\frac{1}{3}}} = {\left( {\cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}} \right)^{\frac{1}{3}}} = \cos \frac{\pi }{6} + {\text{i}}\sin \frac{\pi }{6}\,\,\,\,\,\left( { = \frac{{\sqrt 3 }}{2} + \frac{1}{2}{\text{i}}} \right)$     M1A1

${z_2} = \cos \frac{{5\pi }}{6} + {\text{i}}\sin \frac{{5\pi }}{6}\,\,\,\,\,\left( { = - \frac{{\sqrt 3 }}{2} + \frac{1}{2}{\text{i}}} \right)$     (M1)A1

${z_3} = \cos \left( { - \frac{\pi }{2}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{2}} \right) = - {\text{i}}$     A1

Note: Accept exponential and cis forms for intermediate results, but not the final roots.

Note: Accept the method based on expanding ${({\text{a}} + {\text{b}})^3}$. M1 for attempt, M1 for equating real and imaginary parts, A1 for finding a = 0 and ${\text{b}} = \frac{1}{2}$, then A1A1A1 for the roots.

[6 marks]

A varied response. Many knew how to solve this standard question in the most efficient way. A few candidates expanded ${(a + ib)^3}$ and solved the resulting fairly simple equations. A disappointing minority of candidates did not know how to start.