(a)     Solve the equation ${z^3} = - 2 + 2{\text{i}}$, giving your answers in modulus-argument form.

(b)     Hence show that one of the solutions is 1 + i when written in Cartesian form.

(a)     ${z^3} = 2\sqrt 2 {{\text{e}}^{\frac{{3\pi {\text{i}}}}{4}}}$     (M1)(A1)

${z_1} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4}}}$     A1

adding or subtracting $\frac{{2\pi {\text{i}}}}{3}$     M1

${z_2} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4} + \frac{{2\pi {\text{i}}}}{3}}} = \sqrt 2 {{\text{e}}^{\frac{{11\pi {\text{i}}}}{{12}}}}$     A1

${z_3} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4} - \frac{{2\pi {\text{i}}}}{3}}} = \sqrt 2 {{\text{e}}^{ - \frac{{5\pi {\text{i}}}}{{12}}}}$     A1

Notes: Accept equivalent solutions e.g. ${z_3} = \sqrt 2 {{\text{e}}^{\frac{{19\pi {\text{i}}}}{{12}}}}$

Award marks as appropriate for solving ${(a + b{\text{i}})^3} = - 2 + 2{\text{i}}$.

Accept answers in degrees.

(b)     $\sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4}}}{\text{ }}\left( { = \sqrt 2 \left( {\frac{1}{{\sqrt 2 }} + \frac{{\text{i}}}{{\sqrt 2 }}} \right)} \right)$     A1

= 1 + i     AG

Note: Accept geometrical reasoning.

[7 marks]

Many students incorrectly found the argument of ${z^3}$ to be $\arctan \left( {\frac{2}{{ - 2}}} \right) = - \frac{\pi }{4}$. Of those students correctly finding one solution, many were unable to use symmetry around the origin, to find the other two. In part (b) many students found the cube of 1 + i which could not be awarded marks as it was not “hence”.