If ${z_1} = a + a\sqrt 3 i$ and ${z_2} = 1 - i$, where a is a real constant, express ${z_1}$ and ${z_2}$ in the form $r\,{\text{cis}}\,\theta$, and hence find an expression for ${\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6}$ in terms of a and i.

${z_1} = 2a{\text{cis}}\left( {\frac{\pi }{3}} \right){\text{, }}{z_2} = \sqrt 2 {\text{ cis}}\left( { - \frac{\pi }{4}} \right)$     M1     A1     A1

EITHER

${\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = \frac{{{2^6}{a^6}{\text{cis(0)}}}}{{{{\sqrt 2 }^6}{\text{cis}}\left( {\frac{\pi }{2}} \right)}}\left( { = 8{a^6}{\text{cis}}\left( { - \frac{\pi }{2}} \right)} \right)$     M1     A1     A1

OR

${\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = {\left( {\frac{{2a}}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{7\pi }}{{12}}} \right)} \right)^6}$     M1     A1

$= 8{a^6}{\text{cis}}\left( { - \frac{\pi }{2}} \right)$     A1

THEN

$= - 8{a^6}{\text{i}}$     A1

Note: Accept equivalent angles, in radians or degrees. 

Accept alternate answers without cis e.g. ${\text{ = }}\frac{{8{a^6}}}{{\text{i}}}$

[7 marks]

Most students had an idea of what to do but were frequently let down in their calculations of the modulus and argument. The most common error was to give the argument of ${z_2}$ as $\frac{{3\pi }}{4}$, failing to recognise that it should be in the fourth quadrant. There were also errors seen in the algebraic manipulation, in particular forgetting to raise the modulus to the power 6.