The complex numbers ${z_1}$ and ${z_2}$ have arguments between 0 and $\pi$ radians. Given that ${z_1}{z_2} = - \sqrt 3  + {\text{i}}$ and $\frac{{{z_1}}}{{{z_2}}} = 2{\text{i}}$, find the modulus and argument of ${z_1}$ and of ${z_2}$.

METHOD 1

${\text{arg(}}{z_1}{z_2}) = \frac{{5\pi }}{6}\,\,\,\,\,(150^\circ )$     (A1)

$\arg \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \frac{\pi }{2}\,\,\,\,\,(90^\circ )$     (A1)

$\Rightarrow \arg ({z_1}) + \arg ({z_2}) = \frac{{5\pi }}{6};{\text{ }}\arg ({z_1}) - \arg ({z_2}) = \frac{\pi }{2}$     M1

solving simultaneously

$\arg ({z_1}) = \frac{{2\pi }}{3}{\text{ }}(120^\circ ){\text{ and }}\arg ({z_2}) = \frac{\pi }{6}{\text{ }}(30^\circ )$     A1A1

Note: Accept decimal approximations of the radian measures.

$\left| {{z_1}{z_2}} \right| = 2 \Rightarrow \left| {{z_1}} \right|\left| {{z_2}} \right| = 2;{\text{ }}\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = 2 \Rightarrow \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} = 2$     M1

solving simultaneously

$\left| {{z_1}} \right| = 2;{\text{ }}\left| {{z_2}} \right| = 1$     A1

[7 marks]

METHOD 2

${z_1} = 2i{z_2}\,\,\,\,\,2iz_2^2 = - \sqrt 3  + i$     (M1)

$z_2^2 = \frac{{ - \sqrt 3  + i}}{{2i}}$     A1

${z_2} = \sqrt {\frac{{ - \sqrt 3  + i}}{{2i}}} = \frac{{\sqrt 3 }}{2} + \frac{1}{2}i{\text{ or }}{e^{\frac{\pi }{6}i}}$     (M1)(A1)

(allow $0.866 + 0.5i$ or ${{\text{e}}^{0.524{\text{i}}}}$)

${z_1} = - 1 + \sqrt 3 i{\text{ or }}2{{\text{e}}^{\frac{{2\pi }}{3}i}} -$ (allow −1 + 1.73i or $2{{\text{e}}^{2.09{\text{i}}}}$)     (A1)

${z_1}\,\,\,\,\,$modulus = 2, argument $= \frac{{2\pi }}{3}$     A1

${z_2}\,\,\,\,\,$modulus = 1, argument $= \frac{\pi }{6}$     A1

Note: Accept degrees and decimal approximations to radian measure.

[7 marks]

Candidates generally found this question challenging. Many candidates had difficulty finding the arguments of ${z_1}{z_2}$ and ${z_1}/{z_2}$. Among candidates who attempted to solve for ${z_1}$ and ${z_2}$ in Cartesian form, many had difficulty with the algebraic manipulation involved.