(i)     Write down ${z_1}$ in Cartesian form.

(ii)     Hence determine ${({z_1} + {z_2})^ * }$ in Cartesian form.

(i)     Write ${z_2}$ in modulus-argument form.

(ii)     Hence solve the equation ${z^3} = {z_2}$ .

Let $z = r\,{\text{cis}}\theta$ , where $r \in {\mathbb{R}^ + }$  and $0 \leqslant \theta  < 2\pi$ . Find all possible values of r and $\theta$ ,

(i)     if ${z^2} = {(1 + {z_2})^2}$;

(ii)     if $z = - \frac{1}{{{z_2}}}$.

Find the smallest positive value of n for which ${\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} \in {\mathbb{R}^ + }$ .

(i)     ${z_1} = 2\sqrt 3 {\text{cis}}\frac{{3\pi }}{2} \Rightarrow {z_1} = - 2\sqrt 3 {\text{i}}$     A1

(ii)     ${z_1} + {z_2} = - 2\sqrt 3 {\text{i}} - 1 + \sqrt 3 {\text{i}} = - 1 - \sqrt 3 {\text{i}}$     A1

${({z_1} + {z_2})^ * } = - 1 + \sqrt 3 {\text{i}}$     A1

[3 marks]

(i)     METHOD 1

${z^2} = {\left( {1 - 1 + \sqrt 3 {\text{i}}} \right)^2} = - 3\left( { \Rightarrow z =  \pm \sqrt 3 {\text{i}}} \right)$     M1

$z = \sqrt 3 \,{\text{cis}}\frac{\pi }{2}{\text{ or }}{z_1} = \sqrt 3 \,{\text{cis}}\frac{{3\pi }}{2}\left( { = \sqrt 3 \,{\text{cis}}\left( {\frac{{ - \pi }}{2}} \right)} \right)$     A1A1

so $r = \sqrt 3 {\text{ and }}\theta  = \frac{\pi }{2}{\text{ or }}\theta  = \frac{{3\pi }}{2}\left( { = \frac{{ - \pi }}{2}} \right)$ 

Note: Accept $r\,{\text{cis}}(\theta )$ form.

 METHOD 2

${z^2} = {\left( {1 - 1 + \sqrt 3 {\text{i}}} \right)^2} = - 3 \Rightarrow {z^2} = 3{\text{cis}}\left( {(2n + 1)\pi } \right)$     M1

 ${r^2} = 3 \Rightarrow r = \sqrt 3$     A1

$2\theta  = (2n + 1)\pi  \Rightarrow \theta  = \frac{\pi }{2}{\text{ or }}\theta  = \frac{{3\pi }}{2}{\text{ (as }}0 \leqslant \theta  < 2\pi )$     A1 

Note: Accept $r\,{\text{cis}}(\theta )$ form.

(ii)     METHOD 1

$z = - \frac{1}{{2{\text{cis}}\frac{{2\pi }}{3}}} \Rightarrow z = \frac{{{\text{cis}}\pi }}{{2{\text{cis}}\frac{{2\pi }}{3}}}$     M1

$\Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}$

so $r = \frac{1}{2}{\text{ and }}\theta  = \frac{\pi }{3}$     A1A1

METHOD 2

${z_1} = - \frac{1}{{ - 1 + \sqrt 3 {\text{i}}}} \Rightarrow {z_1} = - \frac{{ - 1 - \sqrt 3 {\text{i}}}}{{\left( { - 1 + \sqrt 3 {\text{i}}} \right)\left( { - 1 - \sqrt 3 {\text{i}}} \right)}}$     M1

$z = \frac{{1 + \sqrt 3 {\text{i}}}}{4} \Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}$

so $r = \frac{1}{2}{\text{ and }}\theta  = \frac{\pi }{3}$     A1A1

[6 marks]

$\frac{{{z_1}}}{{{z_2}}} = \sqrt 3 \,{\text{cis}}\frac{{5\pi }}{6}$     (A1)

${\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} = {\sqrt 3 ^n}{\text{cis}}\frac{{5n\pi }}{6}$     A1

equating imaginary part to zero and attempting to solve     M1

obtain n = 12     A1

Note: Working which only includes the argument is valid.

[4 marks]

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of $2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)$ were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found ${z^2} = - 3$ but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of ${z_1}$ having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. ${\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)$ was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of $2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)$ were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found ${z^2} = - 3$ but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of ${z_1}$ having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. ${\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)$ was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of $2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)$ were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found ${z^2} = - 3$ but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of ${z_1}$ having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. ${\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)$ was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of $2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)$ were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found ${z^2} = - 3$ but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of ${z_1}$ having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. ${\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)$ was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.