Given that ${z_1} = 2$ and ${z_2} = 1 + \sqrt 3 {\text{i}}$ are roots of the cubic equation ${z^3} + b{z^2} + cz + d = 0$

where b, c, $d \in \mathbb{R}$,

(a)     write down the third root, ${z_3}$, of the equation;

(b)     find the values of b, c and d ;

(c)     write ${z_2}$ and ${z_3}$ in the form $r{{\text{e}}^{{\text{i}}\theta }}$.

(a)     $1 - \sqrt 3 {\text{i}}$     A1

(b)     EITHER

$\left( {z - (1 + \sqrt 3 {\text{i)}}} \right)\left( {z - (1 - \sqrt 3 {\text{i)}}} \right) = {z^2} - 2z + 4$     (M1)A1

$p(z) = (z - 2)({z^2} - 2z + 4)$     (M1)

$= {z^3} - 4{z^2} + 8z - 8$     A1

therefore $b = - 4,{\text{ }}c = 8,{\text{ }}d = - 8$

OR

relating coefficients of cubic equations to roots

$- b = 2 + 1 + \sqrt 3 {\text{i}} + 1 - \sqrt 3 {\text{i}} = 4$     M1

$c = 2(1 + \sqrt 3 {\text{i}}) + 2(1 - \sqrt 3 {\text{i}}) + (1 + \sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = 8$

$- d = 2(1 + \sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = 8$

$b = - 4,{\text{ }}c = 8,{\text{ }}d = - 8$     A1A1A1

(c)     ${z_2} = 2{{\text{e}}^{\frac{{{\text{i}}\pi }}{3}}},{\text{ }}{z_3} = 2{{\text{e}}^{ - \frac{{{\text{i}}\pi }}{3}}}$     A1A1A1

Note: Award A1 for modulus,

A1 for each argument.

[8 marks]

Parts a) and c) were done quite well by many but the method used in b) often lead to tedious and long algebraic manipulations in which students got lost and so did not get to the correct solution. Many did not give the principal argument in c).