16N.1.AHL.TZ0.H_12

pestleMathematicsAAHLPaper 116N· ahl-1-12-complex-numbers-cartesian-form-and-argand-diagsource ↗

Let $\omega$ be one of the non-real solutions of the equation ${z^3} = 1$ .

Consider the complex numbers $p = 1 - 3{\text{i}}$ and $q = x + (2x + 1){\text{i}}$ , where $x \in \mathbb{R}$ .

Determine the value of

(i) $1 + \omega + {\omega ^2}$ ;

(ii) $1 + \omega {\text{*}} + {(\omega {\text{*}})^2}$ .

[4]

Show that $(\omega - 3{\omega ^2})({\omega ^2} - 3\omega ) = 13$ .

[4]

Find the values of $x$ that satisfy the equation $\left| p \right| = \left| q \right|$ .

[5]

Solve the inequality $\operatorname{Re} (pq) + 8 < {\left( {\operatorname{Im} (pq)} \right)^2}$ .

[6]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(i) METHOD 1

$1 + \omega + {\omega ^2} = \frac{{1 - {\omega ^3}}}{{1 - \omega }} = 0$ A1

as $\omega \ne 1$ R1

METHOD 2

solutions of $1 - {\omega ^3} = 0$ are $\omega = 1,{\text{ }}\omega {\text{ = }}\frac{{ - 1 \pm \sqrt 3 {\text{i}}}}{2}$ A1

verification that the sum of these roots is 0 R1

(ii) $1 + \omega {\text{*}} + {(\omega {\text{*}})^2} = 0$ A2

[4 marks]

$(\omega - 3{\omega ^2})({\omega ^2} - 3\omega ) = - 3{\omega ^4} + 10{\omega ^3} - 3{\omega ^2}$ M1A1

EITHER

$= - 3{\omega ^2}({\omega ^2} + \omega + 1) + 13{\omega ^3}$ M1

$= - 3{\omega ^2} \times 0 + 13 \times 1$ A1

$= - 3\omega + 10 - 3{\omega ^2} = - 3({\omega ^2} + \omega + 1) + 13$ M1

$= - 3 \times 0 + 13$ A1

substitution by $\omega = \frac{{ - 1 \pm \sqrt 3 {\text{i}}}}{2}$ in any form M1

numerical values of each term seen A1

THEN

$= 13$ AG

[4 marks]

$\left| p \right| = \left| q \right| \Rightarrow \sqrt {{1^2} + {3^2}} = \sqrt {{x^2} + {{(2x + 1)}^2}}$ (M1)(A1)

$5{x^2} + 4x - 9 = 0$ A1

$(5x + 9)(x - 1) = 0$ (M1)

$x = 1,{\text{ }}x = - \frac{9}{5}$ A1

[5 marks]

$pq = (1 - 3{\text{i}})\left( {x + (2x + 1){\text{i}}} \right) = (7x + 3) + (1 - x){\text{i}}$ M1A1

$\operatorname{Re} (pq) + 8 < {\left( {\operatorname{Im} (pq)} \right)^2} \Rightarrow (7x + 3) + 8 < {(1 - x)^2}$ M1

$\Rightarrow {x^2} - 9x - 10 > 0$ A1

$\Rightarrow (x + 1)(x - 10) > 0$ M1

$x < - 1,{\text{ }}x > 10$ A1

[6 marks]

Examiners’ report

[N/A]