Date | November 2016 | Marks available | 4 | Reference code | 16N.1.hl.TZ0.12 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Show that | Question number | 12 | Adapted from | N/A |
Question
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) \(1 + \omega {\text{*}} + {(\omega {\text{*}})^2}\).
Show that \((\omega - 3{\omega ^2})({\omega ^2} - 3\omega ) = 13\).
Find the values of \(x\) that satisfy the equation \(\left| p \right| = \left| q \right|\).
Solve the inequality \(\operatorname{Re} (pq) + 8 < {\left( {\operatorname{Im} (pq)} \right)^2}\).
Markscheme
(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
OR
\( = - 3\omega + 10 - 3{\omega ^2} = - 3({\omega ^2} + \omega + 1) + 13\) M1
\( = - 3 \times 0 + 13\) A1
OR
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]