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

[4]
a.

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

[4]
b.

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

[5]
c.

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

[6]
d.

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]

a.

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

b.

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

c.

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

d.

Examiners report

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Syllabus sections

Topic 1 - Core: Algebra » 1.5 » Complex numbers: the number \({\text{i}} = \sqrt { - 1} \) ; the terms real part, imaginary part, conjugate, modulus and argument.
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