Question
Factorize \({z^3} + 1\) into a linear and quadratic factor.[2]
Let \(\gamma = \frac{{1 + {\text{i}}\sqrt 3 }}{2}\).
(i) Show that \(\gamma \) is one of the cube roots of −1.
(ii) Show that \({\gamma ^2} = \gamma – 1\).
(iii) Hence find the value of \({(1 – \gamma )^6}\).[9]
Answer/Explanation
Markscheme
using the factor theorem z +1 is a factor (M1)
\({z^3} + 1 = (z + 1)({z^2} – z + 1)\) A1
[2 marks]
(i) METHOD 1
\({z^3} = – 1 \Rightarrow {z^3} + 1 = (z + 1)({z^2} – z + 1) = 0\) (M1)
solving \({z^2} – z + 1 = 0\) M1
\(z = \frac{{1 \pm \sqrt {1 – 4} }}{2} = \frac{{1 \pm {\text{i}}\sqrt 3 }}{2}\) A1
therefore one cube root of −1 is \(\gamma \) AG
METHOD 2
\({\gamma ^2} = \left( {{{\frac{{1 + i\sqrt 3 }}{2}}^2}} \right) = \frac{{ – 1 + i\sqrt 3 }}{2}\) M1A1
\({\gamma ^2} = \frac{{ – 1 + i\sqrt 3 }}{2} \times \frac{{1 + i\sqrt 3 }}{2} = \frac{{ – 1 – 3}}{4}\) A1
= −1 AG
METHOD 3
\(\gamma = \frac{{1 + i\sqrt 3 }}{2} = {e^{i\frac{\pi }{3}}}\) M1A1
\({\gamma ^3} = {e^{i\pi }} = – 1\) A1
(ii) METHOD 1
as \(\gamma \) is a root of \({z^2} – z + 1 = 0\) then \({\gamma ^2} – \gamma + 1 = 0\) M1R1
\(\therefore {\gamma ^2} = \gamma – 1\) AG
Note: Award M1 for the use of \({z^2} – z + 1 = 0\) in any way.
Award R1 for a correct reasoned approach.
METHOD 2
\({\gamma ^2} = \frac{{ – 1 + i\sqrt 3 }}{2}\) M1
\(\gamma – 1 = \frac{{1 + i\sqrt 3 }}{2} – 1 = \frac{{ – 1 + i\sqrt 3 }}{2}\) A1
(iii) METHOD 1
\({(1 – \gamma )^6} = {( – {\gamma ^2})^6}\) (M1)
\( = {(\gamma )^{12}}\) A1
\( = {({\gamma ^3})^4}\) (M1)
\( = {( – 1)^4}\)
\( = 1\) A1
METHOD 2
\({(1 – \gamma )^6}\)
\( = 1 – 6\gamma + 15{\gamma ^2} – 20{\gamma ^3} + 15{\gamma ^4} – 6{\gamma ^5} + {\gamma ^6}\) M1A1
Note: Award M1 for attempt at binomial expansion.
use of any previous result e.g. \( = 1 – 6\gamma + 15{\gamma ^2} + 20 – 15\gamma + 6{\gamma ^2} + 1\) M1
= 1 A1
Note: As the question uses the word ‘hence’, other methods that do not use previous results are awarded no marks.
[9 marks]
Question
The same remainder is found when \(2{x^3} + k{x^2} + 6x + 32\) and \({x^4} – 6{x^2} – {k^2}x + 9\) are divided by \(x + 1\) . Find the possible values of k .
Answer/Explanation
Markscheme
let \(f(x) = 2{x^3} + k{x^2} + 6x + 32\)
let \(g(x) = {x^4} – 6{x^2} – {k^2}x + 9\)
\(f( – 1) = – 2 + k – 6 + 32( = 24 + k)\) A1
\(g( – 1) = 1 – 6 + {k^2} + 9( = 4 + {k^2})\) A1
\( \Rightarrow 24 + k = 4 + {k^2}\) M1
\( \Rightarrow {k^2} – k – 20 = 0\)
\( \Rightarrow (k – 5)(k + 4) = 0\) (M1)
\( \Rightarrow k = 5,\, – 4\) A1A1
[6 marks]
Question
(i) Express each of the complex numbers \({z_1} = \sqrt 3 + {\text{i, }}{z_2} = – \sqrt 3 + {\text{i}}\) and \({z_3} = – 2{\text{i}}\) in modulus-argument form.
(ii) Hence show that the points in the complex plane representing \({z_1}\), \({z_2}\) and \({z_3}\) form the vertices of an equilateral triangle.
(iii) Show that \({\text{z}}_1^{3n} + z_2^{3n} = 2z_3^{3n}\) where \(n \in \mathbb{N}\).[9]
(i) State the solutions of the equation \({z^7} = 1\) for \(z \in \mathbb{C}\), giving them in modulus-argument form.
(ii) If w is the solution to \({z^7} = 1\) with least positive argument, determine the argument of 1 + w. Express your answer in terms of \(\pi \).
(iii) Show that \({z^2} – 2z\cos \left( {\frac{{2\pi }}{7}} \right) + 1\) is a factor of the polynomial \({z^7} – 1\). State the two other quadratic factors with real coefficients.[9]
Answer/Explanation
Markscheme
(i) \({z_1} = 2{\text{cis}}\left( {\frac{\pi }{6}} \right),{\text{ }}{z_2} = 2{\text{cis}}\left( {\frac{{5\pi }}{6}} \right),{\text{ }}{z_3} = 2{\text{cis}}\left( { – \frac{\pi }{2}} \right){\text{ or }}2{\text{cis}}\left( {\frac{{3\pi }}{2}} \right)\) A1A1A1
Note: Accept modulus and argument given separately, or the use of exponential (Euler) form.
Note: Accept arguments given in rational degrees, except where exponential form is used.
(ii) the points lie on a circle of radius 2 centre the origin A1
differences are all \(\frac{{2\pi }}{3}(\bmod 2\pi )\) A1
\( \Rightarrow \) points equally spaced \( \Rightarrow \) triangle is equilateral R1AG
Note: Accept an approach based on a clearly marked diagram.
(iii) \({\text{z}}_1^{3n} + z_2^{3n} = {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right) + {2^{3n}}{\text{cis}}\left( {\frac{{5n\pi }}{2}} \right)\) M1
\( = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right)\) A1
\(2z_3^{3n} = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{9n\pi }}{2}} \right) = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right)\) A1AG
[9 marks]
(i) attempt to obtain seven solutions in modulus argument form M1
\(z = {\text{cis}}\left( {\frac{{2k\pi }}{7}} \right),{\text{ }}k = 0,{\text{ }}1 \ldots 6\) A1
(ii) w has argument \(\frac{{2\pi }}{7}\) and 1 + w has argument \(\phi \),
then \(\tan (\phi ) = \frac{{\sin \left( {\frac{{2\pi }}{7}} \right)}}{{1 + \cos \left( {\frac{{2\pi }}{7}} \right)}}\) M1
\( = \frac{{2\sin \left( {\frac{\pi }{7}} \right)\cos \left( {\frac{\pi }{7}} \right)}}{{2{{\cos }^2}\left( {\frac{\pi }{7}} \right)}}\) A1
\( = \tan \left( {\frac{\pi }{7}} \right) \Rightarrow \phi = \frac{\pi }{7}\) A1
Note: Accept alternative approaches.
(iii) since roots occur in conjugate pairs, (R1)
\({z^7} – 1\) has a quadratic factor \(\left( {z – {\text{cis}}\left( {\frac{{2\pi }}{7}} \right)} \right) \times \left( {z – {\text{cis}}\left( { – \frac{{2\pi }}{7}} \right)} \right)\) A1
\( = {z^2} – 2z\cos \left( {\frac{{2\pi }}{7}} \right) + 1\) AG
other quadratic factors are \({z^2} – 2z\cos \left( {\frac{{4\pi }}{7}} \right) + 1\) A1
and \({z^2} – 2z\cos \left( {\frac{{6\pi }}{7}} \right) + 1\) A1
[9 marks]