Question
If \({z_1} = a + a\sqrt 3 i\) and \({z_2} = 1 – i\), where a is a real constant, express \({z_1}\) and \({z_2}\) in the form \(r\,{\text{cis}}\,\theta \), and hence find an expression for \({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6}\) in terms of a and i.
Answer/Explanation
Markscheme
\({z_1} = 2a{\text{cis}}\left( {\frac{\pi }{3}} \right){\text{, }}{z_2} = \sqrt 2 {\text{ cis}}\left( { – \frac{\pi }{4}} \right)\) M1 A1 A1
EITHER
\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = \frac{{{2^6}{a^6}{\text{cis(0)}}}}{{{{\sqrt 2 }^6}{\text{cis}}\left( {\frac{\pi }{2}} \right)}}\left( { = 8{a^6}{\text{cis}}\left( { – \frac{\pi }{2}} \right)} \right)\) M1 A1 A1
OR
\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = {\left( {\frac{{2a}}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{7\pi }}{{12}}} \right)} \right)^6}\) M1 A1
\( = 8{a^6}{\text{cis}}\left( { – \frac{\pi }{2}} \right)\) A1
THEN
\( = – 8{a^6}{\text{i}}\) A1
Note: Accept equivalent angles, in radians or degrees.
Accept alternate answers without cis e.g. \({\text{ = }}\frac{{8{a^6}}}{{\text{i}}}\)
[7 marks]
Question
Given that z is the complex number \(x + {\text{i}}y\) and that \(\left| {\,z\,} \right| + z = 6 – 2{\text{i}}\) , find the value of x
and the value of y .
Answer/Explanation
Markscheme
\(\sqrt {{x^2} + {y^2}} + x + y{\text{i}} = 6 – 2{\text{i}}\) (A1)
equating real and imaginary parts M1
\(y = – 2\) A1
\(\sqrt {{x^2} + 4} + x = 6\) A1
\({x^2} + 4 = {(6 – x)^2}\) M1
\( – 32 = – 12x \Rightarrow x = \frac{8}{3}\) A1
[6 marks]
Question
Given that \((4 – 5{\text{i}})m + 4n = 16 + 15{\text{i}}\) , where \({{\text{i}}^2} = – 1\), find m and n if
a. m and n are real numbers;[3]
Answer/Explanation
Markscheme
a. attempt to equate real and imaginary parts M1
equate real parts: \(4m + 4n = 16\); equate imaginary parts: \( -5m = 15\) A1
\( \Rightarrow m = -3,{\text{ }}n = 7\) A1
[3 marks]
b. let \(m = x + {\text{i}}y,{\text{ }}n = x – {\text{i}}y\) M1
\( \Rightarrow (4 – 5{\text{i}})(x + {\text{i}}y) + 4(x – {\text{i}}y) = 16 + 15{\text{i}}\)
\( \Rightarrow 4x – 5{\text{i}}x + 4{\text{i}}y + 5y + 4x – 4{\text{i}}y = 16 + 15{\text{i}}\)
attempt to equate real and imaginary parts M1
\(8x + 5y = 16,{\text{ }} -5x = 15\) A1
\( \Rightarrow x = -3,{\text{ }}y = 8\) A1
\(( \Rightarrow m = -3 + 8{\text{i}},{\text{ }}n = -3 – 8{\text{i}})\)
[4 marks]
Question
The graph of a polynomial function f of degree 4 is shown below.
Given that \({(x + {\text{i}}y)^2} = – 5 + 12{\text{i}},{\text{ }}x,{\text{ }}y \in \mathbb{R}\) . Show that
(i) \({x^2} – {y^2} = – 5\) ;
(ii) \(xy = 6\) .[2]
Hence find the two square roots of \( – 5 + 12{\text{i}}\) .[5]
For any complex number z , show that \({(z^*)^2} = ({z^2})^*\) .[3]
Hence write down the two square roots of \( – 5 – 12{\text{i}}\) .[2]
Explain why, of the four roots of the equation \(f(x) = 0\) , two are real and two are complex.[2]
The curve passes through the point \(( – 1,\, – 18)\) . Find \(f(x)\) in the form
\(f(x) = (x – a)(x – b)({x^2} + cx + d),{\text{ where }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in \mathbb{Z}\) .[5]
Find the two complex roots of the equation \(f(x) = 0\) in Cartesian form.[2]
Draw the four roots on the complex plane (the Argand diagram).[2]
Express each of the four roots of the equation in the form \(r{{\text{e}}^{{\text{i}}\theta }}\) .[6]
Answer/Explanation
Markscheme
(i) \({(x + {\text{i}}y)^2} = – 5 + 12{\text{i}}\)
\({x^2} + 2{\text{i}}xy + {{\text{i}}^2}{y^2} = – 5 + 12{\text{i}}\) A1
(ii) equating real and imaginary parts M1
\({x^2} – {y^2} = – 5\) AG
\(xy = 6\) AG
[2 marks]
substituting M1
EITHER
\({x^2} – \frac{{36}}{{{x^2}}} = – 5\)
\({x^4} + 5{x^2} – 36 = 0\) A1
\({x^2} = 4,\, – 9\) A1
\(x = \pm 2\) and \(y = \pm 3\) (A1)
OR
\(\frac{{36}}{{{y^2}}} – {y^2} = – 5\)
\({y^4} – 5{y^2} – 36 = 0\) A1
\({y^2} = 9,\, – 4\) A1
\({y^2} = \pm 3\) and \(x = \pm 2\) (A1)
Note: Accept solution by inspection if completely correct.
THEN
the square roots are \((2 + 3{\text{i}})\) and \(( – 2 – 3{\text{i}})\) A1
[5 marks]
EITHER
consider \(z = x + {\text{i}}y\)
\(z^* = x – {\text{i}}y\)
\({(z^*)^2} = {x^2} – {y^2} – 2{\text{i}}xy\) A1
\(({z^2}) = {x^2} – {y^2} + 2{\text{i}}xy\) A1
\(({z^2})^* = {x^2} – {y^2} – 2{\text{i}}xy\) A1
\({(z^*)^2} = ({z^2})^*\) AG
OR
\(z^* = r{{\text{e}}^{ – {\text{i}}\theta }}\)
\({(z^*)^2} = {r^2}{{\text{e}}^{ – 2{\text{i}}\theta }}\) A1
\({z^2} = {r^2}{{\text{e}}^{2{\text{i}}\theta }}\) A1
\(({z^2})^* = {r^2}{{\text{e}}^{ – 2{\text{i}}\theta }}\) A1
\({(z^*)^2} = ({z^2})^*\) AG
[3 marks]
\((2 – 3{\text{i}})\) and \(( – 2 + 3{\text{i}})\) A1A1
[2 marks]
the graph crosses the x-axis twice, indicating two real roots R1
since the quartic equation has four roots and only two are real, the other two roots must be complex R1
[2 marks]
\(f(x) = (x + 4)(x – 2)({x^2} + cx + d)\) A1A1
\(f(0) = – 32 \Rightarrow d = 4\) A1
Since the curve passes through \(( – 1,\, – 18)\),
\( – 18 = 3 \times ( – 3)(5 – c)\) M1
\(c = 3\) A1
Hence \(f(x) = (x + 4)(x – 2)({x^2} + 3x + 4)\)
[5 marks]
\(x = \frac{{ – 3 \pm \sqrt {9 – 16} }}{2}\) (M1)
\( \Rightarrow x = – \frac{3}{2} \pm {\text{i}}\frac{{\sqrt 7 }}{2}\) A1
[2 marks]
A1A1
Note: Accept points or vectors on complex plane.
Award A1 for two real roots and A1 for two complex roots.
[2 marks]
real roots are \(4{{\text{e}}^{{\text{i}}\pi }}\) and \(2{{\text{e}}^{{\text{i}}0}}\) A1A1
considering \( – \frac{3}{2} \pm {\text{i}}\frac{{\sqrt 7 }}{2}\)
\(r = \sqrt {\frac{9}{4} + \frac{7}{4}} = 2\) A1
finding \(\theta \) using \(\arctan \left( {\frac{{\sqrt 7 }}{3}} \right)\) M1
\(\theta = \arctan \left( {\frac{{\sqrt 7 }}{3}} \right) + \pi {\text{ or }}\theta = \arctan \left( { – \frac{{\sqrt 7 }}{3}} \right) + \pi \) A1
\( \Rightarrow z = 2{{\text{e}}^{{\text{i}}\left( {\arctan \left( {\frac{{\sqrt 7 }}{3}} \right) + \pi } \right)}}{\text{ or}} \Rightarrow z = 2{{\text{e}}^{{\text{i}}\left( {\arctan \left( {\frac{{ – \sqrt 7 }}{3}} \right) + \pi } \right)}}\) A1
Note: Accept arguments in the range \( – \pi {\text{ to }}\pi {\text{ or }}0{\text{ to }}2\pi \) .
Accept answers in degrees.
[6 marks]
Question
Consider the complex numbers
\({z_1} = 2\sqrt 3 {\text{cis}}\frac{{3\pi }}{2}\) and \({z_2} = – 1 + \sqrt 3 {\text{i }}\) .
(i) Write down \({z_1}\) in Cartesian form.
(ii) Hence determine \({({z_1} + {z_2})^ * }\) in Cartesian form.[3]
(i) Write \({z_2}\) in modulus-argument form.
(ii) Hence solve the equation \({z^3} = {z_2}\) .[6]
Let \(z = r\,{\text{cis}}\theta \) , where \(r \in {\mathbb{R}^ + }\) and \(0 \leqslant \theta < 2\pi \) . Find all possible values of r and \(\theta \) ,
(i) if \({z^2} = {(1 + {z_2})^2}\);
(ii) if \(z = – \frac{1}{{{z_2}}}\).[6]
Find the smallest positive value of n for which \({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} \in {\mathbb{R}^ + }\) .[4]
Answer/Explanation
Markscheme
(i) \({z_1} = 2\sqrt 3 {\text{cis}}\frac{{3\pi }}{2} \Rightarrow {z_1} = – 2\sqrt 3 {\text{i}}\) A1
(ii) \({z_1} + {z_2} = – 2\sqrt 3 {\text{i}} – 1 + \sqrt 3 {\text{i}} = – 1 – \sqrt 3 {\text{i}}\) A1
\({({z_1} + {z_2})^ * } = – 1 + \sqrt 3 {\text{i}}\) A1
[3 marks]
(i) METHOD 1
\({z^2} = {\left( {1 – 1 + \sqrt 3 {\text{i}}} \right)^2} = – 3\left( { \Rightarrow z = \pm \sqrt 3 {\text{i}}} \right)\) M1
\(z = \sqrt 3 \,{\text{cis}}\frac{\pi }{2}{\text{ or }}{z_1} = \sqrt 3 \,{\text{cis}}\frac{{3\pi }}{2}\left( { = \sqrt 3 \,{\text{cis}}\left( {\frac{{ – \pi }}{2}} \right)} \right)\) A1A1
so \(r = \sqrt 3 {\text{ and }}\theta = \frac{\pi }{2}{\text{ or }}\theta = \frac{{3\pi }}{2}\left( { = \frac{{ – \pi }}{2}} \right)\)
Note: Accept \(r\,{\text{cis}}(\theta )\) form.
METHOD 2
\({z^2} = {\left( {1 – 1 + \sqrt 3 {\text{i}}} \right)^2} = – 3 \Rightarrow {z^2} = 3{\text{cis}}\left( {(2n + 1)\pi } \right)\) M1
\({r^2} = 3 \Rightarrow r = \sqrt 3 \) A1
\(2\theta = (2n + 1)\pi \Rightarrow \theta = \frac{\pi }{2}{\text{ or }}\theta = \frac{{3\pi }}{2}{\text{ (as }}0 \leqslant \theta < 2\pi )\) A1
Note: Accept \(r\,{\text{cis}}(\theta )\) form.
(ii) METHOD 1
\(z = – \frac{1}{{2{\text{cis}}\frac{{2\pi }}{3}}} \Rightarrow z = \frac{{{\text{cis}}\pi }}{{2{\text{cis}}\frac{{2\pi }}{3}}}\) M1
\( \Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}\)
so \(r = \frac{1}{2}{\text{ and }}\theta = \frac{\pi }{3}\) A1A1
METHOD 2
\({z_1} = – \frac{1}{{ – 1 + \sqrt 3 {\text{i}}}} \Rightarrow {z_1} = – \frac{{ – 1 – \sqrt 3 {\text{i}}}}{{\left( { – 1 + \sqrt 3 {\text{i}}} \right)\left( { – 1 – \sqrt 3 {\text{i}}} \right)}}\) M1
\(z = \frac{{1 + \sqrt 3 {\text{i}}}}{4} \Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}\)
so \(r = \frac{1}{2}{\text{ and }}\theta = \frac{\pi }{3}\) A1A1
[6 marks]
\(\frac{{{z_1}}}{{{z_2}}} = \sqrt 3 \,{\text{cis}}\frac{{5\pi }}{6}\) (A1)
\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} = {\sqrt 3 ^n}{\text{cis}}\frac{{5n\pi }}{6}\) A1
equating imaginary part to zero and attempting to solve M1
obtain n = 12 A1
Note: Working which only includes the argument is valid.
[4 marks]
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
Given that \(\frac{z}{{z + 2}} = 2 – {\text{i}}\) , \(z \in \mathbb{C}\) , find z in the form \(a + {\text{i}}b\) .
Answer/Explanation
Markscheme
METHOD 1
\(z = \left( {2 – {\text{i}}} \right)\left( {z + 2} \right)\) M1
\( = 2z + 4 – {\text{i}}z – 2{\text{i}}\)
\(z\left( {1 – {\text{i}}} \right) = – 4 + 2{\text{i}}\)
\(z = \frac{{ – 4 + 2{\text{i}}}}{{1 – {\text{i}}}}\) A1
\(z = \frac{{ – 4 + 2{\text{i}}}}{{1 – {\text{i}}}} \times \frac{{1 + {\text{i}}}}{{1 + {\text{i}}}}\) M1
\( = – 3 – {\text{i}}\) A1
METHOD 2
let \(z = a + {\text{i}}b\)
\(\frac{{a + {\text{i}}b}}{{a + {\text{i}}b + 2}} = 2 – {\text{i}}\) M1
\(a + {\text{i}}b = \left( {2 – i} \right)\left( {\left( {a + 2} \right) + {\text{i}}b} \right)\)
\(a + {\text{i}}b = 2\left( {a + 2} \right) + 2b{\text{i}} – {\text{i}}\left( {a + 2} \right) + b\)
\(a + {\text{i}}b = 2a + b + 4 + \left( {2b – a – 2} \right){\text{i}}\)
attempt to equate real and imaginary parts M1
\(a = 2a + b + 4\left( { \Rightarrow a + b + 4 = 0} \right)\)
and \(b = 2b – a – 2\left( { \Rightarrow – a + b – 2 = 0} \right)\) A1
Note: Award Al for two correct equations.
\(b = – 1\); \(a = – 3\) A1
\(z = – 3 – {\text{i}}\)
[4 marks]
Question
Write down the expansion of \({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3}\) in the form \(a + {\text{i}}b\) , where \(a\) and \(b\) are in terms of \({\sin \theta }\) and \({\cos \theta }\) .[2]
Hence show that \(\cos 3\theta = 4{\cos ^3}\theta – 3\cos \theta \) .[3]
Similarly show that \(\cos 5\theta = 16{\cos ^5}\theta – 20{\cos ^3}\theta + 5\cos \theta \) .[3]
Hence solve the equation \(\cos 5\theta + \cos 3\theta + \cos \theta = 0\) , where \(\theta \in \left[ { – \frac{\pi }{2},\frac{\pi }{2}} \right]\) .[6]
By considering the solutions of the equation \(\cos 5\theta = 0\) , show that \(\cos \frac{\pi }{{10}} = \sqrt {\frac{{5 + \sqrt 5 }}{8}} \) and state the value of \(\cos \frac{{7\pi }}{{10}}\).[8]
Answer/Explanation
Markscheme
\({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3} = {\cos ^3}\theta + 3{\cos ^2}\theta \left( {{\text{i}}\sin \theta } \right) + 3\cos \theta {\left( {{\text{i}}\sin \theta } \right)^2} + {\left( {{\text{i}}\sin \theta } \right)^3}\) (M1)
\( = {\cos ^3}\theta – 3\cos \theta {\sin ^2}\theta + {\text{i}}\left( {3{{\cos }^2}\theta \sin \theta – {{\sin }^3}\theta } \right)\) A1
[2 marks]
from De Moivre’s theorem
\({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3} = \cos 3\theta + {\text{i}}\sin 3\theta \) (M1)
\(\cos 3\theta + {\text{i}}\sin 3\theta = \left( {{{\cos }^3}\theta – 3\cos \theta {{\sin }^2}\theta } \right) + {\text{i}}\left( {3{{\cos }^2}\theta \sin \theta – {{\sin }^3}\theta } \right)\)
equating real parts M1
\(\cos 3\theta = {\cos ^3}\theta – 3\cos \theta {\sin ^2}\theta \)
\( = {\cos ^3}\theta – 3\cos \theta \left( {1 – {{\cos }^2}\theta } \right)\) A1
\( = {\cos ^3}\theta – 3\cos \theta + 3{\cos ^3}\theta \)
\( = 4{\cos ^3}\theta – 3\cos \theta \) AG
Note: Do not award marks if part (a) is not used.
[3 marks]
\({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^5} = \)
\({\cos ^5}\theta + 5{\cos ^4}\theta \left( {{\text{i}}\sin \theta } \right) + 10{\cos ^3}\theta {\left( {{\text{i}}\sin \theta } \right)^2} + 10{\cos ^2}\theta {\left( {{\text{i}}\sin \theta } \right)^3} + 5\cos \theta {\left( {{\text{i}}\sin \theta } \right)^4} + {\left( {{\text{i}}\sin \theta } \right)^5}\) (A1)
from De Moivre’s theorem
\(\cos 5\theta = {\cos ^5}\theta – 10{\cos ^3}\theta {\sin ^2}\theta + 5\cos \theta {\sin ^4}\theta \) M1
\( = {\cos ^5}\theta – 10{\cos ^3}\theta \left( {1 – {{\cos }^2}\theta } \right) + 5\cos \theta {\left( {1 – {{\cos }^2}\theta } \right)^2}\) A1
\( = {\cos ^5}\theta – 10{\cos ^3}\theta + 10{\cos ^5}\theta + 5\cos \theta – 10{\cos ^3}\theta + 5{\cos ^5}\theta \)
\(\therefore \cos 5\theta = 16{\cos ^5}\theta – 20{\cos ^3}\theta + 5\cos \theta \) AG
Note: If compound angles used in (b) and (c), then marks can be allocated in (c) only.
[3 marks]
\(\cos 5\theta + \cos 3\theta + \cos \theta \)
\( = \left( {16{{\cos }^5}\theta – 20{{\cos }^3}\theta + 5\cos \theta } \right) + \left( {4{{\cos }^3}\theta – 3\cos \theta } \right) + \cos \theta = 0\) M1
\(16{\cos ^5}\theta – 16{\cos ^3}\theta + 3\cos \theta = 0\) A1
\(\cos \theta \left( {16{{\cos }^4}\theta – 16{{\cos }^2}\theta + 3} \right) = 0\)
\(\cos \theta \left( {4{{\cos }^2}\theta – 3} \right)\left( {4{{\cos }^2}\theta – 1} \right) = 0\) A1
\(\therefore \cos \theta = 0\); \( \pm \frac{{\sqrt 3 }}{2}\); \( \pm \frac{1}{2}\) A1
\(\therefore \theta = \pm \frac{\pi }{6}\); \(\pm \frac{\pi }{3}\); \( \pm \frac{\pi }{2}\) A2
[6 marks]
\(\cos 5\theta = 0\)
\(5\theta = …\frac{\pi }{2}\); \(\left( {\frac{{3\pi }}{2};\frac{{5\pi }}{2}} \right)\); \(\frac{{7\pi }}{2}\); \(…\) (M1)
\(\theta = …\frac{\pi }{{10}}\); \(\left( {\frac{{3\pi }}{{10}};\frac{{5\pi }}{{10}}} \right)\); \(\frac{{7\pi }}{10}\); \(…\) (M1)
Note: These marks can be awarded for verifications later in the question.
now consider \(16{\cos ^5}\theta – 20{\cos ^3}\theta + 5\cos \theta = 0\) M1
\(\cos \theta \left( {16{{\cos }^4}\theta – 20{{\cos }^2}\theta + 5} \right) = 0\)
\({\cos ^2}\theta = \frac{{20 \pm \sqrt {400 – 4\left( {16} \right)\left( 5 \right)} }}{{32}}\); \(\cos \theta = 0\) A1
\(\cos \theta = \pm \sqrt {\frac{{20 \pm \sqrt {400 – 4\left( {16} \right)\left( 5 \right)} }}{{32}}} \)
\(\cos \frac{\pi }{{10}} = \sqrt {\frac{{20 + \sqrt {400 – 4\left( {16} \right)\left( 5 \right)} }}{{32}}} \) since max value of cosine \( \Rightarrow \) angle closest to zero R1
\(\cos \frac{\pi }{{10}} = \sqrt {\frac{{4.5 + 4\sqrt {25 – 4\left( 5 \right)} }}{{4.8}}} = \sqrt {\frac{{5 + \sqrt 5 }}{8}} \) A1
\(\cos \frac{{7\pi }}{{10}} = – \sqrt {\frac{{5 – \sqrt 5 }}{8}} \) A1A1
[8 marks]
Question
Solve the equation \({z^3} = 8{\text{i}},{\text{ }}z \in \mathbb{C}\) giving your answers in the form \(z = r(\cos \theta + {\text{i}}\sin \theta )\) and in the form \(z = a + b{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{R}\).[6]
Consider the complex numbers \({z_1} = 1 + {\text{i}}\) and \({z_2} = 2\left( {\cos \left( {\frac{\pi }{2}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6}} \right)} \right)\).
(i) Write \({z_1}\) in the form \(r(\cos \theta + {\text{i}}\sin \theta )\).
(ii) Calculate \({z_1}{z_2}\) and write in the form \(z = a + b{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{R}\).
(iii) Hence find the value of \(\tan \frac{{5\pi }}{{12}}\) in the form \(c + d\sqrt 3 \), where \(c,{\text{ }}d \in \mathbb{Z}\).
(iv) Find the smallest value \(p > 0\) such that \({({z_2})^p}\) is a positive real number.[11]
Answer/Explanation
Markscheme
Note: Accept answers and working in degrees, throughout.
\({z^3} = 8\left( {\cos \left( {\frac{\pi }{2} + 2\pi k} \right) + {\text{i}}\sin \left( {\frac{\pi }{2} + 2\pi k} \right)} \right)\) (A1)
attempt the use of De Moivre’s Theorem in reverse M1
\(z = 2\left( {\cos \left( {\frac{\pi }{6}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6}} \right)} \right);{\text{ }}2\left( {\cos \left( {\frac{{5\pi }}{6}} \right) + {\text{i}}\sin \left( {\frac{{5\pi }}{6}} \right)} \right);\)
\(2\left( {\cos \left( {\frac{{9\pi }}{6}} \right) + {\text{i}}\sin \left( {\frac{{9\pi }}{6}} \right)} \right)\) A2
Note: Accept cis form.
\(z = \pm \sqrt 3 + {\text{i}},{\text{ }} – 2{\text{i}}\) A2
Note: Award A1 for two correct solutions in each of the two lines above.
[6 marks]
Note: Accept answers and working in degrees, throughout.
(i) \({z_1} = \sqrt 2 \left( {\cos \left( {\frac{\pi }{4}} \right) + {\text{i}}\sin \left( {\frac{\pi }{4}} \right)} \right)\) A1A1
(ii) \(\left( {{z_2} = \left( {\sqrt 3 + {\text{i}}} \right)} \right)\)
\({z_1}{z_2} = (1 + {\text{i}})\left( {\sqrt 3 + {\text{i}}} \right)\) M1
\( = \left( {\sqrt 3 – 1} \right) + {\text{i}}\left( {1 + \sqrt 3 } \right)\) A1
(iii) \({z_1}{z_2} = 2\sqrt 2 \left( {\cos \left( {\frac{\pi }{6} + \frac{\pi }{4}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6} + \frac{\pi }{4}} \right)} \right)\) M1A1
Note: Interpret “hence” as “hence or otherwise”.
\(\tan \frac{{5\pi }}{{12}} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 – 1}}\) A1
\( = 2 + \sqrt 3 \) M1A1
Note: Award final M1 for an attempt to rationalise the fraction.
(iv) \({z_2}^p = {2^p}\left( {{\text{cis}}\left( {\frac{{p\pi }}{6}} \right)} \right)\) (M1)
\({z_2}^p\) is a positive real number when \(p = 12\) A1
Note: Accept a solution based on part (a).
[11 marks]
Total [17 marks]