Question
Let \(\omega = \cos \theta + {\text{i}}\sin \theta \) . Find, in terms of \(\theta \) , the modulus and argument of \({(1 – {\omega ^2})^ * }\) .
Answer/Explanation
Markscheme
METHOD 1
\({(1 – {\omega ^2})^ * } = {(1 – {\text{cis}}\,2\theta )^ * } = {\left( {(1 – \cos 2\theta ) – {\text{i}}\sin 2\theta } \right)^ * }\) M1A1
\( = (1 – \cos 2\theta ) + {\text{i}}\sin 2\theta \) A1
\(\left| {{{(1 – {\omega ^2})}^ * }} \right| = \sqrt {{{(1 – \cos 2\theta )}^2} + {{\sin }^2}2\theta } \left( { = \sqrt {{{(2{{\sin }^2}\theta )}^2} + {{(2\sin \theta \cos \theta )}^2}} } \right)\) M1
\( = \left| {2\sin \theta } \right|\) A1
\(\arg \left( {{{(1 – {\omega ^2})}^ * }} \right) = \alpha \Rightarrow \tan \alpha = \cot (\theta )\) M1
\(\alpha = \frac{\pi }{2} – \theta \) A1
therefore:
modulus is \(2\left| {\sin \theta } \right|\) and argument is \(\frac{\pi }{2} – \theta {\text{ or }}\frac{\pi }{2} – \theta \pm \pi \)
Note: Accept modulus is 2sinθ and argument is \(\frac{\pi }{2} – \theta \)
METHOD 2
EITHER
\({(1 – {\varpi ^2})^ * } = {(1 – {\text{cis}}\,2\theta )^ * } = {\left( {(1 – \cos 2\theta ) – {\text{i}}\sin 2\theta } \right)^ * }\) M1A1
\( = (1 – \cos 2\theta ) + {\text{i}}\sin 2\theta \) A1
\( = (1 – 1 + 2{\sin ^2}\theta ) + 2{\text{i}}\sin \theta \cos \theta \) M1
OR
\({(1 – {\varpi ^2})^ * } = {\left( {1 – {{(\cos \theta + {\text{i}}\sin \theta )}^2}} \right)^ * }\) M1A1
\( = {(1 – {\cos ^2}\theta + {\sin ^2}\theta – 2{\text{i}}\sin \theta \cos \theta )^ * }\) A1
\( = 2{\sin ^2}\theta + 2{\text{i}}\sin \theta \cos \theta \) M1
THEN
\( = 2\sin \theta (\sin \theta + {\text{i}}\cos \theta )\) (M1)
\( = 2\sin \theta \left( {\cos \left( {\frac{\pi }{2} – \theta } \right) + {\text{i}}\sin \left( {\frac{\pi }{2} – \theta } \right)} \right)\) A1A1
\( = 2\sin \theta \,{\text{cis}}\left( {\frac{\pi }{2} – \theta } \right)\)
therefore:
modulus is \(2\left| {\sin \theta } \right|\) and argument is \(\frac{\pi }{2} – \theta {\text{ or }}\frac{\pi }{2} – \theta \pm \pi \)
Note: Accept modulus is 2sinθ and argument is \(\frac{\pi }{2} – \theta \) .
[7 marks]
Examiners report
This was the most challenging question in part A with just a few candidates scoring full marks. This question showed that many candidates have difficulties with algebraic manipulations, application of De Moivre’s theorem and use of trigonometric identities. Although some candidates managed to calculate the square of a complex number, many failed to write down its conjugate or made algebraic errors which lead to wrong results in many cases. Just a few candidates were able to calculate the modulus and the argument of the complex number.
Question
Find, in its simplest form, the argument of \({\left( {\sin \theta + {\text{i}}(1 – \cos \theta )} \right)^2}\) where \(\theta \) is an acute angle.
Answer/Explanation
Markscheme
\({\left( {\sin \theta + {\text{i}}(1 – \cos \theta )} \right)^2} = {\sin ^2}\theta – {(1 – \cos \theta )^2} + {\text{i}}2\sin \theta (1 – \cos \theta )\) M1A1
Let \(\alpha \) be the required argument.
\(\tan \alpha = \frac{{2\sin \theta (1 – \cos \theta )}}{{{{\sin }^2}\theta – {{(1 – \cos \theta )}^2}}}\) M1
\( = \frac{{2\sin \theta (1 – \cos \theta )}}{{(1 – {{\cos }^2}\theta ) – (1 – 2\cos \theta + {{\cos }^2}\theta )}}\) (M1)
\( = \frac{{2\sin \theta (1 – \cos \theta )}}{{2\cos \theta (1 – \cos \theta )}}\) A1
\( = \tan \theta \) A1
\(\alpha = \theta \) A1
[7 marks]
Examiners report
Very few candidates scored more than the first two marks in this question. Some candidates had difficulty manipulating trigonometric identities. Most candidates did not get as far as defining the argument of the complex expression.
Question
\({z_1} = {(1 + {\text{i}}\sqrt 3 )^m}{\text{ and }}{z_2} = {(1 – {\text{i}})^n}\) .
(a) Find the modulus and argument of \({z_1}\) and \({z_2}\) in terms of m and n, respectively.
(b) Hence, find the smallest positive integers m and n such that \({z_1} = {z_2}\) .
Answer/Explanation
Markscheme
(a) \(\left| {1 + {\text{i}}\sqrt 3 } \right| = 2{\text{ or }}\left| {1 – {\text{i}}} \right| = \sqrt 2 \) (A1)
\(\arg (1 + {\text{i}}\sqrt 3 ) = \frac{\pi }{3}{\text{ or }}\arg (1 – {\text{i}}) = – \frac{\pi }{4}\,\,\,\,\,\left( {{\text{accept }}\frac{{7\pi }}{4}} \right)\) (A1)
\(\left| {{z_1}} \right| = {2^m}\) A1
\(\left| {{z_2}} \right| = {\sqrt 2 ^n}\) A1
\(\arg ({z_1}) = m\arctan \sqrt 3 = m\frac{\pi }{3}\) A1
\(\arg ({z_2}) = n\arctan ( – 1) = n\frac{{ – \pi }}{4}\,\,\,\,\,\left( {{\text{accept }}n\frac{{7\pi }}{4}} \right)\) A1 N2
[6 marks]
(b) \({2^m} = {\sqrt 2 ^n} \Rightarrow n = 2m\) (M1)A1
\(m\frac{\pi }{3} = n\frac{{ – \pi }}{4} + 2\pi k\) , where k is an integer M1A1
\( \Rightarrow m\frac{\pi }{3} + n\frac{\pi }{4} = 2\pi k\)
\( \Rightarrow m\frac{\pi }{3} + 2m\frac{\pi }{4} = 2\pi k\) (M1)
\(\frac{5}{6}m\pi = 2\pi k\)
\( \Rightarrow m = \frac{{12}}{5}k\) A1
The smallest value of k such that m is an integer is 5, hence
m =12 A1
n = 24. A1 N2
[8 marks]
Total [14 marks]
Examiners report
Part (a) of this question was answered fairly well by candidates who attempted this question. The main error was the sign of the argument of \({z_2}\). Few candidates attempted part (b), and of those who did, most scored the first two marks for equating the modulii. Only a very small number equated the arguments correctly using \(2\pi k\).
Question
Consider \(w = \frac{z}{{{z^2} + 1}}{\text{ where }}z = x + {\text{i}}y{\text{ , }}y \ne 0{\text{ and }}{z^2} + 1 \ne 0\) .
Given that \(\operatorname{Im} w = 0\), show that \(\left| z \right| = 1\).
Answer/Explanation
Markscheme
METHOD 1
Substituting \(z = x + {\text{i}}y\) to obtain \(w = \frac{{x + y{\text{i}}}}{{{{(x + y{\text{i}})}^2} + 1}}\) (A1)
\(w = \frac{{x + y{\text{i}}}}{{{x^2} – {y^2} + 1 + 2xy{\text{i}}}}\) A1
Use of \(({x^2} – {y^2} + 1 + 2xy{\text{i)}}\) to make the denominator real. M1
\({\text{ = }}\frac{{(x + y{\text{i)}}({x^2} – {y^2} + 1 – 2xy{\text{i)}}}}{{{{({x^2} – {y^2} + 1)}^2} + 4{x^2}{y^2}}}\) A1
\(\operatorname{Im} w = \frac{{y({x^2} – {y^2} + 1) – 2{x^2}y}}{{{{({x^2} – {y^2} + 1)}^2} + 4{x^2}{y^2}}}\) (A1)
\( = \frac{{y(1 – {x^2} – {y^2})}}{{{{({x^2} – {y^2} + 1)}^2} + 4{x^2}{y^2}}}\) A1
\(\operatorname{Im} w = 0 \Rightarrow 1 – {x^2} – {y^2} = 0\) i.e. \(\left| z \right| = 1{\text{ as }}y \ne 0\) R1AG N0
[7 marks]
METHOD 2
\(w({z^2} + 1) = z\) (A1)
\(w({x^2} – {y^2} + 1 + 2{\text{i}}xy) = x + yi\) A1
Equating real and imaginary parts
\(w({x^2} – {y^2} + 1) = x{\text{ and }}2wx = 1,{\text{ }}y \ne 0\) M1A1
Substituting \(w = \frac{1}{{2x}}\) to give \(\frac{x}{2} – \frac{{{y^2}}}{{2x}} + \frac{1}{{2x}} = x\) A1
\( – \frac{1}{{2x}}({y^2} – 1) = \frac{x}{2}\) or equivalent (A1)
\({x^2} + {y^2} = 1\), i.e. \(\left| z \right| = 1{\text{ as }}y \ne 0\) R1AG
[7 marks]
Examiners report
This was a difficult question that troubled most candidates. Most candidates were able to substitute z = x + yi into w but were then unable to make any further meaningful progress. Common errors included not expanding \({(x + {\text{i}}y)^2}\) correctly or not using a correct complex conjugate to make the denominator real. A small number of candidates produced correct solutions by using \(w = \frac{1}{{z + {z^{ – 1}}}}\).
Question
(a) Solve the equation \({z^3} = – 2 + 2{\text{i}}\), giving your answers in modulus-argument form.
(b) Hence show that one of the solutions is 1 + i when written in Cartesian form.
Answer/Explanation
Markscheme
(a) \({z^3} = 2\sqrt 2 {{\text{e}}^{\frac{{3\pi {\text{i}}}}{4}}}\) (M1)(A1)
\({z_1} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4}}}\) A1
adding or subtracting \(\frac{{2\pi {\text{i}}}}{3}\) M1
\({z_2} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4} + \frac{{2\pi {\text{i}}}}{3}}} = \sqrt 2 {{\text{e}}^{\frac{{11\pi {\text{i}}}}{{12}}}}\) A1
\({z_3} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4} – \frac{{2\pi {\text{i}}}}{3}}} = \sqrt 2 {{\text{e}}^{ – \frac{{5\pi {\text{i}}}}{{12}}}}\) A1
Notes: Accept equivalent solutions e.g. \({z_3} = \sqrt 2 {{\text{e}}^{\frac{{19\pi {\text{i}}}}{{12}}}}\)
Award marks as appropriate for solving \({(a + b{\text{i}})^3} = – 2 + 2{\text{i}}\).
Accept answers in degrees.
(b) \(\sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4}}}{\text{ }}\left( { = \sqrt 2 \left( {\frac{1}{{\sqrt 2 }} + \frac{{\text{i}}}{{\sqrt 2 }}} \right)} \right)\) A1
= 1 + i AG
Note: Accept geometrical reasoning.
[7 marks]
Examiners report
Many students incorrectly found the argument of \({z^3}\) to be \(\arctan \left( {\frac{2}{{ – 2}}} \right) = – \frac{\pi }{4}\). Of those students correctly finding one solution, many were unable to use symmetry around the origin, to find the other two. In part (b) many students found the cube of 1 + i which could not be awarded marks as it was not “hence”.
Question
Given that \(z = \cos \theta + {\text{i}}\sin \theta \) show that
(a) \(\operatorname{Im} \left( {{z^n} + \frac{1}{{{z^n}}}} \right) = 0,{\text{ }}n \in {\mathbb{Z}^ + }\);
(b) \(\operatorname{Re} \left( {\frac{{z – 1}}{{z + 1}}} \right) = 0,{\text{ }}z \ne – 1\).
Answer/Explanation
Markscheme
(a) using de Moivre’s theorem
\({z^n} + \frac{1}{{{z^n}}} = \cos n\theta + {\text{i}}\sin n\theta + \cos n\theta – {\text{i}}\sin n\theta {\text{ }}( = 2\cos n\theta )\), imaginary part of which is 0 M1A1
so \(\operatorname{Im} \left( {{z^n} + \frac{1}{{{z^n}}}} \right) = 0\) AG
(b) \(\frac{{z – 1}}{{z + 1}} = \frac{{\cos \theta + {\text{i}}\sin \theta – 1}}{{\cos \theta + {\text{i}}\sin \theta + 1}}\)
\( = \frac{{(\cos \theta – 1 + {\text{i}}\sin \theta )(\cos \theta + 1 – {\text{i}}\sin \theta )}}{{(\cos \theta + 1 + {\text{i}}\sin \theta )(\cos \theta + 1 – {\text{i}}\sin \theta )}}\) M1A1
Note: Award M1 for an attempt to multiply numerator and denominator by the complex conjugate of their denominator.
\( \Rightarrow \operatorname{Re} \left( {\frac{{z – 1}}{{z + 1}}} \right) = \frac{{(\cos \theta – 1)(\cos \theta + 1) + {{\sin }^2}\theta }}{{{\text{real denominator}}}}\) M1A1
Note: Award M1 for multiplying out the numerator.
\( = \frac{{{{\cos }^2}\theta + {{\sin }^2}\theta – 1}}{{{\text{real denominator}}}}\) A1
\( = 0\) AG
[7 marks]
Examiners report
Part(a) – The majority either obtained full marks or no marks here.
Part(b) – This question was algebraically complex and caused some candidates to waste their efforts for little credit.
Question
The complex numbers \({z_1}\) and \({z_2}\) have arguments between 0 and \(\pi \) radians. Given that \({z_1}{z_2} = – \sqrt 3 + {\text{i}}\) and \(\frac{{{z_1}}}{{{z_2}}} = 2{\text{i}}\), find the modulus and argument of \({z_1}\) and of \({z_2}\).
Answer/Explanation
Markscheme
METHOD 1
\({\text{arg(}}{z_1}{z_2}) = \frac{{5\pi }}{6}\,\,\,\,\,(150^\circ )\) (A1)
\(\arg \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \frac{\pi }{2}\,\,\,\,\,(90^\circ )\) (A1)
\( \Rightarrow \arg ({z_1}) + \arg ({z_2}) = \frac{{5\pi }}{6};{\text{ }}\arg ({z_1}) – \arg ({z_2}) = \frac{\pi }{2}\) M1
solving simultaneously
\(\arg ({z_1}) = \frac{{2\pi }}{3}{\text{ }}(120^\circ ){\text{ and }}\arg ({z_2}) = \frac{\pi }{6}{\text{ }}(30^\circ )\) A1A1
Note: Accept decimal approximations of the radian measures.
\(\left| {{z_1}{z_2}} \right| = 2 \Rightarrow \left| {{z_1}} \right|\left| {{z_2}} \right| = 2;{\text{ }}\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = 2 \Rightarrow \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} = 2\) M1
solving simultaneously
\(\left| {{z_1}} \right| = 2;{\text{ }}\left| {{z_2}} \right| = 1\) A1
[7 marks]
METHOD 2
\({z_1} = 2i{z_2}\,\,\,\,\,2iz_2^2 = – \sqrt 3 + i\) (M1)
\(z_2^2 = \frac{{ – \sqrt 3 + i}}{{2i}}\) A1
\({z_2} = \sqrt {\frac{{ – \sqrt 3 + i}}{{2i}}} = \frac{{\sqrt 3 }}{2} + \frac{1}{2}i{\text{ or }}{e^{\frac{\pi }{6}i}}\) (M1)(A1)
(allow \(0.866 + 0.5i\) or \({{\text{e}}^{0.524{\text{i}}}}\))
\({z_1} = – 1 + \sqrt 3 i{\text{ or }}2{{\text{e}}^{\frac{{2\pi }}{3}i}} – \) (allow −1 + 1.73i or \(2{{\text{e}}^{2.09{\text{i}}}}\)) (A1)
\({z_1}\,\,\,\,\,\)modulus = 2, argument \( = \frac{{2\pi }}{3}\) A1
\({z_2}\,\,\,\,\,\)modulus = 1, argument \( = \frac{\pi }{6}\) A1
Note: Accept degrees and decimal approximations to radian measure.
[7 marks]
Examiners report
Candidates generally found this question challenging. Many candidates had difficulty finding the arguments of \({z_1}{z_2}\) and \({z_1}/{z_2}\). Among candidates who attempted to solve for \({z_1}\) and \({z_2}\) in Cartesian form, many had difficulty with the algebraic manipulation involved.
Question
Given that \(z = \frac{{2 – {\text{i}}}}{{1 + {\text{i}}}} – \frac{{6 + 8{\text{i}}}}{{u + {\text{i}}}}\), find the values of u, u \( \in \mathbb{R}\), such that \(\operatorname{Re} z = \operatorname{Im} z\).
Answer/Explanation
Markscheme
METHOD 1
\(\frac{{2 – {\text{i}}}}{{1 + {\text{i}}}} = \frac{{1 – 3{\text{i}}}}{2}\) A1
\(\frac{{6 + 8{\text{i}}}}{{u + {\text{i}}}} \times \frac{{u – {\text{i}}}}{{u – {\text{i}}}} = \frac{{6u + 8 + (8u – 6){\text{i}}}}{{{u^2} + 1}}\) M1A1
\( \Rightarrow \frac{{2 – {\text{i}}}}{{1 + {\text{i}}}} – \frac{{6 + 8u}}{{u + {\text{i}}}} = \frac{1}{2} – \frac{{6u + 8}}{{{u^2} + 1}} – \left( {\frac{3}{2} + \frac{{8u – 6}}{{{u^2} + 1}}} \right){\text{i}}\)
\(\operatorname{Im} z = \operatorname{Re} z\)
\( \Rightarrow \frac{1}{2} – \frac{{6u + 8}}{{{u^2} + 1}} = – \frac{3}{2} – \frac{{8u – 6}}{{{u^2} + 1}}\) A1
(sketch from gdc, or algebraic method) (M1)
u = −3; u = 2 A1A1 N2
[7 marks]
METHOD 2
\(\frac{{2 – {\text{i}}}}{{1 + {\text{i}}}} – \frac{{6 + 8{\text{i}}}}{{u + {\text{i}}}} = \frac{{(2 – {\text{i}})(u + {\text{i}}) – (1 + {\text{i}})(6 + 8{\text{i}})}}{{(u – 1) + {\text{i}}(u + 1)}}\) M1A1
\( = \frac{{(2 – {\text{i}})(u + {\text{i}}) – (1 + {\text{i}})(6 + 8{\text{i}})}}{{(u – 1) + {\text{i}}(u + 1)}} \cdot \frac{{(u – 1) – {\text{i}}(u + 1)}}{{(u – 1) – {\text{i}}(u + 1)}}\) M1
\( = \frac{{{u^2} – 12u – 15 + {\text{i}}( – 3{u^2} – 16u + 9)}}{{2({u^2} + 1)}}\) A1
\(\operatorname{Re} z = \operatorname{Im} z \Rightarrow {u^2} – 12u – 15 = – 3{u^2} – 16u + 9\) M1
u = −3; u = 2 A1A1 N2
[7 marks]
Examiners report
Many candidates failed to access their GDC early enough to avoid huge algebraic manipulations, often carried out with many errors. Some candidates failed to separate and equate the real and imaginary parts of the expression obtained.
Question
Show that \(\left| {{{\text{e}}^{{\text{i}}\theta }}} \right| = 1\).
Consider the geometric series \(1 + \frac{1}{3}{{\text{e}}^{{\text{i}}\theta }} + \frac{1}{9}{{\text{e}}^{2{\text{i}}\theta }} + \ldots {\text{ .}}\)
Write down the common ratio, z, of the series, and show that \(\left| z \right| = \frac{1}{3}\).
Find an expression for the sum to infinity of this series.
Hence, show that \(\sin \theta + \frac{1}{3}\sin 2\theta + \frac{1}{9}\sin 3\theta + \ldots = \frac{{9\sin \theta }}{{10 – 6\cos \theta }}\).
Answer/Explanation
Markscheme
\(\left| {{{\text{e}}^{{\text{i}}\theta }}} \right|{\text{ }}\left( { = \left| {\cos \theta + {\text{i}}\sin \theta } \right|} \right) = \sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } = 1\) M1AG
[1 mark]
\(z = \frac{1}{3}{{\text{e}}^{{\text{i}}\theta }}\) A1
\(\left| z \right| = \left| {\frac{1}{3}{{\text{e}}^{{\text{i}}\theta }}} \right| = \frac{1}{3}\) A1AG
[2 marks]
\({S_\infty } = \frac{a}{{1 – r}} = \frac{1}{{1 – \frac{1}{3}{{\text{e}}^{{\text{i}}\theta }}}}\) (M1)A1
[2 marks]
EITHER
\({S_\infty } = \frac{1}{{1 – \frac{1}{3}\cos \theta – \frac{1}{3}{\text{i}}\sin \theta }}\) A1
\( = \frac{{1 – \frac{1}{3}\cos \theta + \frac{1}{3}{\text{i}}\sin \theta }}{{\left( {1 – \frac{1}{3}\cos \theta – \frac{1}{3}{\text{i}}\sin \theta } \right)\left( {1 – \frac{1}{3}\cos \theta + \frac{1}{3}{\text{i}}\sin \theta } \right)}}\) M1A1
\( = \frac{{1 – \frac{1}{3}\cos \theta + \frac{1}{3}{\text{i}}\sin \theta }}{{{{\left( {1 – \frac{1}{3}\cos \theta } \right)}^2} + \frac{1}{9}{{\sin }^2}\theta }}\) A1
\( = \frac{{1 – \frac{1}{3}\cos \theta + \frac{1}{3}{\text{i}}\sin \theta }}{{1 – \frac{2}{3}\cos \theta + \frac{1}{9}}}\) A1
OR
\({S_\infty } = \frac{1}{{1 – \frac{1}{3}{{\text{e}}^{{\text{i}}\theta }}}}\)
\( = \frac{{1 – \frac{1}{3}{{\text{e}}^{ – {\text{i}}\theta }}}}{{\left( {1 – \frac{1}{3}{{\text{e}}^{{\text{i}}\theta }}} \right)\left( {1 – \frac{1}{3}{{\text{e}}^{ – {\text{i}}\theta }}} \right)}}\) M1A1
\( = \frac{{1 – \frac{1}{3}{{\text{e}}^{ – {\text{i}}\theta }}}}{{1 – \frac{1}{3}({{\text{e}}^{{\text{i}}\theta }} + {{\text{e}}^{ – {\text{i}}\theta }}) + \frac{1}{9}}}\) A1
\( = \frac{{1 – \frac{1}{3}{{\text{e}}^{ – {\text{i}}\theta }}}}{{\frac{{10}}{9} – \frac{2}{3}\cos \theta }}\) A1
\( = \frac{{1 – \frac{1}{3}(\cos \theta – {\text{i}}\sin \theta )}}{{\frac{{10}}{9} – \frac{2}{3}\cos \theta }}\) A1
THEN
taking imaginary parts on both sides
\(\frac{1}{3}\sin \theta + \frac{1}{9}\sin 2\theta + \ldots = \frac{{\frac{1}{3}\sin \theta }}{{\frac{{10}}{9} – \frac{2}{3}\cos \theta }}\) M1A1A1
\( = \frac{{\sin \theta }}{{\frac{{10}}{9} – \frac{2}{3}\cos \theta }}\)
\( \Rightarrow \sin \theta + \frac{1}{3}\sin 2\theta + \ldots = \frac{{9\sin \theta }}{{10 – 6\cos \theta }}\) AG
[8 marks]
Examiners report
Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.
Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.
Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.
Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.
Question
A complex number z is given by \(z = \frac{{a + {\text{i}}}}{{a – {\text{i}}}},{\text{ }}a \in \mathbb{R}\).
(a) Determine the set of values of a such that
(i) z is real;
(ii) z is purely imaginary.
(b) Show that \(\left| z \right|\) is constant for all values of a.
Answer/Explanation
Markscheme
(a) \(\frac{{a + {\text{i}}}}{{a – {\text{i}}}} \times \frac{{a + {\text{i}}}}{{a + {\text{i}}}}\) M1
\( = \frac{{{a^2} – 1 + 2a{\text{i}}}}{{{a^2} + 1}}{\text{ }}\left( { = \frac{{{a^2} – 1}}{{{a^2} + 1}} + \frac{{2a}}{{{a^2} + 1}}{\text{i}}} \right)\) A1
(i) z is real when \(a = 0\) A1
(ii) z is purely imaginary when \(a = \pm 1\) A1
Note: Award M1A0A1A0 for \(\frac{{{a^2} – 1 + 2a{\text{i}}}}{{{a^2} – 1}}{\text{ }}\left( { = 1 + \frac{{2a}}{{{a^2} – 1}}{\text{i}}} \right)\) leading to \(a = 0\) in (i).
[4 marks]
(b) METHOD 1
attempting to find either \(\left| z \right|\) or \({\left| z \right|^2}\) by expanding and simplifying
eg \({\left| z \right|^2} = \frac{{{{\left( {{a^2} – 1} \right)}^2} + 4{a^2}}}{{{{\left( {{a^2} + 1} \right)}^2}}} = \frac{{{a^4} + 2{a^2} + 1}}{{{{\left( {{a^2} + 1} \right)}^2}}}\) M1
\( = \frac{{{{\left( {{a^2} + 1} \right)}^2}}}{{{{\left( {{a^2} + 1} \right)}^2}}}\)
\({\left| z \right|^2} = 1 \Rightarrow \left| z \right| = 1\) A1
METHOD 2
\(\left| z \right| = \frac{{\left| {a + {\text{i}}} \right|}}{{\left| {a – {\text{i}}} \right|}}\) M1
\(\left| z \right| = \frac{{\sqrt {{a^2} + 1} }}{{\sqrt {{a^2} + 1} }} \Rightarrow \left| z \right| = 1\) A1
[2 marks]
Total [6 marks]
Examiners report
Part (a) was reasonably well done. When multiplying and dividing by the conjugate of \(a – {\text{i}}\), some candidates incorrectly determined their denominator as \({a^2} – 1\).
In part (b), a significant number of candidates were able to correctly expand and simplify \(\left| z \right|\) although many candidates appeared to not understand the definition of \(\left| z \right|\).
Question
Let \(z = r(\cos \alpha + {\text{i}}\sin \alpha )\), where \(\alpha \) is measured in degrees, be the solution of \({z^5} – 1 = 0\) which has the smallest positive argument.
(i) Use the binomial theorem to expand \({(\cos \theta + {\text{i}}\sin \theta )^5}\).
(ii) Hence use De Moivre’s theorem to prove
\[\sin 5\theta = 5{\cos ^4}\theta \sin \theta – 10{\cos ^2}\theta {\sin ^3}\theta + {\sin ^5}\theta .\]
(iii) State a similar expression for \(\cos 5\theta \) in terms of \(\cos \theta \) and \(\sin \theta \).
Find the value of \(r\) and the value of \(\alpha \).
Using (a) (ii) and your answer from (b) show that \(16{\sin ^4}\alpha – 20{\sin ^2}\alpha + 5 = 0\).
Hence express \(\sin 72^\circ \) in the form \(\frac{{\sqrt {a + b\sqrt c } }}{d}\) where \(a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in \mathbb{Z}\).
Answer/Explanation
Markscheme
(i) \({(\cos \theta + {\text{i}}\sin \theta )^5}\)
\( = {\cos ^5}\theta + 5{\text{i}}{\cos ^4}\theta \sin \theta + 10{{\text{i}}^2}{\cos ^3}\theta {\sin ^2}\theta + \)
\(10{{\text{i}}^3}{\cos ^2}\theta {\sin ^3}\theta + 5{{\text{i}}^4}\cos \theta {\sin ^4}\theta + {{\text{i}}^5}{\sin ^5}\theta \) A1A1
\(( = {\cos ^5}\theta + 5{\text{i}}{\cos ^4}\theta \sin \theta – 10{\cos ^3}\theta {\sin ^2}\theta – \)
\(10{\text{i}}{\cos ^2}\theta {\sin ^3}\theta + 5\cos \theta {\sin ^4}\theta + {\text{i}}{\sin ^5}\theta )\)
Note: Award first A1 for correct binomial coefficients.
(ii) \({({\text{cis}}\theta )^5} = {\text{cis}}5\theta = \cos 5\theta + {\text{i}}\sin 5\theta \) M1
\( = {\cos ^5}\theta + 5{\text{i}}{\cos ^4}\theta \sin \theta – 10{\cos ^3}\theta {\sin ^2}\theta – 10{\text{i}}{\cos ^2}\theta {\sin ^3}\theta + \)
\(5\cos \theta {\sin ^4}\theta + {\text{i}}{\sin ^5}\theta \) A1
Note: Previous line may be seen in (i)
equating imaginary terms M1
\(\sin 5\theta = 5{\cos ^4}\theta \sin \theta – 10{\cos ^2}\theta {\sin ^3}\theta + {\sin ^5}\theta \) AG
(iii) equating real terms
\(\cos 5\theta = {\cos ^5}\theta – 10{\cos ^3}\theta {\sin ^2}\theta + 5\cos \theta {\sin ^4}\theta \) A1
[6 marks]
\({(r{\text{cis}}\alpha )^5} = 1 \Rightarrow {r^5}{\text{cis}}5\alpha = 1{\text{cis}}0\) M1
\({r^5} = 1 \Rightarrow r = 1\) A1
\(5\alpha = 0 \pm 360k,{\text{ }}k \in \mathbb{Z} \Rightarrow a = 72k\) (M1)
\(\alpha = 72^\circ \) A1
Note: Award M1A0 if final answer is given in radians.
[4 marks]
use of \(\sin (5 \times 72) = 0\) OR the imaginary part of \(1\) is \(0\) (M1)
\(0 = 5{\cos ^4}\alpha \sin \alpha – 10{\cos ^2}\alpha {\sin ^3}\alpha + {\sin ^5}\alpha \) A1
\(\sin \alpha \ne 0 \Rightarrow 0 = 5{(1 – {\sin ^2}\alpha )^2} – 10(1 – {\sin ^2}\alpha ){\sin ^2}\alpha + {\sin ^4}\alpha \) M1
Note: Award M1 for replacing \({\cos ^2}\alpha \).
\(0 = 5(1 – 2{\sin ^2}\alpha + {\sin ^4}\alpha ) – 10{\sin ^2}\alpha + 10{\sin ^4}\alpha + {\sin ^4}\alpha \) A1
Note: Award A1 for any correct simplification.
so \(16{\sin ^4}\alpha – 20{\sin ^2}\alpha + 5 = 0\) AG
[4 marks]
\({\sin ^2}\alpha = \frac{{20 \pm \sqrt {400 – 320} }}{{32}}\) M1A1
\(\sin \alpha = \pm \sqrt {\frac{{20 \pm \sqrt {80} }}{{32}}} \)
\(\sin \alpha = \frac{{ \pm \sqrt {10 \pm 2\sqrt 5 } }}{4}\) A1
Note: Award A1 regardless of signs. Accept equivalent forms with integral denominator, simplification may be seen later.
as \(72 > 60\), \(\sin 72 > \frac{{\sqrt 3 }}{2} = 0.866 \ldots \) we have to take both positive signs (or equivalent argument) R1
Note: Allow verification of correct signs with calculator if clearly stated
\(\sin 72 = \frac{{\sqrt {10 + 2\sqrt 5 } }}{4}\) A1
[5 marks]
Total [19 marks]
Examiners report
In part (i) many candidates tried to multiply it out the binomials rather than using the binomial theorem. In parts (ii) and (iii) many candidates showed poor understanding of complex numbers and made no attempt to equate real and imaginary parts. In a some cases the correct answer to part (iii) was seen although it was unclear how it was obtained.
This question was poorly done. Very few candidates made a good attempt to apply De Moivre’s theorem and most of them could not even equate the moduli to obtain \(r\).
This question was poorly done. From the few candidates that attempted it, many candidates started by writing down what they were trying to prove and made no progress.
Very few made a serious attempt to answer this question. Also very few realised that they could use the answers given in part (c) to attempt this part.