Use de Moivre’s theorem to find the value of \({\left( {\cos \left( {\frac{\pi }{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)^3}\).

Use mathematical induction to prove that

\[{(\cos \theta  - {\text{i}}\sin \theta )^n} = \cos n\theta  - {\text{i}}\sin n\theta {\text{ for }}n \in {\mathbb{Z}^ + }.\]

Find an expression in terms of \(\theta \) for \({(z)^n} + {(z{\text{*}})^n},{\text{ }}n \in {\mathbb{Z}^ + }\) where \(z{\text{*}}\) is the complex conjugate of \(z\).

(i)     Show that \(zz{\text{*}} = 1\).

(ii)     Write down the binomial expansion of \({(z + z{\text{*}})^3}\) in terms of \(z\) and \(z{\text{*}}\).

(iii)     Hence show that \(\cos 3\theta  = 4{\cos ^3}\theta  - 3\cos \theta \).

Hence solve \(4{\cos ^3}\theta  - 2{\cos ^2}\theta  - 3\cos \theta  + 1 = 0\) for \(0 \leqslant \theta  < \pi \).

\({\left( {\cos \left( {\frac{\pi }{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)^3} = \cos \pi  + {\text{i}}\sin \pi \)    M1

\( =  - 1\)    A1

[2 marks]

show the expression is true for \(n = 1\)     R1

assume true for \(n = k,{\text{ }}{(\cos \theta  - {\text{i}}\sin \theta )^k} = \cos k\theta  - {\text{i}}\sin k\theta \)     M1

Note:     Do not accept “let \(n = k\)” or “assume \(n = k\)”, assumption of truth must be present.

\({(\cos \theta  - {\text{i}}\sin \theta )^{k + 1}} = {(\cos \theta  - {\text{i}}\sin \theta )^k}(\cos \theta  - {\text{i}}\sin \theta )\)

\( = (\cos k\theta  - {\text{i}}\sin k\theta )(\cos \theta  - {\text{i}}\sin \theta )\)    M1

\( = \cos k\theta \cos \theta  - \sin k\theta \sin \theta  - {\text{i}}(\cos k\theta \sin \theta  + \sin k\theta \cos \theta )\)    A1

Note:     Award A1 for any correct expansion.

\( = \cos \left( {(k + 1)\theta } \right) - {\text{i}}\sin \left( {(k + 1)\theta } \right)\)    A1

therefore if true for \(n = k\) true for \(n = k + 1\), true for \(n = 1\), so true for all \(n( \in {\mathbb{Z}^ + })\)     R1

Note:     To award the final R mark the first 4 marks must be awarded.

[6 marks]

\({(z)^n} + {(z{\text{*}})^n} = {(\cos \theta  + {\text{i}}\sin \theta )^n} + {(\cos \theta  - {\text{i}}\sin \theta )^n}\)

\( = \cos n\theta  + {\text{i}}\sin n\theta  + \cos n\theta  - {\text{i}}\sin n\theta  = 2\cos (n\theta )\)    (M1)A1

[2 marks]

(i)     \(zz* = (\cos \theta  + {\text{i}}\sin \theta )(\cos \theta  - {\text{i}}\sin \theta )\)

\( = {\cos ^2}\theta  + {\sin ^2}\theta \)    A1

\( = 1\)    AG

Note:     Allow justification starting with \(|z| = 1\).

(ii)     \({(z + z{\text{*}})^3} = {z^3} + 3{z^2}z{\text{*}} + 3z{({z^*})^2} + (z{\text{*}})3\left( { = {z^3} + 3z + 3z{\text{*}} + {{(z{\text{*}})}^3}} \right)\)     A1

(iii)     \({(z + z{\text{*}})^3} = {(2\cos \theta )^3}\)     A1

\({z^3} + 3z + 3z{\text{*}} + {(z{\text{*}})^3} = 2\cos 3\theta  + 6\cos \theta \)    M1A1

\(\cos 3\theta  = 4{\cos ^3}\theta  - 3\cos \theta \)   AG

Note:     M1 is for using \(zz{\text{*}} = 1\), this might be seen in d(ii).

[5 marks]

\(4{\cos ^3}\theta  - 2{\cos ^2}\theta  - 3\cos \theta  + 1 = 0\)

\(4{\cos ^3}\theta  - 3\cos \theta  = 2{\cos ^2}\theta  - 1\)

\(\cos (3\theta ) = \cos (2\theta )\)    A1A1

Note:     A1 for \(\cos (3\theta )\) and A1 for \(\cos (2\theta )\).

\(\theta  = 0\)    A1

or \(3\theta  = 2\pi  - 2\theta {\text{ }}({\text{or }}3\theta  = 4\pi  - 2\theta )\)     M1

\(\theta  = \frac{{2\pi }}{5},{\text{ }}\frac{{4\pi }}{5}\)    A1A1

Note:     Do not accept solutions via factor theorem or other methods that do not follow “hence”.

[6 marks]

This was well done by most candidates who correctly applied de Moivre’s theorem.

This question was poorly done, which was surprising as it is very similar to the proof of de Moivre’s theorem which is stated as being required in the course guide. Many candidates spotted that they needed to use trigonometric identities but fell down through not being able to set out the proof in a logical form.

This was well done by the majority of candidates.

(d) parts (i) and (ii) were well done by the candidates, who were able to successfully use trigonometrical identities and the binomial theorem.

(d)(iii) This is a familiar technique that has appeared in several recent past papers and was successfully completed by many of the better candidates. Some candidates though neglected the instruction ‘hence’ and tried to derive the expression using trigonometric identities.

Again some candidates ignored ‘hence’ and tried to form a polynomial equation. Many candidates obtained the solution \(\cos (2\theta ) = \cos (3\theta )\) and hence the solution \(\theta  = 0\). Few were able to find the other solutions which can be obtained from consideration of the unit circle or similar methods.