Let \(z = x + {\text{i}}y\) be any non-zero complex number.

(i)     Express \(\frac{1}{z}\) in the form \(u + {\text{i}}v\) .

(ii)     If \(z + \frac{1}{z} = k\) , \(k \in \mathbb{R}\) , show that either y = 0 or \({x^2} + {y^2} = 1\).

(iii)     Show that if \({x^2} + {y^2} = 1\) then \(\left| k \right| \leqslant 2\) .

Let \(w = \cos \theta + {\text{i}}\sin \theta \) .

(i)     Show that \({w^n} + {w^{ - n}} = 2\cos n\theta \) , \(n \in \mathbb{Z}\) .

(ii)     Solve the equation \(3{w^2} - w + 2 - {w^{ - 1}} + 3{w^{ - 2}} = 0\), giving the roots in the form \(x + {\text{i}}y\) .

(i)     \(\frac{1}{z} = \frac{1}{{x + {\text{i}}y}} \times \frac{{x - {\text{i}}y}}{{x - {\text{i}}y}} = \frac{x}{{{x^2} + {y^2}}} - {\text{i}}\frac{y}{{{x^2} + {y^2}}}\)     (M1)A1

(ii)     \(z + \frac{1}{z} = x + \frac{x}{{{x^2} + {y^2}}} + {\text{i}}\left( {y - \frac{y}{{{x^2} + {y^2}}}} \right) = k\)     (A1)

for k to be real, \(y - \frac{y}{{{x^2} + {y^2}}} = 0 \Rightarrow y({x^2} + {y^2} - 1) = 0\)     M1A1

hence, \(y = 0{\text{ or }}{x^2} + {y^2} - 1 = 0 \Rightarrow {x^2} + {y^2} = 1\)     AG

(iii)     when \({x^2} + {y^2} = 1,{\text{ }}z + \frac{1}{z} = 2x\)     (M1)A1

\(\left| x \right| \leqslant 1\)     R1

\( \Rightarrow \left| k \right| \leqslant 2\)     AG

[8 marks]

(i)     \({w^{ - n}} = \cos ( - n\theta ) + {\text{i}}\sin ( - n\theta ) = \cos n\theta - {\text{i}}\sin n\theta \)     M1A1

\( \Rightarrow {w^n} + {w^{ - n}} = (\cos n\theta + {\text{i}}\sin n\theta ) + (\cos n\theta - {\text{i}}\sin n\theta ) = 2\cos n\theta \)     M1AG

(ii)     (rearranging)

\(3({w^2} + {w^{ - 2}}) - (w + {w^{ - 1}}) + 2 = 0\)     (M1)

\( \Rightarrow 3(2\cos 2\theta ) - 2\cos \theta + 2 = 0\)     A1

\( \Rightarrow 2(3\cos 2\theta - \cos \theta + 1) = 0\)

\( \Rightarrow 3(2{\cos ^2}\theta - 1) - \cos \theta + 1 = 0\)     M1

\( \Rightarrow 6{\cos ^2}\theta - \cos \theta - 2 = 0\)     A1

\( \Rightarrow (3\cos \theta - 2)(2\cos \theta + 1) = 0\)     M1

\(\therefore \cos \theta = \frac{2}{3},{\text{ }}\cos \theta = - \frac{1}{2}\)     A1A1

\(\cos \theta = \frac{2}{3} \Rightarrow \sin \theta = \pm \frac{{\sqrt 5 }}{3}\)     A1

\(\cos \theta = - \frac{1}{2} \Rightarrow \sin \theta = \pm \frac{{\sqrt 3 }}{2}\)     A1

\(\therefore w = \frac{2}{3} \pm \frac{{{\text{i}}\sqrt 5 }}{3}, - \frac{1}{2} \pm \frac{{{\text{i}}\sqrt 3 }}{2}\)     A1A1

Note: Allow FT from incorrect \(\cos \theta \) and/or \(\sin \theta \) .

[14 marks]

A large number of candidates did not attempt part (a), or did so unsuccessfully.

It was obvious that many candidates had been trained to answer questions of the type in part (b), and hence of those who attempted it, many did so successfully. Quite a few however failed to find all solutions.