(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 \(n \in \mathbb{N}\).

(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.

(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]

(i) A disappointingly large number of candidates were unable to give the correct arguments for the three complex numbers. Such errors undermined their efforts to tackle parts (ii) and (iii).

Many candidates were successful in part (i), but failed to capitalise on that – in particular, few used the fact that roots of \({z^7} - 1 = 0\) come in complex conjugate pairs.