Date | May 2013 | Marks available | 9 | Reference code | 13M.1.hl.TZ2.13 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Determine, Express, Show that, and State | Question number | 13 | Adapted from | N/A |
Question
(i) Express each of the complex numbers z1=√3+i, z2=−√3+i and z3=−2i in modulus-argument form.
(ii) Hence show that the points in the complex plane representing z1, z2 and z3 form the vertices of an equilateral triangle.
(iii) Show that z3n1+z3n2=2z3n3 where n∈N.
(i) State the solutions of the equation z7=1 for z∈C, giving them in modulus-argument form.
(ii) If w is the solution to z7=1 with least positive argument, determine the argument of 1 + w. Express your answer in terms of π.
(iii) Show that z2−2zcos(2π7)+1 is a factor of the polynomial z7−1. State the two other quadratic factors with real coefficients.
Markscheme
(i) z1=2cis(π6), z2=2cis(5π6), z3=2cis(−π2) or 2cis(3π2) 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 2π3(mod 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]
Examiners report
(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.