(a)     Find the six roots of the equation ${z^6} - 1 = 0$ , giving your answers in the form $r\,{\text{cis}}\,\theta {\text{, }}r \in {\mathbb{R}^ + }{\text{, }}0 \leqslant \theta  < 2\pi$ .

(b)     (i)     Show that these six roots form a group G under multiplication of complex numbers.

          (ii)     Show that G is cyclic and find all the generators.

          (iii)     Give an example of another group that is isomorphic to G, stating clearly the corresponding elements in the two groups.

(a)     ${z^6} = 1 = {\text{cis}}\,2n\pi$     (M1)

The six roots are

${\text{cis}}\,0(1),{\text{ cis}}\frac{\pi }{3},{\text{ cis}}\frac{{2\pi }}{3},{\text{ cis}}\,\pi ( - 1),{\text{ cis}}\frac{{4\pi }}{3},{\text{ cis}}\frac{{5\pi }}{3}$     A3

Note: Award A2 for 4 or 5 correct roots, A1 for 2 or 3 correct roots.

[4 marks]

(b)     (i)     Closure: Consider any two roots ${\text{cis}}\frac{{m\pi }}{3},{\text{ cis}}\frac{{n\pi }}{3}$.     M1

${\text{cis}}\frac{{m\pi }}{3} \times {\text{cis}}\frac{{n\pi }}{3} = {\text{cis}}\,(m + n){\text{(mod6)}}\frac{\pi }{3} \in G$     A1

Note: Award M1A1 for a correct Cayley table showing closure.

Identity: The identity is 1.     A1

Inverse: The inverse of ${\text{cis}}\frac{{m\pi }}{3}{\text{ is cis}}\frac{{(6 - m)\pi }}{3} \in G$ .     A2

Associative: This follows from the associativity of multiplication.     R1

The 4 group axioms are satisfied.     R1

(ii)     Successive powers of ${\text{cis}}\frac{\pi }{3}\left( {{\text{or cis}}\frac{{5\pi }}{3}} \right)$

generate the group which is therefore cyclic.     R2

The (only) other generator is ${\text{cis}}\frac{{5\pi }}{3}\left( {{\text{or cis}}\frac{\pi }{3}} \right)$ .     A1

Note: Award A0 for any additional answers.

(iii)     The group of the integers 0, 1, 2, 3, 4, 5 under addition modulo 6.     R2

The correspondence is

$m \to {\text{cis}}\frac{{m\pi }}{3}$     R1

Note: Accept any other cyclic group of order 6.

[13 marks]

Total [17 marks]

This question was reasonably well answered by many candidates, although in (b)(iii), some candidates were unable to give another group isomorphic to G.