${z_1} = {(1 + {\text{i}}\sqrt 3 )^m}{\text{ and }}{z_2} = {(1 - {\text{i}})^n}$ .

(a)     Find the modulus and argument of ${z_1}$ and ${z_2}$ in terms of m and n, respectively.

(b)     Hence, find the smallest positive integers m and n such that ${z_1} = {z_2}$ .

(a)     $\left| {1 + {\text{i}}\sqrt 3 } \right| = 2{\text{ or }}\left| {1 - {\text{i}}} \right| = \sqrt 2$     (A1)

$\arg (1 + {\text{i}}\sqrt 3 ) = \frac{\pi }{3}{\text{ or }}\arg (1 - {\text{i}}) =  - \frac{\pi }{4}\,\,\,\,\,\left( {{\text{accept }}\frac{{7\pi }}{4}} \right)$     (A1)

$\left| {{z_1}} \right| = {2^m}$     A1

$\left| {{z_2}} \right| = {\sqrt 2 ^n}$     A1

$\arg ({z_1}) = m\arctan \sqrt 3  = m\frac{\pi }{3}$     A1

$\arg ({z_2}) = n\arctan ( - 1) = n\frac{{ - \pi }}{4}\,\,\,\,\,\left( {{\text{accept }}n\frac{{7\pi }}{4}} \right)$     A1     N2

[6 marks]

(b)     ${2^m} = {\sqrt 2 ^n} \Rightarrow n = 2m$     (M1)A1

$m\frac{\pi }{3} = n\frac{{ - \pi }}{4} + 2\pi k$ , where k is an integer     M1A1

$\Rightarrow m\frac{\pi }{3} + n\frac{\pi }{4} = 2\pi k$

$\Rightarrow m\frac{\pi }{3} + 2m\frac{\pi }{4} = 2\pi k$     (M1)

$\frac{5}{6}m\pi  = 2\pi k$

$\Rightarrow m = \frac{{12}}{5}k$     A1

The smallest value of k such that m is an integer is 5, hence

m =12     A1

n = 24.     A1     N2

[8 marks]

Total [14 marks]

Part (a) of this question was answered fairly well by candidates who attempted this question. The main error was the sign of the argument of ${z_2}$. Few candidates attempted part (b), and of those who did, most scored the first two marks for equating the modulii. Only a very small number equated the arguments correctly using $2\pi k$.