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

(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 \(2\pi k\).