Date | November 2011 | Marks available | 7 | Reference code | 11N.2.hl.TZ0.6 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 6 | Adapted from | N/A |
Question
The complex numbers \({z_1}\) and \({z_2}\) have arguments between 0 and \(\pi \) radians. Given that \({z_1}{z_2} = - \sqrt 3 + {\text{i}}\) and \(\frac{{{z_1}}}{{{z_2}}} = 2{\text{i}}\), find the modulus and argument of \({z_1}\) and of \({z_2}\).
Markscheme
METHOD 1
\({\text{arg(}}{z_1}{z_2}) = \frac{{5\pi }}{6}\,\,\,\,\,(150^\circ )\) (A1)
\(\arg \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \frac{\pi }{2}\,\,\,\,\,(90^\circ )\) (A1)
\( \Rightarrow \arg ({z_1}) + \arg ({z_2}) = \frac{{5\pi }}{6};{\text{ }}\arg ({z_1}) - \arg ({z_2}) = \frac{\pi }{2}\) M1
solving simultaneously
\(\arg ({z_1}) = \frac{{2\pi }}{3}{\text{ }}(120^\circ ){\text{ and }}\arg ({z_2}) = \frac{\pi }{6}{\text{ }}(30^\circ )\) A1A1
Note: Accept decimal approximations of the radian measures.
\(\left| {{z_1}{z_2}} \right| = 2 \Rightarrow \left| {{z_1}} \right|\left| {{z_2}} \right| = 2;{\text{ }}\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = 2 \Rightarrow \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} = 2\) M1
solving simultaneously
\(\left| {{z_1}} \right| = 2;{\text{ }}\left| {{z_2}} \right| = 1\) A1
[7 marks]
METHOD 2
\({z_1} = 2i{z_2}\,\,\,\,\,2iz_2^2 = - \sqrt 3 + i\) (M1)
\(z_2^2 = \frac{{ - \sqrt 3 + i}}{{2i}}\) A1
\({z_2} = \sqrt {\frac{{ - \sqrt 3 + i}}{{2i}}} = \frac{{\sqrt 3 }}{2} + \frac{1}{2}i{\text{ or }}{e^{\frac{\pi }{6}i}}\) (M1)(A1)
(allow \(0.866 + 0.5i\) or \({{\text{e}}^{0.524{\text{i}}}}\))
\({z_1} = - 1 + \sqrt 3 i{\text{ or }}2{{\text{e}}^{\frac{{2\pi }}{3}i}} - \) (allow −1 + 1.73i or \(2{{\text{e}}^{2.09{\text{i}}}}\)) (A1)
\({z_1}\,\,\,\,\,\)modulus = 2, argument \( = \frac{{2\pi }}{3}\) A1
\({z_2}\,\,\,\,\,\)modulus = 1, argument \( = \frac{\pi }{6}\) A1
Note: Accept degrees and decimal approximations to radian measure.
[7 marks]
Examiners report
Candidates generally found this question challenging. Many candidates had difficulty finding the arguments of \({z_1}{z_2}\) and \({z_1}/{z_2}\). Among candidates who attempted to solve for \({z_1}\) and \({z_2}\) in Cartesian form, many had difficulty with the algebraic manipulation involved.