A2 October 2021 Paper 2 Q2
2 In this question you must show detailed reasoning.
The complex numbers \(z_1\) and \(z_2\) are given by \(z_1 = 3 - 7\mathrm{i}\) and \(z_2 = 2 + 4\mathrm{i}\).
| Scheme | Marks | AO |
|---|---|---|
| DR (i) \(3z_1 + 4z_2 = 3(3 - 7\mathrm{i}) + 4(2 + 4\mathrm{i}) = 17 - 5\mathrm{i}\) | B1 | 1.1 |
| [1] | ||
| DR (ii) \(z_1 z_2 = (3 - 7\mathrm{i})(2 + 4\mathrm{i}) = 6 + 12\mathrm{i} - 14\mathrm{i} - 28(-1)\) | M1 | 1.1 |
| \(= 34 - 2\mathrm{i}\) | A1 | 1.1 |
| [2] | ||
| DR (iii) \(\dfrac{z_1}{z_2} = \dfrac{3 - 7\mathrm{i}}{2 + 4\mathrm{i}} = \dfrac{3 - 7\mathrm{i}}{2 + 4\mathrm{i}} \times \dfrac{2 - 4\mathrm{i}}{2 - 4\mathrm{i}}\) | M1 | 1.1 |
| \(= \dfrac{6 - 12\mathrm{i} - 14\mathrm{i} - 28}{4 + 16} = \dfrac{-22 - 26\mathrm{i}}{20} = -\dfrac{11}{10} - \dfrac{13}{10}\mathrm{i}\) | A1 | 1.1 |
| [2] |
Notes
(a)(ii)
M1: Attempted expansion with \(\mathrm{i}^2 = -1\) used and at least 3 correctly expanded terms
\(-28(-1)\) can be simply \(+28\)
(a)(iii)
M1: Multiplying top and bottom by (real multiple of) conjugate of bottom
A1: Must see some evidence of expansion
Allow \(\dfrac{-11 - 13\mathrm{i}}{10}\) or \(-\dfrac{11 + 13\mathrm{i}}{10}\)
| Scheme | Marks | AO |
|---|---|---|
| DR \(\sqrt{3^2 + (-7)^2}\) or \(\tan^{-1}\left(\dfrac{-7}{3}\right)\) | M1 | 1.1 |
| \(|z_1| = \sqrt{58}\) or awrt 7.62 or \(\arg z_1 =\) awrt \(-1.17\) or 5.12 rads | A1 | 1.1 |
| \(z_1 = \sqrt{58}\,\mathrm{cis}(-1.17)\) or \(z_1 = \sqrt{58}\,\mathrm{e}^{-1.17\mathrm{i}}\) or \(z_1 = \sqrt{58}\left(\cos(-1.17) + \mathrm{i}\sin(-1.17)\right)\) or \(\left[\sqrt{58}, -1.17\right]\) | A1 | 2.5 |
| [3] |
Notes
M1: Explicit working must be seen
Other trig calculations could be sufficient for M1 provided that these are being used to find the argument.
A1: (second) Must be in correct form with \(\sqrt{58}\) exact and could be awrt 5.12 instead of \(-1.17\).
Do not condone degrees
Condone round brackets