AS June 2018 Paper 1 Q3
3 Find real numbers \(a\) and \(b\) such that \((a - 3\mathrm{i})(5 - \mathrm{i}) = b - 17\mathrm{i}\). [5]
| Scheme | Marks | AO |
|---|---|---|
| \((a - 3\mathrm{i})(5 - \mathrm{i}) = 5a - 3 - 15\mathrm{i} - a\mathrm{i}\) | M1 | 1.1a |
| M1 | 1.1 | |
| \(\Rightarrow 5a - 3 = b\) | A1 | 3.1a |
| \(15 + a = 17\) | A1 | 1.1 |
| \(\Rightarrow a = 2,\ b = 7\) | A1 | 1.1 |
| [5] |
Notes
M1: (1st) expanding and \(\mathrm{i}^2 = -1\)
allow 1 sign error (e.g. +3)
M1: (2nd) equating Re and Im parts
A1: (1st) soi
A1: (2nd) o.e. eg \(15\mathrm{i} + a\mathrm{i} = 17\mathrm{i}\)
A1: (3rd) both correct