AS June 2022 Paper 1 Q3
3 The complex number \(z\) satisfies the equation \(5(z - \mathrm{i}) = (-1 + 2\mathrm{i})z^*\).
Determine \(z\), giving your answer in the form \(a + b\mathrm{i}\), where \(a\) and \(b\) are real. [5]
| Scheme | Marks | AO |
|---|---|---|
| \(5[a + (b - 1)\mathrm{i}] = (-1 + 2\mathrm{i})(a - b\mathrm{i})\) | M1 | 1.2 |
| \(= -a + 2\mathrm{i}a + b\mathrm{i} - 2b\mathrm{i}^2\) | A1 | 1.1 |
| \(\Rightarrow 5a = -a + 2b,\ 5b - 5 = 2a + b\) | M1 | 1.1 |
| \(\Rightarrow b = 3a,\ 12a - 5 = 2a\) [so \(a = \tfrac{1}{2},\ b = \tfrac{3}{2}\)] | M1 | 1.1 |
| \(\Rightarrow z = \dfrac{1}{2} + \dfrac{3}{2}\mathrm{i}\) | A1cao | 1.1 |
| [5] |
Notes
M1: (1st) \(z^* = a - b\mathrm{i}\)
A1: expanding brackets correctly
M1: (2nd) equating Re and Im parts
M1: (3rd) solving simultaneous eqns by elimination or substitution
Alternative solution
| Scheme | Marks |
|---|---|
| \(z^* = \dfrac{5(z - \mathrm{i})}{-1 + 2\mathrm{i}} = \dfrac{5(z - \mathrm{i})(-1 - 2\mathrm{i})}{(-1 + 2\mathrm{i})(-1 - 2\mathrm{i})}\) | M1 |
| \(= \dfrac{5(-z + \mathrm{i} - 2z\mathrm{i} + 2\mathrm{i}^2)}{5}\) | A1 |
| \(\Rightarrow z^* = -z - 2 + \mathrm{i} - 2z\mathrm{i}\) \(\Rightarrow a - b\mathrm{i} = -a - b\mathrm{i} - 2 + \mathrm{i} - 2a\mathrm{i} + 2b\) | M1 |
| \(\Rightarrow a = -a + 2b - 2,\ -b = -b + 1 - 2a\) | M1 |
| \(\Rightarrow a = \tfrac{1}{2},\ b = \tfrac{3}{2}\) \(\Rightarrow z = \dfrac{1}{2} + \dfrac{3}{2}\mathrm{i}\) | A1cao |
M1: (1st) realising denominator
M1: (2nd) \(z^* = a - b\mathrm{i}\)
M1: (3rd) equating Re and Im parts