A2 June 2019 Paper 1 Q4
4 Solve the equation \(2z - 5\mathrm{i}z^* = 12\) [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses correct conjugate of \(z\) and expresses equation in terms of \(x\) and \(y\) where \(x\) and \(y\) are real | M1 | 1.1a |
| Equates real and imaginary parts of their equation (conjugate might be wrong). | M1 | 1.1a |
| Solves their equations correctly for \(x\) and \(y\) having used the correct conjugate of \(z\) | M1 | 1.1a |
| States a fully correct solution, must be \(z = \ldots\) | A1 | 1.1b |
| (4 marks) |
Typical solution
\[z = x + \mathrm{i}y\]\[2(x + \mathrm{i}y) - 5\mathrm{i}(x - \mathrm{i}y) = 12\]Re: \(2x - 5y = 12\)
Im: \(2y - 5x = 0 \Rightarrow y = 2.5x\)
\[2x - 12.5x = 12\]\[x = -\frac{8}{7} \text{ and } y = -\frac{20}{7}\]\[z = -\frac{8}{7} - \frac{20}{7}\mathrm{i}\]