FP1 June 2011 Q6
6. Given that \(z = x + \mathrm{i}y\), find the value of \(x\) and the value of \(y\) such that \[z + 3\mathrm{i}z^* = -1 + 13\mathrm{i}\] where \(z^*\) is the complex conjugate of \(z\). (7)
| Scheme | Marks |
|---|---|
| \(z + 3\mathrm{i}z^* = -1 + 13\mathrm{i}\) | |
| \((x + \mathrm{i}y) + 3\mathrm{i}(x - \mathrm{i}y)\) \(z^* = x - \mathrm{i}y\) Substituting \(z = x + \mathrm{i}y\) and their \(z^*\) into \(z + 3\mathrm{i}z^*\) | B1 M1 |
| \(x + \mathrm{i}y + 3\mathrm{i}x + 3y = -1 + 13\mathrm{i}\) Correct equation in \(x\) and \(y\) with \(\mathrm{i}^2 = -1\). Can be implied. | A1 |
| \((x + 3y) + \mathrm{i}(y + 3x) = -1 + 13\mathrm{i}\) | |
| Re part: \(\quad x + 3y = -1\) Im part: \(\quad y + 3x = 13\) An attempt to equate real and imaginary parts. Correct equations. | M1 A1 |
| \(3x + 9y = -3\) \(3x + y = 13\) | |
| \(8y = -16 \Rightarrow y = -2\) Attempt to solve simultaneous equations to find one of \(x\) or \(y\). At least one of the equations must contain both \(x\) and \(y\) terms. | M1 |
| \(x + 3y = -1 \Rightarrow x - 6 = -1 \Rightarrow x = 5\) Both \(x = 5\) and \(y = -2\). | A1 |
| \(\{z = 5 - 2\mathrm{i}\}\) | |
| (7) | |
| (7 marks) |