A2 October 2020 Paper 1 Q6
6 The complex number \(z\) satisfies the equation \(z^2 - 4\mathrm{i}z^* + 11 = 0\).
Given that \(\operatorname{Re}(z) \gt 0\), find \(z\) in the form \(a + b\mathrm{i}\), where \(a\) and \(b\) are real numbers. [4]
| Scheme | Marks | AO |
|---|---|---|
| \(a^2 - b^2 + 2ab\mathrm{i} - 4\mathrm{i}(a - \mathrm{i}b) + 11 = 0\) | B1 | 3.1a |
| \(\Rightarrow a^2 - b^2 - 4b + 11 = 0,\ 2ab - 4a = 0\) | M1 | 1.1 |
| \(\Rightarrow b = 2\) | B1 | 1.1 |
| \(a^2 = 1,\ a = 1\) so \(z = 1 + 2\mathrm{i}\) | A1 | 3.2a |
| [4] |
Notes
B1: substitution for \(z\) and \(z^*\) soi
M1: put Re and Im parts equal to 0