AS June 2024 Paper 1 Q8
8 In an Argand diagram, the point P representing the complex number \(w\) lies on the locus defined by \(\left\{z : \arg(z - 7) = \tfrac{3}{4}\pi\right\}\). You are given that \(\mathrm{Re}(w) = 1\).
The point P also lies on the locus defined by \(\{z : |z + 3 - 9\mathrm{i}| = k\}\), where \(k\) is a constant.
| Scheme | Marks | AO |
|---|---|---|
| line is \(y = 7 - x\) when \(x = 1\), \(y = 6\) | M1 | 3.1a |
| so \(w = 1 + 6\mathrm{i}\) | A1 | 3.2a |
| [2] |
Notes
M1: oe, e.g. \(w = 1 + b\mathrm{i}\), \(\arg(-6 + b\mathrm{i}) = \dfrac{3\pi}{4} \Rightarrow \dfrac{b}{-6} = -1\)
| Scheme | Marks | AO |
|---|---|---|
| \(|1 + 6\mathrm{i} + 3 - 9\mathrm{i}| = k \Rightarrow k^2 = 4^2 + (-3)^2\) | M1 | 1.1 |
| \(\Rightarrow k = 5\) | A1 | 1.1 |
| Circle equation is \((x + 3)^2 + (y - 9)^2 = k^2\) | M1 | 3.1a |
| \((x + 3)^2 + (7 - x - 9)^2 = 25\) | M1 | 1.1 |
| \(\Rightarrow x =\) [1 and] \(-6\) | A1 | 1.1 |
| when \(x = -6\), \(y = 13\) | A1 | 1.1 |
| other complex number is \(-6 + 13\mathrm{i}\) | A1 | 3.2a |
| [7] |
Notes
M1: (1st) finding \(k\) using modulus or substituting their \(x = 1\), \(y = 6\) into circle equation (see below)
M1: (2nd) oe condone sign errors or \(k\) for \(k^2\) (but not both)
M1: (3rd) solving simultaneously with \(y = 7 - x\)
A1: (2nd) or (eliminating \(x\)) \(y =\) [6 and] 13
A1: (3rd) or when \(y = 13\), \(x = -6\)