AS October 2021 Paper 1 Q9
9
(a) On a single Argand diagram, sketch the loci defined by
- \(\arg(z - 2) = \tfrac{3}{4}\pi\),
- \(|z| = |z + 2 - \mathrm{i}|\). [4]
(b) In this question you must show detailed reasoning.
The point of intersection of the two loci in part (a) represents the complex number \(w\).
Find \(w\), giving your answer in exact form. [5]
The point of intersection of the two loci in part (a) represents the complex number \(w\).
Find \(w\), giving your answer in exact form. [5]
| Scheme | Marks | AO |
|---|---|---|
| 1st locus: line at \(135^\circ\) to +ve real axis | M1 | 1.1 |
| half line only starting at 2 | A1 | 1.1 |
| 2nd locus: perp bisector of OP | M1 | 1.1 |
| where P represents \(-2 + \mathrm{i}\) | A1 | 1.1 |
| [4] |
Notes
M1: (2nd) Allow M1 for errors in P
| Scheme | Marks | AO |
|---|---|---|
| DR 1st locus is line \(y = 2 - x\) | B1 | 3.1a |
| Midpoint of OP is \(\left(-1, \tfrac{1}{2}\right)\) Gradient of perpendicular bisector is 2 | B1FT | 1.1 |
| Equation is \(y - \tfrac{1}{2} = 2(x + 1)\) \(\Rightarrow y = 2x + 2\tfrac{1}{2}\) | B1FT | 3.1a |
| Hence \(2 - x = 2x + 2\tfrac{1}{2}\) \(\Rightarrow x = -\tfrac{1}{6},\ y = \tfrac{13}{6}\) | M1 | 1.1 |
| \(\Rightarrow w = -\tfrac{1}{6} + \tfrac{13}{6}\mathrm{i}\) | A1 | 3.2a |
| [5] |
Notes
B1: oe
B1FT: (1st) midpoint and gradient ft their \((-2, 1)\)
B1FT: (2nd) oe
M1: Solving simultaneously