AS June 2025 Paper 1 Q5
5 In this question you must show detailed reasoning.
The complex number \(w\) is given by \(w = -4\sqrt{2} + \left(4\sqrt{2}\right)\mathrm{i}\).
The complex numbers \(z_1\) and \(z_2\) are given by \(z_1 = a + \mathrm{i}\) and \(z_2 = 4(\cos\theta + \mathrm{i}\sin\theta)\), where \(a\) is a positive real constant and \(-\pi \lt \theta \leqslant \pi\).
| Scheme | Marks | AO |
|---|---|---|
| (i) DR \(|w| = \sqrt{(-4\sqrt{2})^2 + (4\sqrt{2})^2}\) | M1 | 1.1 |
| \(= 8\) | A1 | 1.1 |
| [2] | ||
| (ii) DR \([\arg(w) =]\arctan\left(\dfrac{4\sqrt{2}}{-4\sqrt{2}}\right)\) | M1 | 1.1 |
| \(= \dfrac{3\pi}{4}\) [as 2nd quadrant] | A1 | 3.2a |
| [2] |
Notes
(a)(i)
M1: modulus formula
A1: no working scores 0 as DR
(a)(ii)
M1: argument formula (oe, e.g. using triangle)
A1: no working scores 0 as DR
| Scheme | Marks | AO |
|---|---|---|
| (i) DR \(|z_1| = \sqrt{a^2 + 1}\) | B1 | 1.1 |
| so \(4\sqrt{a^2 + 1} = 8\) | M1 | 3.1a |
| \(\Rightarrow a^2 = 3\), \(a = \sqrt{3}\) | A1 | 1.1 |
| [3] | ||
| (ii) DR \(\arg(z_1) = \arctan\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}\) | B1ft | 1.1 |
| so \(\dfrac{\pi}{6} + \theta = \dfrac{3\pi}{4}\) | M1 | 3.1a |
| \(\Rightarrow \theta = \dfrac{7\pi}{12}\) | A1 | 1.1 |
| [3] |
Notes
(b)(i)
M1: \(|z_1z_2| = |z_1||z_2|\) used
(b)(ii)
B1ft: ft their \(a\)
M1: \(\arg(z_1z_2) = \arg z_1 + \arg z_2\) used
A1: must be as exact multiple of \(\pi\).
Alternative method (mark 5(b)(ii) first)
| Scheme | Marks |
|---|---|
| (ii) \(4(a + \mathrm{i})(\cos\theta + \mathrm{i}\sin\theta) = -4\sqrt{2} + 4\sqrt{2}\,\mathrm{i}\) \(\Rightarrow a\cos\theta - \sin\theta = -\sqrt{2},\quad a\sin\theta + \cos\theta = \sqrt{2}\) \(\cos\theta + \sin\theta = \dfrac{\sqrt{2}}{2}\) | B1 |
| \(\sqrt{2}\cos\left(\theta - \dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}\) or \(\sqrt{2}\sin\left(\theta + \dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}\) \(\Rightarrow \cos\left(\theta - \dfrac{\pi}{4}\right) = \dfrac{1}{2}\), \(\theta - \dfrac{\pi}{4} = \dfrac{\pi}{3}\) | M1 |
| \(\Rightarrow \theta = \dfrac{7\pi}{12}\) | A1cao |
| (i) \(a = \dfrac{\sin\theta - \sqrt{2}}{\cos\theta}\) | B1 M1 |
| \(a = \sqrt{3}\) | A1 |
M1: (ii) \(R\cos(\theta - \alpha)\) or \(R\sin(\theta + \alpha)\) method
or \(\sin\left(\theta + \dfrac{\pi}{4}\right) = \dfrac{1}{2}\), \(\theta + \dfrac{\pi}{4} = \dfrac{5\pi}{6}\)
M1: (i) Substituting \(\theta = 7\pi/12\) into either equation
Alternative method (mark 5(b)(ii) first)
| Scheme | Marks |
|---|---|
| (ii) \(z_1 = a + \mathrm{i} = 2\left[\cos\left(\dfrac{3\pi}{4} - \theta\right) + \mathrm{i}\sin\left(\dfrac{3\pi}{4} - \theta\right)\right]\) | B1 |
| \(\Rightarrow \sin\left(\dfrac{3\pi}{4} - \theta\right) = \dfrac{1}{2}\) | M1 |
| \(\dfrac{3\pi}{4} - \theta = \dfrac{\pi}{6} \Rightarrow \theta = \dfrac{7\pi}{12}\) | A1 |
| (i) \(a = 2\cos\left(\dfrac{3\pi}{4} - \theta\right)\) | M1 |
| \(= 2\cos\left(\dfrac{\pi}{6}\right) = 2 \times \dfrac{\sqrt{3}}{2}\) | B1 |
| \(a = \sqrt{3}\) | A1 |
B1: (ii) Ft their \(3\pi/4\)
M1: (ii) Equating Im parts
M1: (i) Equating Re parts (ft their \(3\pi/4\))
B1: (i) \(\cos(\pi/6) = \sqrt{3}/2\)
A1: (i) but must show exact working for final A1
(Corrected from the printed mark scheme: the first line of (i) is printed as \(a = 2\cos\left(\frac{3\pi}{4} - \theta\right) = \frac{1}{2}\); the “\(= \frac{1}{2}\)” is a slip and has been removed.)