AS October 2020 Paper 1 Q2
2. Given that
\[\begin{aligned}z_1 &= 2 + 3\mathrm{i}\\ |z_1z_2| &= 39\sqrt{2}\\ \arg(z_1z_2) &= \frac{\pi}{4}\end{aligned}\]where \(z_1\) and \(z_2\) are complex numbers,
Give the exact value of \(r\) and give the value of \(\theta\) in radians to 4 significant figures. (2)
| Scheme | Marks | AO |
|---|---|---|
| \(|z_1| = \sqrt{13}\) and \(\arg z_1 = \tan^{-1}\left(\dfrac{3}{2}\right)\) | B1 | 1.1b |
| \(z_1 = \sqrt{13}(\cos 0.9828 + \mathrm{i}\sin 0.9828)\) | B1ft | 1.1b |
| (2) |
Notes
B1: Correct exact value for \(|z_1| = \sqrt{13}\) and \(\arg z_1 = \tan^{-1}\left(\dfrac{3}{2}\right)\). The value for \(\arg z_1\) can be implied by sight of awrt 0.98 or awrt 56.3°
B1ft: Follow through on \(r = |z_1|\) and \(\theta = \arg z_1\) and writes \(z_1 = r(\cos\theta + \mathrm{i}\sin\theta)\) where \(r\) is exact and \(\theta\) is correct to 4 s.f. do not follow through on rounding errors.
| Scheme | Marks | AO |
|---|---|---|
| A complete method to find the modulus of \(z_2\) e.g. \(|z_1| = \sqrt{13}\) and uses \(|z_1z_2| = |z_1| \times |z_2| = 39\sqrt{2} \Rightarrow |z_2| = 3\sqrt{26}\) or \(\sqrt{234}\) | M1 A1 | 3.1a 1.1b |
| A complete method to find the argument of \(z_2\) e.g. \(\arg(z_1z_2) = \arg(z_1) + \arg(z_2) = \dfrac{\pi}{4} \Rightarrow \arg(z_2) = \ldots\) \(\arg(z_2) = \dfrac{\pi}{4} - \tan^{-1}\left(\dfrac{3}{2}\right)\) or \(\dfrac{\pi}{4} - 0.9828\) or \(-0.1974\ldots\) | M1 A1 | 3.1a 1.1b |
| \(z_2 = 3\sqrt{26}\left(\cos(\text{‘}{-0.1974\ldots}\text{’}) + \mathrm{i}\sin(\text{‘}{-0.1974\ldots}\text{’})\right)\) or \(z_2 = a + b\mathrm{i} \Rightarrow a^2 + b^2 = 234\) and \(\tan(-0.1974) = \dfrac{b}{a} \Rightarrow \dfrac{b}{a} = -0.2\) \(\Rightarrow a = \ldots\) and \(b = \ldots\) | ddM1 | 1.1b |
| Deduces that \(z_2 = 15 - 3\mathrm{i}\) only | A1 | 2.2a |
| (6) | ||
| (8 marks) |
Notes
(Corrected from the printed mark scheme: the printed scheme has \(\tan^{-1}(-0.1974) = \dfrac{b}{a}\); it should be \(\tan(-0.1974) = \dfrac{b}{a}\), which gives \(\dfrac{b}{a} = -0.2\).)
M1: A complete method to find the modulus of \(z_2\)
A1: \(|z_2| = 3\sqrt{26}\)
M1: A complete method to find the argument of \(z_2\)
A1: \(\arg(z_2) = \dfrac{\pi}{4} - \tan^{-1}\left(\dfrac{3}{2}\right)\) or \(\dfrac{\pi}{4} - 0.9828\) or \(-0.1974\ldots\)
ddM1: Writes \(z_2\) in the form \(r(\cos\theta + \mathrm{i}\sin\theta)\), dependent on both previous M marks.
Alternative forms two equations involving \(a\) and \(b\) using the modulus and argument of \(z_2\) and solve to find values for \(a\) and \(b\)
A1: Deduces that \(z_2 = 15 - 3\mathrm{i}\) only
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(z_1z_2 = (a + b\mathrm{i})(2 + 3\mathrm{i}) = (2a - 3b) + (3a + 2b)\mathrm{i}\) | ||
| \((2a - 3b)^2 + (3a + 2b)^2 = \left(39\sqrt{2}\right)^2\) or 3042 \(\Rightarrow a^2 + b^2 = 234\) or \(|z_1z_2| = |z_1| \times |z_2| = 39\sqrt{2} \Rightarrow |z_2| = 3\sqrt{26}\) or \(\sqrt{234}\) \(\Rightarrow a^2 + b^2 = 234\) | M1 A1 | 3.1a 1.1b |
| \(\arg\left[(2a - 3b) + (3a + 2b)\mathrm{i}\right] = \dfrac{\pi}{4} \Rightarrow \tan^{-1}\left(\dfrac{3a + 2b}{2a - 3b}\right) = \dfrac{\pi}{4} \Rightarrow \dfrac{3a + 2b}{2a - 3b} = 1\) \(\Rightarrow a = -5b\) | M1 A1 | 3.1a 1.1b |
| Solves \(a = -5b\) and \(a^2 + b^2 = 234\) to find values for \(a\) and \(b\) | ddM1 | 1.1b |
| Deduces that \(z_2 = 15 - 3\mathrm{i}\) only | A1 | 2.2a |
| (6) |
(b) Alternative: \(z_1z_2 = (a + b\mathrm{i})(2 + 3\mathrm{i}) = (2a - 3b) + (3a + 2b)\mathrm{i}\)
M1: A complete method to find an equation involving \(a\) and \(b\) using the modulus
A1: Correct simplified equation \(a^2 + b^2 = 234\) o.e.
M1: A complete method to find an equation involving \(a\) and \(b\) using the argument.
Note \(\tan^{-1}\left(\dfrac{2a - 3b}{3a + 2b}\right) = \dfrac{\pi}{4}\) this would score M0 A0 ddM0 A0
A1: Correct simplified equation \(a = -5b\) o.e.
ddM1: Dependent on both the previous method marks. Solves their equations to find values for \(a\) and \(b\)
A1: Deduces that \(z_2 = 15 - 3\mathrm{i}\) only