FP1 June 2013 Q7
7. \[z_1 = 2 + 3\mathrm{i}, \quad z_2 = 3 + 2\mathrm{i}, \quad z_3 = a + b\mathrm{i}, \quad a, b \in \mathbb{R}\]
(a) Find the exact value of \(|z_1 + z_2|\). (2)
Given that \(w = \dfrac{z_1 z_3}{z_2}\),
(b) find \(w\) in terms of \(a\) and \(b\), giving your answer in the form \(x + \mathrm{i}y,\quad x, y \in \mathbb{R}\) (4)
Given also that \(w = \dfrac{17}{13} - \dfrac{7}{13}\mathrm{i}\),
(c) find the value of \(a\) and the value of \(b\), (3)
(d) find \(\arg w\), giving your answer in radians to 3 decimal places. (2)
| Scheme | Marks |
|---|---|
| \(z_1 = 2 + 3\mathrm{i},\quad z_2 = 3 + 2\mathrm{i}\) | |
| \(z_1 + z_2 = 5 + 5\mathrm{i} \Rightarrow |z_1 + z_2| = \sqrt{5^2 + 5^2}\) Adds \(z_1\) and \(z_2\) and correct use of Pythagoras. i under square root award M0. | M1 |
| \(\sqrt{50}\ (= 5\sqrt{2})\) | A1 cao |
| (2) |
| Scheme | Marks |
|---|---|
| \(\dfrac{z_1 z_3}{z_2} = \dfrac{(2 + 3\mathrm{i})(a + b\mathrm{i})}{3 + 2\mathrm{i}}\) \(= \dfrac{(2 + 3\mathrm{i})(a + b\mathrm{i})(3 - 2\mathrm{i})}{(3 + 2\mathrm{i})(3 - 2\mathrm{i})}\) Substitutes for \(z_1, z_2\) and \(z_3\) and multiplies by \(\dfrac{3 - 2\mathrm{i}}{3 - 2\mathrm{i}}\) | M1 |
| \((3 + 2\mathrm{i})(3 - 2\mathrm{i}) = 13\) 13 seen. | B1 |
| \(\dfrac{z_1 z_3}{z_2} = \dfrac{(12a - 5b) + (5a + 12b)\mathrm{i}}{13}\) M1: Obtains a numerator with 2 real and 2 imaginary parts. A1: As stated or \(\dfrac{(12a - 5b)}{13} + \dfrac{(5a + 12b)}{13}\mathrm{i}\) ONLY. | dM1A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(12a - 5b = 17\) \(5a + 12b = -7\) Compares real and imaginary parts to obtain 2 equations which both involve \(a\) and \(b\). Condone sign errors only. | M1 |
| \(\begin{aligned} 60a - 25b &= 85 \\ 60a + 144b &= -84 \end{aligned} \Rightarrow b = -1\) Solves as far as \(a = \) or \(b = \) | dM1 |
| \(a = 1\), \(b = -1\) Both correct | A1 |
| (3) |
Notes
Correct answers with no working award 3/3.
| Scheme | Marks |
|---|---|
| \(\arg(w) = -\tan^{-1}\left(\dfrac{7}{17}\right)\) Accept use of \(\pm\tan^{-1}\) or \(\pm\tan\). awrt \(\pm 0.391\) or \(\pm 5.89\) implies M1. | M1 |
| \(= \) awrt \(-0.391\) or awrt 5.89 | A1 |
| (2) | |
| Total 11 |