A2 June 2019 Paper 1 Q9
9
(a) Solve the equation \(z^3 = \sqrt{2} - \sqrt{6}\mathrm{i}\), giving your answers in the form \(r\mathrm{e}^{\mathrm{i}\theta}\) where \(r \gt 0\) and \(0 \leqslant \theta \lt 2\pi\) [5 marks]
(b) The transformation represented by the matrix \(\mathbf{M} = \begin{bmatrix} 5 & 1 \\ 1 & 3 \end{bmatrix}\) acts on the points on an Argand Diagram which represent the roots of the equation in part (a).
Find the exact area of the shape formed by joining the transformed points. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes complex number in Eulerian form or equivalent. PI correct \(r\) & \(\theta\) | B1 | 1.1b |
| Obtains \(r\) by taking cube root of their modulus of \(z^3\), accept AWRT 1.41 or \(\left(2\sqrt{2}\right)^{\frac{1}{3}}\) OE | B1F | 1.1b |
| Divides their argument by 3 | M1 | 1.1a |
| Finds three correct angles \(\theta = \frac{5\pi}{9},\ \frac{11\pi}{9}\left(\text{or } \frac{-7\pi}{9}\right),\ \frac{17\pi}{9}\left(\text{or } \frac{-\pi}{9}\right)\) | A1 | 2.2a |
| Finds fully correct solution, accept decimal equivalents & \(\left(2\sqrt{2}\right)^{\frac{1}{3}}\) OE Accept \(\theta = \frac{5\pi}{9},\ \frac{11\pi}{9}\left(\text{or } \frac{-7\pi}{9}\right),\ \frac{17\pi}{9}\left(\text{or } \frac{-\pi}{9}\right)\) | A1 | 2.2a |
Typical solution
\[z^3 = 2\sqrt{2}\mathrm{e}^{\frac{-\pi\mathrm{i}}{3}}\]\[r = \sqrt{2}\]\[\theta = \frac{-\pi}{9}\]\[\theta = \frac{5\pi}{9},\ \frac{11\pi}{9},\ \frac{17\pi}{9}\]\[z = \sqrt{2}\mathrm{e}^{\frac{5\pi\mathrm{i}}{9}},\ \sqrt{2}\mathrm{e}^{\frac{11\pi\mathrm{i}}{9}},\ \sqrt{2}\mathrm{e}^{\frac{17\pi\mathrm{i}}{9}}\]| Scheme | Marks | AO |
|---|---|---|
| Finds the area of their triangle from part (a) - ft their \(r\) only Or Applies Matrix \(\mathbf{M}\) to the three points | M1 | 3.1a |
| Finds correct area of original triangle Or Finds three correct new points | A1 | 1.1b |
| Finds correct \(|\mathbf{M}|\) and uses as area scale factor with their area of original triangle Or Works out area of new triangle | M1 | 2.2a |
| Finds correct answer from correct reasoning in exact form only | A1 | 1.1b |
| (9 marks) |