A2 June 2019 Paper 1 Q10
10 In this question you must show detailed reasoning.
(a) You are given that \(-1 + \mathrm{i}\) is a root of the equation \(z^3 = a + b\mathrm{i}\), where \(a\) and \(b\) are real numbers. Find \(a\) and \(b\). [3]
(b) Find all the roots of the equation in part (a), giving your answers in the form \(r\mathrm{e}^{\mathrm{i}\theta}\), where \(r\) and \(\theta\) are exact. [4]
(c) Chris says “the complex roots of a polynomial equation come in complex conjugate pairs”. Explain why this does not apply to the polynomial equation in part (a). [1]
| Scheme | Marks | AO |
|---|---|---|
| DR \((-1 + \mathrm{i})^3 = (-1)^3 + 3(-1)^2\mathrm{i} + 3(-1)\mathrm{i}^2 + \mathrm{i}^3\) | M1 A1 | 3.1a 1.1b |
| \(= 2 + 2\mathrm{i}\) [so \(a = 2\) and \(b = 2\)] | A1 | 2.2a |
| [3] |
Notes
M1: expanding \((-1 + \mathrm{i})^3\)
A1: correct expression; or \(-2\mathrm{i}(-1 + \mathrm{i})\)
Alternative solution
| Scheme | Marks |
|---|---|
| \(-1 + \mathrm{i} = \sqrt{2}(\cos 3\pi/4 + \mathrm{i}\sin 3\pi/4)\) | B1 |
| \((-1 + \mathrm{i})^3 = 2\sqrt{2}(\cos 9\pi/4 + \mathrm{i}\sin 9\pi/4)\) | M1 |
| \(= 2\sqrt{2}(1/\sqrt{2} + \mathrm{i}/\sqrt{2}) = 2 + 2\mathrm{i}\) | A1 |
| [3] |
B1: \(|z| = \sqrt{2}\), \(\arg z = 3\pi/4\); or \(\sqrt{2}\,\mathrm{e}^{3\mathrm{i}\pi/4}\)
M1: \(|z^3| = |z|^3\), \(\arg(z^3) = 3\arg z\)
(corrected from the printed mark scheme: the last line is printed as \(2\sqrt{2}(1/\sqrt{2} + 1/\sqrt{2})\), missing the \(\mathrm{i}\).)
| Scheme | Marks | AO |
|---|---|---|
| \(z = \sqrt{2}\mathrm{e}^{\mathrm{i}\pi/12},\ \sqrt{2}\mathrm{e}^{3\mathrm{i}\pi/4},\ \sqrt{2}\mathrm{e}^{-7\mathrm{i}\pi/12}\) | B1 B1B1B1 | 2.5 1.1b |
| [4] |
Notes
B1: 3 roots with modulus \(\sqrt{2}\) oe; oe eg \(8^{1/6}\)
B1B1B1: oe, e.g. \(8^{1/6}\mathrm{e}^{17\mathrm{i}\pi/12}\), etc; \(0 \lt \arg \lt 2\pi\) or \(-\pi \lt \arg \lt \pi\)
| Scheme | Marks | AO |
|---|---|---|
| The statement only applies to polynomial equations with real coefficients. | B1 | 2.3 |
| [1] |