| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Complex number arithmetic and simplification |
| Difficulty | Standard +0.8 This is a Further Maths question requiring manipulation of complex exponentials and factorization using roots of unity. Part (i) is straightforward expansion using Euler's formula, but part (ii) requires recognizing the pattern, finding appropriate θ and n values, then computing four sixth roots of unity—a multi-step problem requiring insight beyond routine techniques. |
| Spec | 4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| \((z^n - e^{i\theta})(z^n - e^{-i\theta}) = z^{2n} - 2z^n\left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right) + 1 \equiv z^{2n} - (2\cos\theta)z^n + 1\) | B1 | For multiplying out to AG with evidence of \(\cos\theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})\) (Can be implied by \(2\cos\theta = (e^{i\theta} + e^{-i\theta})\)) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1 \(2\cos\theta = 1 \Rightarrow \theta = \frac{1}{3}\pi\) | M1 | For using (i) to find \(\theta\) |
| \(\Rightarrow z^4 - z^2 + 1 = (z^2 - e^{i\pi/3})(z^2 - e^{-i\pi/3})\) | A1 | For correct quadratic factors (Or \(\frac{5\pi}{3}i\) in place of \(-\frac{\pi}{3}i\)) |
| \(= (z + e^{i\pi/6})(z - e^{i\pi/6})(z + e^{-i\pi/6})(z - e^{-i\pi/6})\) | M1 | For factorising \((z^2 - a^2)\) |
| \(= (z - e^{i\pi/6})(z - e^{i2\pi/3})(z - e^{i\pi/6})(z - e^{i\pi/3})\) | A1 | For correct linear factors |
| M1 | For adjusting arguments (must attempt correct range and "\((z - \) root\()\)") | |
| A1 | For correct factors CAO Correct answer www gets 6 | |
| [6] | ||
| METHOD 2 | ||
| \(z^4 - z^2 + 1 = 0 \Rightarrow z^2 = \frac{1 \pm \sqrt{3}i}{2} = e^{\pm i\pi/3}, e^{-i\pi}\) | M1 A1 M1 A1 | For solving quadratic For correct roots in exp form For attempt to find 4 roots For correct roots \(\pm e^{i\pi/6}, \pm e^{-i\pi/6}\) |
| \(\Rightarrow (z - e^{i\pi/6})(z - e^{i2\pi/3})(z - e^{i2\pi/3})(z - e^{i11\pi/6})\) | M1 A1 | For adjusting arguments For correct factors CAO |
| [6] |
**(i)**
| $(z^n - e^{i\theta})(z^n - e^{-i\theta}) = z^{2n} - 2z^n\left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right) + 1 \equiv z^{2n} - (2\cos\theta)z^n + 1$ | B1 | For multiplying out to AG with evidence of $\cos\theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$ (Can be implied by $2\cos\theta = (e^{i\theta} + e^{-i\theta})$) |
| | [1] | |
**(ii)**
| METHOD 1 $2\cos\theta = 1 \Rightarrow \theta = \frac{1}{3}\pi$ | M1 | For using (i) to find $\theta$ |
| $\Rightarrow z^4 - z^2 + 1 = (z^2 - e^{i\pi/3})(z^2 - e^{-i\pi/3})$ | A1 | For correct quadratic factors (Or $\frac{5\pi}{3}i$ in place of $-\frac{\pi}{3}i$) |
| $= (z + e^{i\pi/6})(z - e^{i\pi/6})(z + e^{-i\pi/6})(z - e^{-i\pi/6})$ | M1 | For factorising $(z^2 - a^2)$ |
| $= (z - e^{i\pi/6})(z - e^{i2\pi/3})(z - e^{i\pi/6})(z - e^{i\pi/3})$ | A1 | For correct linear factors |
| | M1 | For adjusting arguments (must attempt correct range and "$(z - $ root$)$") |
| | A1 | For correct factors CAO Correct answer www gets 6 |
| | [6] | |
| METHOD 2 | | |
| $z^4 - z^2 + 1 = 0 \Rightarrow z^2 = \frac{1 \pm \sqrt{3}i}{2} = e^{\pm i\pi/3}, e^{-i\pi}$ | M1 A1 M1 A1 | For solving quadratic For correct roots in exp form For attempt to find 4 roots For correct roots $\pm e^{i\pi/6}, \pm e^{-i\pi/6}$ |
| $\Rightarrow (z - e^{i\pi/6})(z - e^{i2\pi/3})(z - e^{i2\pi/3})(z - e^{i11\pi/6})$ | M1 A1 | For adjusting arguments For correct factors CAO |
| | [6] | |2 (i) Show that $\left( z ^ { n } - \mathrm { e } ^ { \mathrm { i } \theta } \right) \left( z ^ { n } - \mathrm { e } ^ { - \mathrm { i } \theta } \right) \equiv z ^ { 2 n } - ( 2 \cos \theta ) z ^ { n } + 1$.\\
(ii) Express $z ^ { 4 } - z ^ { 2 } + 1$ as the product of four factors of the form $\left( z - e ^ { \mathrm { i } \alpha } \right)$, where $0 \leqslant \alpha < 2 \pi$.
\hfill \mbox{\textit{OCR FP3 2012 Q2 [7]}}