| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Trigonometric method of differences |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear scaffolding through parts (a)(i)-(iii) leading to part (b). While it requires knowledge of complex exponentials and De Moivre's theorem, the steps are heavily guided. The key insight—dividing the quartic by z² to create a symmetric form—is essentially given by the structure. Solving the resulting quadratic in cos θ and finding complex roots is standard FP2 technique. More challenging than typical A-level but routine for Further Maths students who have practiced this method. |
| Spec | 4.02d Exponential form: re^(i*theta)4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(z + \frac{1}{z} = \cos\theta + i\sin\theta + \cos(-\theta) + i\sin(-\theta)\) | M1 | Or \(z + \frac{1}{z} = e^{i\theta} + e^{-i\theta}\) |
| \(= 2\cos\theta\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(z^2 + \frac{1}{z^2} = \cos 2\theta + i\sin 2\theta + \cos(-2\theta) + i\sin(-2\theta)\) | M1 | |
| \(= 2\cos 2\theta\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(z^2 - z + 2 - \frac{1}{z} + \frac{1}{z^2} = 2\cos 2\theta - 2\cos\theta + 2\) | M1 | |
| Use of \(\cos 2\theta = 2\cos^2\theta - 1\) | m1 | |
| \(= 4\cos^2\theta - 2\cos\theta\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(z + \frac{1}{z} = 0 \Rightarrow z = \pm i\) | M1A1 | |
| \(z + \frac{1}{z} = 1 \Rightarrow z^2 - z + 1 = 0\) | M1A1 | |
| \(z = \frac{1 \pm i\sqrt{3}}{2}\) | A1F | Accept solution to (b) if done otherwise |
| Alternative: \(\cos\theta = 0 \Rightarrow \theta = \pm\frac{1}{2}\pi\) | M1 | |
| \(z = \pm i\) | A1 | |
| \(\cos\theta = \frac{1}{2} \Rightarrow \theta = \pm\frac{1}{3}\pi\) | M1 | |
| \(z = e^{\pm\frac{1}{3}\pi i} = \frac{1}{2}(1 \pm i\sqrt{3})\) | A1 A1 | |
| Any 2 correct answers | A1 | |
| One correct answer only | B1 |
# Question 6:
## Part 6(a)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $z + \frac{1}{z} = \cos\theta + i\sin\theta + \cos(-\theta) + i\sin(-\theta)$ | M1 | Or $z + \frac{1}{z} = e^{i\theta} + e^{-i\theta}$ |
| $= 2\cos\theta$ | A1 | AG |
## Part 6(a)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $z^2 + \frac{1}{z^2} = \cos 2\theta + i\sin 2\theta + \cos(-2\theta) + i\sin(-2\theta)$ | M1 | |
| $= 2\cos 2\theta$ | A1 | OE |
## Part 6(a)(iii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $z^2 - z + 2 - \frac{1}{z} + \frac{1}{z^2} = 2\cos 2\theta - 2\cos\theta + 2$ | M1 | |
| Use of $\cos 2\theta = 2\cos^2\theta - 1$ | m1 | |
| $= 4\cos^2\theta - 2\cos\theta$ | A1 | AG |
## Part 6(b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $z + \frac{1}{z} = 0 \Rightarrow z = \pm i$ | M1A1 | |
| $z + \frac{1}{z} = 1 \Rightarrow z^2 - z + 1 = 0$ | M1A1 | |
| $z = \frac{1 \pm i\sqrt{3}}{2}$ | A1F | Accept solution to (b) if done otherwise |
| **Alternative:** $\cos\theta = 0 \Rightarrow \theta = \pm\frac{1}{2}\pi$ | M1 | |
| $z = \pm i$ | A1 | |
| $\cos\theta = \frac{1}{2} \Rightarrow \theta = \pm\frac{1}{3}\pi$ | M1 | |
| $z = e^{\pm\frac{1}{3}\pi i} = \frac{1}{2}(1 \pm i\sqrt{3})$ | A1 A1 | |
| Any 2 correct answers | A1 | |
| One correct answer only | B1 | |
---6 It is given that $z = \mathrm { e } ^ { \mathrm { i } \theta }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that
$$z + \frac { 1 } { z } = 2 \cos \theta$$
(2 marks)
\item Find a similar expression for
$$z ^ { 2 } + \frac { 1 } { z ^ { 2 } }$$
(2 marks)
\item Hence show that
$$z ^ { 2 } - z + 2 - \frac { 1 } { z } + \frac { 1 } { z ^ { 2 } } = 4 \cos ^ { 2 } \theta - 2 \cos \theta$$
(3 marks)
\end{enumerate}\item Hence solve the quartic equation
$$z ^ { 4 } - z ^ { 3 } + 2 z ^ { 2 } - z + 1 = 0$$
giving the roots in the form $a + \mathrm { i } b$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2006 Q6 [12]}}