| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Standard +0.3 This is a standard Further Maths question testing De Moivre's theorem through routine steps: (a) straightforward induction proof of a well-known result, (b) direct application using z^n and 1/z^n, (c) finding cos θ from the given condition and substituting. All parts follow predictable patterns with no novel insight required, making it slightly easier than average for FP2. |
| Spec | 4.01a Mathematical induction: construct proofs4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \((\cos\theta + i\sin\theta)^{+1}\) | M1 | |
| \(= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)\) | A1 | Any form. Clearly shown |
| Multiply out | A1 | |
| \(= \cos(k+1)\theta + i\sin(k+1)\theta\) | B1 | |
| True for \(n = 1\) shown | E1 | 5 marks |
| (b) \(\frac{1}{z^n} = \frac{1}{\cos n\theta + i\sin n\theta}\) | M1A1 | SC \(\frac{(\cos\theta + i\sin\theta)^{-n}}{\text{quoted as } \cos n\theta - i\sin n\theta}\) earns M1A1 only |
| \(z^n + \frac{1}{z^n} = 2\cos n\theta\) | A1 | 3 marks |
| (c) \(z + \frac{1}{z} = \sqrt{2}\) | M1 | |
| \(2\cos\theta = \sqrt{2}\) | A1 | |
| \(\theta = \frac{\pi}{4}\) | M1 | |
| \(z^{10} + \frac{1}{z^{10}} = 2\cos\left(\frac{10\pi}{4}\right)\) | M1 | |
| \(= 0\) | A1F | 4 marks |
**(a)** $(\cos\theta + i\sin\theta)^{+1}$ | M1 |
$= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)$ | A1 | Any form. Clearly shown
Multiply out | A1 |
$= \cos(k+1)\theta + i\sin(k+1)\theta$ | B1 |
True for $n = 1$ shown | E1 | 5 marks | provided previous 4 marks earned
**(b)** $\frac{1}{z^n} = \frac{1}{\cos n\theta + i\sin n\theta}$ | M1A1 | SC $\frac{(\cos\theta + i\sin\theta)^{-n}}{\text{quoted as } \cos n\theta - i\sin n\theta}$ earns M1A1 only
$z^n + \frac{1}{z^n} = 2\cos n\theta$ | A1 | 3 marks | AG
**(c)** $z + \frac{1}{z} = \sqrt{2}$ | M1 |
$2\cos\theta = \sqrt{2}$ | A1 |
$\theta = \frac{\pi}{4}$ | M1 |
$z^{10} + \frac{1}{z^{10}} = 2\cos\left(\frac{10\pi}{4}\right)$ | M1 |
$= 0$ | A1F | 4 marks | M0 for merely writing $z^{10} + \frac{1}{z^{10}} = 2\cos 10\theta$
**Total: 12 marks**5
\begin{enumerate}[label=(\alph*)]
\item Prove by induction that, if $n$ is a positive integer,
$$( \cos \theta + \mathrm { i } \sin \theta ) ^ { n } = \cos n \theta + \mathrm { i } \sin n \theta$$
\item Hence, given that
$$z = \cos \theta + \mathrm { i } \sin \theta$$
show that
$$z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta$$
\item Given further that $z + \frac { 1 } { z } = \sqrt { 2 }$, find the value of
$$z ^ { 10 } + \frac { 1 } { z ^ { 10 } }$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2009 Q5 [12]}}