| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths question combining two routine techniques: (a) applying de Moivre's theorem to derive a trigonometric identity (mechanical expansion and algebraic manipulation), and (b) finding nth roots and basic Argand diagram work. While it requires multiple steps and careful algebra, both parts follow well-established procedures taught in FP2 with no novel insight required. The difficulty is elevated above average A-level due to the Further Maths content and algebraic complexity, but remains a textbook-style exercise. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cos 5\theta + j\sin 5\theta = (\cos\theta+j\sin\theta)^5 = c^5+5c^4js-10c^3s^2-10c^2js^3+5cs^4+js^5\) | M1 | Expanding |
| M1 | Separating real and imaginary parts; dependent on first M1 | |
| \(\cos 5\theta = c^5-10c^3s^2+5cs^4\) | A1 | Alternative: \(16c^5-20c^3+5c\) |
| \(\sin 5\theta = 5c^4s-10c^2s^3+s^5\) | A1 | Alternative: \(16s^5-20s^3+5s\) |
| \(\tan 5\theta = \frac{5c^4s-10c^2s^3+s^5}{c^5-10c^3s^2+5cs^4}\) | M1 | Using \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) and simplifying |
| \(= \frac{t(t^4-10t^2+5)}{5t^4-10t^2+1}\) | A1 (ag) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\arg(-4\sqrt{2})=\pi \Rightarrow\) fifth roots have \(r=\sqrt{2}\) | B1 | |
| and \(\theta = \frac{\pi}{5}\) | B1 | No credit for arguments in degrees |
| \(z = \sqrt{2}e^{\frac{1}{5}j\pi}, \sqrt{2}e^{\frac{3}{5}j\pi}, \sqrt{2}e^{j\pi}, \sqrt{2}e^{\frac{7}{5}j\pi}, \sqrt{2}e^{\frac{9}{5}j\pi}\) | M1, A1 | Adding (or subtracting) \(\frac{2\pi}{5}\); all correct, allow \(-\pi\leq\theta<\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Points at vertices of "regular" pentagon, with one on negative real axis | G1 | |
| Points correctly labelled | G1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\arg(w) = \frac{1}{2}\left(\frac{\pi}{5}+\frac{3\pi}{5}\right) = \frac{2\pi}{5}\) | B1 | |
| \( | w | = \sqrt{2}\cos\frac{\pi}{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(w = \sqrt{2}\cos\frac{\pi}{5}e^{\frac{2}{5}\pi j} \Rightarrow w^n = \left(\sqrt{2}\cos\frac{\pi}{5}\right)^n e^{\frac{2}{5}\pi nj}\) | ||
| which is real if \(\sin\frac{2\pi n}{5}=0 \Rightarrow n=5\) | B1 | |
| \( | w^5 | = \left(\sqrt{2}\cos\frac{\pi}{5}\right)^5\) |
| \(\Rightarrow a = 2^{\frac{5}{2}}\cos^5\frac{\pi}{5}\) | A1 | Accept 1.96 or better |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos 5\theta + j\sin 5\theta = (\cos\theta+j\sin\theta)^5 = c^5+5c^4js-10c^3s^2-10c^2js^3+5cs^4+js^5$ | M1 | Expanding |
| | M1 | Separating real and imaginary parts; dependent on first M1 |
| $\cos 5\theta = c^5-10c^3s^2+5cs^4$ | A1 | Alternative: $16c^5-20c^3+5c$ |
| $\sin 5\theta = 5c^4s-10c^2s^3+s^5$ | A1 | Alternative: $16s^5-20s^3+5s$ |
| $\tan 5\theta = \frac{5c^4s-10c^2s^3+s^5}{c^5-10c^3s^2+5cs^4}$ | M1 | Using $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and simplifying |
| $= \frac{t(t^4-10t^2+5)}{5t^4-10t^2+1}$ | A1 (ag) | |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\arg(-4\sqrt{2})=\pi \Rightarrow$ fifth roots have $r=\sqrt{2}$ | B1 | |
| and $\theta = \frac{\pi}{5}$ | B1 | No credit for arguments in degrees |
| $z = \sqrt{2}e^{\frac{1}{5}j\pi}, \sqrt{2}e^{\frac{3}{5}j\pi}, \sqrt{2}e^{j\pi}, \sqrt{2}e^{\frac{7}{5}j\pi}, \sqrt{2}e^{\frac{9}{5}j\pi}$ | M1, A1 | Adding (or subtracting) $\frac{2\pi}{5}$; all correct, allow $-\pi\leq\theta<\pi$ |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Points at vertices of "regular" pentagon, with one on negative real axis | G1 | |
| Points correctly labelled | G1 | |
## Part (b)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\arg(w) = \frac{1}{2}\left(\frac{\pi}{5}+\frac{3\pi}{5}\right) = \frac{2\pi}{5}$ | B1 | |
| $|w| = \sqrt{2}\cos\frac{\pi}{5}$ | M1, A1ft | Attempting to find length; f.t. (positive) $r$ from (i) |
## Part (b)(iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $w = \sqrt{2}\cos\frac{\pi}{5}e^{\frac{2}{5}\pi j} \Rightarrow w^n = \left(\sqrt{2}\cos\frac{\pi}{5}\right)^n e^{\frac{2}{5}\pi nj}$ | | |
| which is real if $\sin\frac{2\pi n}{5}=0 \Rightarrow n=5$ | B1 | |
| $|w^5| = \left(\sqrt{2}\cos\frac{\pi}{5}\right)^5$ | M1 | Attempting the $n$th power of his modulus in (iii), or attempting the modulus of the $n$th power |
| $\Rightarrow a = 2^{\frac{5}{2}}\cos^5\frac{\pi}{5}$ | A1 | Accept 1.96 or better |
---2
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to find expressions for $\sin 5 \theta$ and $\cos 5 \theta$ in terms of $\sin \theta$ and $\cos \theta$.\\
Hence show that, if $t = \tan \theta$, then
$$\tan 5 \theta = \frac { t \left( t ^ { 4 } - 10 t ^ { 2 } + 5 \right) } { 5 t ^ { 4 } - 10 t ^ { 2 } + 1 }$$
\item \begin{enumerate}[label=(\roman*)]
\item Find the 5th roots of $- 4 \sqrt { 2 }$ in the form $r \mathrm { e } ^ { \mathrm { j } \theta }$, where $r > 0$ and $0 \leqslant \theta < 2 \pi$.
These 5th roots are represented in the Argand diagram, in order of increasing $\theta$, by the points A , $\mathrm { B } , \mathrm { C } , \mathrm { D } , \mathrm { E }$.
\item Draw the Argand diagram, making clear which point is which.
The mid-point of AB is the point P which represents the complex number $w$.
\item Find, in exact form, the modulus and argument of $w$.
\item $w$ is an $n$th root of a real number $a$, where $n$ is a positive integer. State the least possible value of $n$ and find the corresponding value of $a$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2011 Q2 [18]}}