| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 16 |
| 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) uses De Moivre's theorem to derive a trig identity through algebraic manipulation of binomial expansion—methodical but well-practiced; (b) involves finding nth roots and basic geometric reasoning on an Argand diagram. Both parts require multiple steps and Further Maths content, placing it above average difficulty, but the techniques are textbook applications without requiring novel insight. |
| Spec | 4.02d Exponential form: re^(i*theta)4.02k Argand diagrams: geometric interpretation4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z^n + \frac{1}{z^n} = 2\cos n\theta\), \(z^n - \frac{1}{z^n} = 2j\sin n\theta\) | B1 | Both |
| \(\left(z-\frac{1}{z}\right)^5 = z^5 - 5z^3 + 10z - \frac{10}{z} + \frac{5}{z^3} - \frac{1}{z^5}\) | M1 | Expanding \(\left(z-\frac{1}{z}\right)^5\) |
| \(= z^5 - \frac{1}{z^5} - 5\left(z^3-\frac{1}{z^3}\right) + 10\left(z-\frac{1}{z}\right)\) | M1 | Introducing sines (and possibly cosines) of multiple angles |
| \(\Rightarrow 32j\sin^5\theta = 2j\sin 5\theta - 10j\sin 3\theta + 20j\sin\theta\) | A1 | RHS |
| \(\Rightarrow \sin^5\theta = \frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin\theta\) | A1ft | Division by \(32j\) |
| \(A=\frac{5}{8}\), \(B=-\frac{5}{16}\), \(C=\frac{1}{16}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 4th roots of \(-9j = 9e^{\frac{3}{2}\pi j}\), \(r=\sqrt{3}\) | B1 | Accept \(9^{\frac{1}{4}}\) |
| \(\theta = \frac{3\pi}{8}\) | B1 | |
| \(\theta = \frac{3\pi}{8} + \frac{2k\pi}{4}\) | M1 | Implied by at least two correct (ft) further values |
| \(\theta = \frac{7\pi}{8}, \frac{11\pi}{8}, \frac{15\pi}{8}\) (or equivalent in \([-\pi,\pi]\)) | A1 | Allow arguments in range \(-\pi \leq \theta \leq \pi\) |
| Diagram: points at vertices of a square centre O | M1 | Or 3 correct points (ft) or 1 point in each quadrant |
| Correct diagram | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mid-point of SP has argument \(\frac{\pi}{8}\) | B1 | |
| and modulus \(\sqrt{\frac{3}{2}}\) | B1 | |
| Argument of \(w = 4 \times \frac{\pi}{8} = \frac{\pi}{2}\) | ||
| Modulus \(= \left(\sqrt{\frac{3}{2}}\right)^4 = \frac{9}{4}\) | M1, A1 | Multiplying argument by 4 and modulus raised to power of 4; Both correct |
| \(w\) plotted on imaginary axis above level of P | G1 |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z^n + \frac{1}{z^n} = 2\cos n\theta$, $z^n - \frac{1}{z^n} = 2j\sin n\theta$ | B1 | Both |
| $\left(z-\frac{1}{z}\right)^5 = z^5 - 5z^3 + 10z - \frac{10}{z} + \frac{5}{z^3} - \frac{1}{z^5}$ | M1 | Expanding $\left(z-\frac{1}{z}\right)^5$ |
| $= z^5 - \frac{1}{z^5} - 5\left(z^3-\frac{1}{z^3}\right) + 10\left(z-\frac{1}{z}\right)$ | M1 | Introducing sines (and possibly cosines) of multiple angles |
| $\Rightarrow 32j\sin^5\theta = 2j\sin 5\theta - 10j\sin 3\theta + 20j\sin\theta$ | A1 | RHS |
| $\Rightarrow \sin^5\theta = \frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin\theta$ | A1ft | Division by $32j$ |
| $A=\frac{5}{8}$, $B=-\frac{5}{16}$, $C=\frac{1}{16}$ | | |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 4th roots of $-9j = 9e^{\frac{3}{2}\pi j}$, $r=\sqrt{3}$ | B1 | Accept $9^{\frac{1}{4}}$ |
| $\theta = \frac{3\pi}{8}$ | B1 | |
| $\theta = \frac{3\pi}{8} + \frac{2k\pi}{4}$ | M1 | Implied by at least two correct (ft) further values |
| $\theta = \frac{7\pi}{8}, \frac{11\pi}{8}, \frac{15\pi}{8}$ (or equivalent in $[-\pi,\pi]$) | A1 | Allow arguments in range $-\pi \leq \theta \leq \pi$ |
| Diagram: points at vertices of a square centre O | M1 | Or 3 correct points (ft) or 1 point in each quadrant |
| Correct diagram | A1 | |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mid-point of SP has argument $\frac{\pi}{8}$ | B1 | |
| and modulus $\sqrt{\frac{3}{2}}$ | B1 | |
| Argument of $w = 4 \times \frac{\pi}{8} = \frac{\pi}{2}$ | | |
| Modulus $= \left(\sqrt{\frac{3}{2}}\right)^4 = \frac{9}{4}$ | M1, A1 | Multiplying argument by 4 and modulus raised to power of 4; Both correct |
| $w$ plotted on imaginary axis above level of P | G1 | |
---2
\begin{enumerate}[label=(\alph*)]
\item Given that $z = \cos \theta + \mathrm { j } \sin \theta$, express $z ^ { n } + \frac { 1 } { z ^ { n } }$ and $z ^ { n } - \frac { 1 } { z ^ { n } }$ in simplified trigonometric form.\\
Hence find the constants $A , B , C$ in the identity
$$\sin ^ { 5 } \theta \equiv A \sin \theta + B \sin 3 \theta + C \sin 5 \theta$$
\item \begin{enumerate}[label=(\roman*)]
\item Find the 4th roots of - 9 j in the form $r \mathrm { e } ^ { \mathrm { j } \theta }$, where $r > 0$ and $0 < \theta < 2 \pi$. Illustrate the roots on an Argand diagram.
\item Let the points representing these roots, taken in order of increasing $\theta$, be $\mathrm { P } , \mathrm { Q } , \mathrm { R } , \mathrm { S }$. The mid-points of the sides of PQRS represent the 4th roots of a complex number $w$. Find the modulus and argument of $w$. Mark the point representing $w$ on your Argand diagram.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2010 Q2 [16]}}