| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Sum geometric series with complex terms |
| Difficulty | Challenging +1.2 Part (a) is routine application of de Moivre's theorem for finding roots. Part (b)(i) is straightforward algebraic manipulation. Part (b)(ii) requires recognizing that C and S form the real and imaginary parts of a geometric series with complex terms, then applying the geometric series formula and separating real/imaginary parts—this is a standard Further Maths technique but requires more insight than typical A-level questions. The multi-step nature and need to connect complex exponentials to trigonometric series elevates it slightly above average difficulty. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02r nth roots: of complex numbers4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| 4th roots of \(16i = 16e^{i\frac{\pi}{2}}\) are \(re^{i\theta}\) where \(r=2\), \(\theta = \frac{\pi}{8}\) | B1 | Accept \(16^{\frac{1}{4}}\) |
| \(\theta = \frac{\pi}{8} + \frac{2k\pi}{4}\) | M1 | Implied by at least two correct (ft) further values or stating \(k=-2,-1,(0),1\) |
| \(\theta = -\frac{7}{8}\pi, -\frac{3}{8}\pi, \frac{3}{8}\pi\) | A1 | Points at vertices of a square centre O or 3 correct points (ft) or 1 point in each quadrant |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| \((1-2e^{i\theta})(1-2e^{-i\theta})=1-2e^{i\theta}-2e^{-i\theta}+4 = 5-2(e^{i\theta}+e^{-i\theta}) = 5-4\cos\theta\) | M1, A1 ag | For \(e^{i\theta}e^{-i\theta}=1\) |
| OR \((1-2\cos\theta-2j\sin\theta)(1-2\cos\theta+2j\sin\theta) = (1-2\cos\theta)^2+4\sin^2\theta = 1-4\cos\theta+4(\cos^2\theta+\sin^2\theta) = 5-4\cos\theta\) | M1, A1, A1 ag | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(c+jS = 2e^{i\theta}+4e^{2i\theta}+8e^{3i\theta}+\ldots+2^ne^{ni\theta} = \frac{2e^{i\theta}(1-(2e^{i\theta})^n)}{1-2e^{i\theta}}\) | M1, M1, A1 | Obtaining a geometric series; Summing (M0 for sum to infinity) |
| \(= \frac{2e^{i\theta}(1-2^ne^{ni\theta})(1-2e^{-i\theta})}{(1-2e^{i\theta})(1-2e^{-i\theta})} = \frac{2e^{i\theta}-4-2^{n+1}e^{(n+1)i\theta}+2^{n+2}e^{ni\theta}}{5-4\cos\theta}\) | M1, A2 | Give A1 for two correct terms in numerator |
| \(C = \frac{2\cos\theta-4-2^{n+1}\cos(n+1)\theta+2^{n+2}\cos n\theta}{5-4\cos\theta}\) | M1, A1 ag | Equating real (or imaginary) parts |
| \(S = \frac{2\sin\theta-2^{n+1}\sin(n+1)\theta+2^{n+2}\sin n\theta}{5-4\cos\theta}\) | A1 | |
| 9 |
## 2(a)
| 4th roots of $16i = 16e^{i\frac{\pi}{2}}$ are $re^{i\theta}$ where $r=2$, $\theta = \frac{\pi}{8}$ | B1 | Accept $16^{\frac{1}{4}}$ |
|---|---|---|
| $\theta = \frac{\pi}{8} + \frac{2k\pi}{4}$ | M1 | Implied by at least two correct (ft) further values or stating $k=-2,-1,(0),1$ |
| $\theta = -\frac{7}{8}\pi, -\frac{3}{8}\pi, \frac{3}{8}\pi$ | A1 | Points at vertices of a square centre O or 3 correct points (ft) or 1 point in each quadrant |
| | | 6 |
## 2(b)(i)
| $(1-2e^{i\theta})(1-2e^{-i\theta})=1-2e^{i\theta}-2e^{-i\theta}+4 = 5-2(e^{i\theta}+e^{-i\theta}) = 5-4\cos\theta$ | M1, A1 ag | For $e^{i\theta}e^{-i\theta}=1$ |
|---|---|---|
| OR $(1-2\cos\theta-2j\sin\theta)(1-2\cos\theta+2j\sin\theta) = (1-2\cos\theta)^2+4\sin^2\theta = 1-4\cos\theta+4(\cos^2\theta+\sin^2\theta) = 5-4\cos\theta$ | M1, A1, A1 ag | |
| | 3 | |
## 2(b)(ii)
| $c+jS = 2e^{i\theta}+4e^{2i\theta}+8e^{3i\theta}+\ldots+2^ne^{ni\theta} = \frac{2e^{i\theta}(1-(2e^{i\theta})^n)}{1-2e^{i\theta}}$ | M1, M1, A1 | Obtaining a geometric series; Summing (M0 for sum to infinity) |
|---|---|---|
| $= \frac{2e^{i\theta}(1-2^ne^{ni\theta})(1-2e^{-i\theta})}{(1-2e^{i\theta})(1-2e^{-i\theta})} = \frac{2e^{i\theta}-4-2^{n+1}e^{(n+1)i\theta}+2^{n+2}e^{ni\theta}}{5-4\cos\theta}$ | M1, A2 | Give A1 for two correct terms in numerator |
|---|---|---|
| $C = \frac{2\cos\theta-4-2^{n+1}\cos(n+1)\theta+2^{n+2}\cos n\theta}{5-4\cos\theta}$ | M1, A1 ag | Equating real (or imaginary) parts |
|---|---|---|
| $S = \frac{2\sin\theta-2^{n+1}\sin(n+1)\theta+2^{n+2}\sin n\theta}{5-4\cos\theta}$ | A1 | |
|---|---|---|
| | 9 | |2
\begin{enumerate}[label=(\alph*)]
\item Find the 4th roots of 16j, in the form $r \mathrm { e } ^ { \mathrm { j } \theta }$ where $r > 0$ and $- \pi < \theta \leqslant \pi$. Illustrate the 4th roots on an Argand diagram.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\left( 1 - 2 \mathrm { e } ^ { \mathrm { j } \theta } \right) \left( 1 - 2 \mathrm { e } ^ { - \mathrm { j } \theta } \right) = 5 - 4 \cos \theta$.
Series $C$ and $S$ are defined by
$$\begin{aligned}
& C = 2 \cos \theta + 4 \cos 2 \theta + 8 \cos 3 \theta + \ldots + 2 ^ { n } \cos n \theta \\
& S = 2 \sin \theta + 4 \sin 2 \theta + 8 \sin 3 \theta + \ldots + 2 ^ { n } \sin n \theta
\end{aligned}$$
\item Show that $C = \frac { 2 \cos \theta - 4 - 2 ^ { n + 1 } \cos ( n + 1 ) \theta + 2 ^ { n + 2 } \cos n \theta } { 5 - 4 \cos \theta }$, and find a similar expression for $S$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2008 Q2 [18]}}