| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Sum geometric series with complex terms |
| Difficulty | Challenging +1.2 This is a standard FP3 complex numbers question requiring geometric series summation and rationalization of complex denominators. While it involves multiple steps (recognizing the series pattern, applying geometric series formula, multiplying by conjugate, separating real/imaginary parts), these are well-practiced techniques at this level. The series pattern is clearly presented, and the method is algorithmic rather than requiring novel insight. Slightly above average difficulty due to the multi-step nature and FP3 content, but remains a textbook-style exercise. |
| Spec | 1.04j Sum to infinity: convergent geometric series |r|<14.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| \(C + iS = 1 + \frac{1}{2}e^{i\theta} + \frac{1}{4}e^{2i\theta} + \frac{1}{8}e^{3i\theta} + \ldots\) | M1 | For using \(\cos n\theta + i\sin n\theta = e^{in\theta}\) at least once for \(n \geq 2\) |
| A1 | For correct series | |
| \(= \frac{1}{1 - \frac{1}{2}e^{i\theta}} - \frac{2}{2 - e^{i\theta}}\) | M1 | For using sum of infinite GP |
| A1 4 | For correct expression AG | |
| SR For omission of 1st stage award up to M0 A0 M1 A1 OEW |
| Answer | Marks | Guidance |
|---|---|---|
| \(C + iS = \frac{2(2 - e^{-i\theta})}{(2 - e^{i\theta})(2 - e^{-i\theta})}\) | M1 | For multiplying top and bottom by complex conjugate |
| \(= \frac{4 - 2e^{-i\theta}}{4 - 2(e^{i\theta} + e^{-i\theta}) + 1} = \frac{4 - 2\cos\theta + 2i\sin\theta}{4 - 4\cos\theta + 1}\) | M1 | For reverting to \(\cos\theta\) and \(\sin\theta\) and equating Re OR Im parts |
| \(\Rightarrow C = \frac{4 - 2\cos\theta}{5 - 4\cos\theta}\), \(S = \frac{2\sin\theta}{5 - 4\cos\theta}\) | A1 | For correct expression for \(C\) AG |
| A1 4 | For correct expression for \(S\) |
## (i)
$C + iS = 1 + \frac{1}{2}e^{i\theta} + \frac{1}{4}e^{2i\theta} + \frac{1}{8}e^{3i\theta} + \ldots$ | M1 | For using $\cos n\theta + i\sin n\theta = e^{in\theta}$ at least once for $n \geq 2$
| A1 | For correct series
$= \frac{1}{1 - \frac{1}{2}e^{i\theta}} - \frac{2}{2 - e^{i\theta}}$ | M1 | For using sum of infinite GP
| A1 4 | For correct expression AG
| | SR For omission of 1st stage award up to M0 A0 M1 A1 OEW
## (ii)
$C + iS = \frac{2(2 - e^{-i\theta})}{(2 - e^{i\theta})(2 - e^{-i\theta})}$ | M1 | For multiplying top and bottom by complex conjugate
$= \frac{4 - 2e^{-i\theta}}{4 - 2(e^{i\theta} + e^{-i\theta}) + 1} = \frac{4 - 2\cos\theta + 2i\sin\theta}{4 - 4\cos\theta + 1}$ | M1 | For reverting to $\cos\theta$ and $\sin\theta$ and equating Re OR Im parts
$\Rightarrow C = \frac{4 - 2\cos\theta}{5 - 4\cos\theta}$, $S = \frac{2\sin\theta}{5 - 4\cos\theta}$ | A1 | For correct expression for $C$ AG
| A1 4 | For correct expression for $S$
---Convergent infinite series $C$ and $S$ are defined by
\begin{align}
C &= 1 + \frac{1}{4} \cos \theta + \frac{1}{4} \cos 2\theta + \frac{1}{8} \cos 3\theta + \ldots, \\
S &= \frac{1}{2} \sin \theta + \frac{1}{4} \sin 2\theta + \frac{1}{8} \sin 3\theta + \ldots.
\end{align}
\begin{enumerate}[label=(\roman*)]
\item Show that $C + iS = \frac{2}{2 - e^{i\theta}}$. [4]
\item Hence show that $C = \frac{4 - 2\cos \theta}{5 - 4\cos \theta}$ and find a similar expression for $S$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2010 Q5 [8]}}