| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| 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 Further Maths geometric series question with complex exponentials requiring systematic application of the GP formula, Euler's formula, and algebraic manipulation. While it involves multiple parts and careful bookkeeping with complex exponentials, the techniques are well-rehearsed in FP3 syllabi with no novel insights required—making it moderately above average difficulty but routine for Further Maths students. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae4.04a Line equations: 2D and 3D, cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{i\theta}+e^{2i\theta}+\ldots+e^{10i\theta}=\frac{e^{i\theta}\left((e^{i\theta})^{10}-1\right)}{e^{i\theta}-1}\) | M1, A1 | Sum of a GP |
| \(=\frac{e^{\frac{1}{2}i\theta}(e^{10i\theta}-1)}{e^{\frac{1}{2}i\theta}-e^{-\frac{1}{2}i\theta}}\) | M1 | |
| \(=\frac{e^{\frac{1}{2}i\theta}(e^{10i\theta}-1)}{2i\sin(\frac{1}{2}\theta)}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\theta=2n\pi \Rightarrow\) sum \(=10\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\theta + \cos 2\theta + \ldots + \cos 10\theta = \text{Re}\left(\dfrac{e^{\frac{1}{2}i\theta}(e^{10i\theta}-1)}{2i\sin(\frac{1}{2}\theta)}\right)\) | M1 | Take real parts |
| \(= \dfrac{\text{Re}\left(-ie^{\frac{1}{2}i\theta}(e^{10i\theta}-1)\right)}{2\sin(\frac{1}{2}\theta)} = \dfrac{\text{Re}\left(-ie^{\frac{21}{2}i\theta}+ie^{\frac{1}{2}i\theta}\right)}{2\sin(\frac{1}{2}\theta)}\) | M1 | Manipulate expression; must at least make genuine progress in sorting real part of numerator, or in converting numerator to trig terms |
| \(= \dfrac{\sin\left(\frac{21}{2}\theta\right) - \sin\left(\frac{1}{2}\theta\right)}{2\sin\left(\frac{1}{2}\theta\right)}\) | ||
| \(= \dfrac{\sin\left(\frac{21}{2}\theta\right)}{2\sin\left(\frac{1}{2}\theta\right)} - \dfrac{1}{2}\) | A1 | AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\frac{1}{11}\pi + \cos\frac{2}{11}\pi + \ldots + \cos\frac{10}{11}\pi = \dfrac{\sin\left(\frac{21}{22}\pi\right)}{2\sin\left(\frac{1}{22}\pi\right)} - \dfrac{1}{2}\) | M1 | For second M1, must convince that solution is exact and not simply from calculator |
| But \(\sin\frac{21}{22}\pi = \sin\left(\pi - \frac{21}{22}\pi\right) = \sin\frac{1}{22}\pi\) | M1 | |
| So RHS \(= \frac{1}{2} - \frac{1}{2} = 0\), so \(\frac{1}{11}\pi\) is a root | A1 | AG |
| Using \(\sin(2\pi + x) = \sin x\) gives \(2\pi + \frac{1}{2}\theta = \frac{21}{2}\theta \Rightarrow \theta = \frac{1}{5}\pi\) | A1 | |
| [4] |
# Question 7:
## Part (i)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{i\theta}+e^{2i\theta}+\ldots+e^{10i\theta}=\frac{e^{i\theta}\left((e^{i\theta})^{10}-1\right)}{e^{i\theta}-1}$ | M1, A1 | Sum of a GP |
| $=\frac{e^{\frac{1}{2}i\theta}(e^{10i\theta}-1)}{e^{\frac{1}{2}i\theta}-e^{-\frac{1}{2}i\theta}}$ | M1 | |
| $=\frac{e^{\frac{1}{2}i\theta}(e^{10i\theta}-1)}{2i\sin(\frac{1}{2}\theta)}$ | A1 | AG |
## Part (i)(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\theta=2n\pi \Rightarrow$ sum $=10$ | B1 | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta + \cos 2\theta + \ldots + \cos 10\theta = \text{Re}\left(\dfrac{e^{\frac{1}{2}i\theta}(e^{10i\theta}-1)}{2i\sin(\frac{1}{2}\theta)}\right)$ | M1 | Take real parts |
| $= \dfrac{\text{Re}\left(-ie^{\frac{1}{2}i\theta}(e^{10i\theta}-1)\right)}{2\sin(\frac{1}{2}\theta)} = \dfrac{\text{Re}\left(-ie^{\frac{21}{2}i\theta}+ie^{\frac{1}{2}i\theta}\right)}{2\sin(\frac{1}{2}\theta)}$ | M1 | Manipulate expression; must at least make genuine progress in sorting real part of numerator, or in converting numerator to trig terms |
| $= \dfrac{\sin\left(\frac{21}{2}\theta\right) - \sin\left(\frac{1}{2}\theta\right)}{2\sin\left(\frac{1}{2}\theta\right)}$ | | |
| $= \dfrac{\sin\left(\frac{21}{2}\theta\right)}{2\sin\left(\frac{1}{2}\theta\right)} - \dfrac{1}{2}$ | A1 | AG |
| **[3]** | | |
---
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\frac{1}{11}\pi + \cos\frac{2}{11}\pi + \ldots + \cos\frac{10}{11}\pi = \dfrac{\sin\left(\frac{21}{22}\pi\right)}{2\sin\left(\frac{1}{22}\pi\right)} - \dfrac{1}{2}$ | M1 | For second M1, must convince that solution is exact and not simply from calculator |
| But $\sin\frac{21}{22}\pi = \sin\left(\pi - \frac{21}{22}\pi\right) = \sin\frac{1}{22}\pi$ | M1 | |
| So RHS $= \frac{1}{2} - \frac{1}{2} = 0$, so $\frac{1}{11}\pi$ is a root | A1 | AG |
| Using $\sin(2\pi + x) = \sin x$ gives $2\pi + \frac{1}{2}\theta = \frac{21}{2}\theta \Rightarrow \theta = \frac{1}{5}\pi$ | A1 | |
| **[4]** | | |
---7 Let $S = \mathrm { e } ^ { \mathrm { i } \theta } + \mathrm { e } ^ { 2 \mathrm { i } \theta } + \mathrm { e } ^ { 3 \mathrm { i } \theta } + \ldots + \mathrm { e } ^ { 10 \mathrm { i } \theta }$.
\begin{enumerate}[label=(\roman*)]
\item (a) Show that, for $\theta \neq 2 n \pi$, where $n$ is an integer,
$$S = \frac { \mathrm { e } ^ { \frac { 1 } { 2 } \mathrm { i } \theta } \left( \mathrm { e } ^ { 10 \mathrm { i } \theta } - 1 \right) } { 2 \mathrm { i } \sin \left( \frac { 1 } { 2 } \theta \right) }$$
(b) State the value of $S$ for $\theta = 2 n \pi$, where $n$ is an integer.
\item Hence show that, for $\theta \neq 2 n \pi$, where $n$ is an integer,
$$\cos \theta + \cos 2 \theta + \cos 3 \theta + \ldots + \cos 10 \theta = \frac { \sin \left( \frac { 21 } { 2 } \theta \right) } { 2 \sin \left( \frac { 1 } { 2 } \theta \right) } - \frac { 1 } { 2 }$$
\item Hence show that $\theta = \frac { 1 } { 11 } \pi$ is a root of $\cos \theta + \cos 2 \theta + \cos 3 \theta + \ldots + \cos 10 \theta = 0$ and find another root in the interval $0 < \theta < \frac { 1 } { 4 } \pi$.
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2013 Q7 [12]}}