| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | March |
| Marks | 14 |
| Topic | Complex numbers 2 |
| Type | Sum geometric series with complex terms |
| Difficulty | Challenging +1.2 This is a Further Maths FP2 question on complex exponentials and series summation. Parts (i)-(ii) are routine bookwork and algebraic manipulation. Part (iii) requires recognizing a geometric series in complex form and applying the result from (ii), which is a standard technique but requires careful execution across multiple steps. Part (iv) is a verification with geometric insight. The question is well-scaffolded and uses standard FP2 methods, making it moderately above average difficulty but not requiring novel problem-solving approaches. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{i\theta} - e^{-i\theta} = (\cos\theta + i\sin\theta) - (\cos\theta - i\sin\theta) = 2i\sin\theta\) | B1 (AG) [1] | Must be clear evidence of use of Euler's equation and both \(\cos(-\theta) = \cos\theta\) and \(\sin(-\theta) = -\sin\theta\) seen or clearly implied. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2e^{i\theta}}{e^{i\theta} - e^{-i\theta}}\) or \(\frac{2}{e^{i\theta}(e^{i\theta} - e^{-i\theta})}\) | B1 [2.1] | Condone omission of 2 |
| \(\frac{2ie^{i\theta}}{2i\sin\theta}\) | M1 | Use of (i) and \(1/i = -i\). Condone incorrect 2. |
| \(\frac{i(\cos\theta - i\sin\theta)}{sin\theta} = -(1 + i\cot\theta)\) | A1 [2.1] | AG At least one intermediate step must be shown. |
| Answer | Marks | Guidance |
|---|---|---|
| \(C + iS = \sum 2e^{i\frac{\pi x}{10}}\) | B1 | Expressing C + iS as the sum of (multiples of) positive integer powers of \(e^{i\frac{\pi}{10}}\) [3.1a] |
| \(GP\) with and \(r = e^{i\frac{\pi}{10}}\) seen or implied | B1 [2.2a] | Could be implied by eg formula even if a and/or r incorrect. |
| \(2\left(\frac{(e^{i\frac{\pi}{10}})^5 - 1}{e^{i\frac{\pi}{10}} - 1}\right)\) | M1 [2.2a] | Use of formula for their series. |
| \(\frac{e^{i\frac{\pi}{10}} - 1}{e^{\frac{i\pi}{10}} - 1}\) | A1 [1.1] | All correct |
| \((e^{i\frac{\pi}{10}})^5 = i\) | A1 [1.1] | B1 [1.1] |
| \(\frac{2}{e^{i\frac{\pi}{10}} - 1} = -(1 + i\cot\frac{\pi}{20})\) | M1 [2.2a] | Correct use of (ii) for their expression. |
| \(-(1 + i\cot\frac{\pi}{20})(i - 1) = \cot\frac{\pi}{20} + 1 + i(\cot\frac{\pi}{20} - 1)\) | M1 [3.1a] | Expanding brackets and collecting |
| \(C = \cot\frac{\pi}{20} + 1\) | A1 [2.2a] |
| Answer | Marks | Guidance |
|---|---|---|
| \(S = \cot\frac{\pi}{20} - 1 = \cot\frac{\pi}{20} + 1 - 2 = C - 2\) | B1 [2.2a] | Must be from fully correct working and solution in (iii). |
| All of the trigonometric terms are, in effect, the same; | B1 [2.4] | Must be at least one explicit example or an appeal to the statement eg \(\cos\frac{\pi}{10} = \sin\frac{2\pi}{5}\) oe. |
## 9(i)
$e^{i\theta} - e^{-i\theta} = (\cos\theta + i\sin\theta) - (\cos\theta - i\sin\theta) = 2i\sin\theta$ | B1 (AG) [1] | Must be clear evidence of use of Euler's equation and both $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$ seen or clearly implied. | eg just $(\cos(-\theta) + i\sin(-\theta))$ is not sufficient.
## 9(ii)
$\frac{2e^{i\theta}}{e^{i\theta} - e^{-i\theta}}$ or $\frac{2}{e^{i\theta}(e^{i\theta} - e^{-i\theta})}$ | B1 [2.1] | Condone omission of 2
$\frac{2ie^{i\theta}}{2i\sin\theta}$ | M1 | Use of (i) and $1/i = -i$. Condone incorrect 2. | [2.1]
$\frac{i(\cos\theta - i\sin\theta)}{sin\theta} = -(1 + i\cot\theta)$ | A1 [2.1] | AG At least one intermediate step must be shown.
## 9(iii)
$C + iS = \sum 2e^{i\frac{\pi x}{10}}$ | B1 | Expressing C + iS as the sum of (multiples of) positive integer powers of $e^{i\frac{\pi}{10}}$ [3.1a]
$GP$ with and $r = e^{i\frac{\pi}{10}}$ seen or implied | B1 [2.2a] | Could be implied by eg formula even if a and/or r incorrect.
$2\left(\frac{(e^{i\frac{\pi}{10}})^5 - 1}{e^{i\frac{\pi}{10}} - 1}\right)$ | M1 [2.2a] | Use of formula for their series.
$\frac{e^{i\frac{\pi}{10}} - 1}{e^{\frac{i\pi}{10}} - 1}$ | A1 [1.1] | All correct
$(e^{i\frac{\pi}{10}})^5 = i$ | A1 [1.1] | B1 [1.1]
$\frac{2}{e^{i\frac{\pi}{10}} - 1} = -(1 + i\cot\frac{\pi}{20})$ | M1 [2.2a] | Correct use of (ii) for their expression.
$-(1 + i\cot\frac{\pi}{20})(i - 1) = \cot\frac{\pi}{20} + 1 + i(\cot\frac{\pi}{20} - 1)$ | M1 [3.1a] | Expanding brackets and collecting
$C = \cot\frac{\pi}{20} + 1$ | A1 [2.2a]
[8]
## 9(iv)
$S = \cot\frac{\pi}{20} - 1 = \cot\frac{\pi}{20} + 1 - 2 = C - 2$ | B1 [2.2a] | Must be from fully correct working and solution in (iii).
All of the trigonometric terms are, in effect, the same; | B1 [2.4] | Must be at least one explicit example or an appeal to the statement eg $\cos\frac{\pi}{10} = \sin\frac{2\pi}{5}$ oe.
eg $\cos\frac{\pi}{10} = \sin\frac{2\pi}{5}$
[2]In this question you must show detailed reasoning.
\begin{enumerate}[label=(\roman*)]
\item Show that $e^{i\theta} - e^{-i\theta} = 2i\sin\theta$. [1]
\item Hence, show that $\frac{2}{e^{2i\theta} - 1} = -(1 + i\cot\theta)$. [3]
\item Two series, $C$ and $S$, are defined as follows.
$$C = 2 + 2\cos\frac{\pi}{10} + 2\cos\frac{\pi}{5} + 2\cos\frac{3\pi}{10} + 2\cos\frac{2\pi}{5}$$
$$S = 2\sin\frac{\pi}{10} + 2\sin\frac{\pi}{5} + 2\sin\frac{3\pi}{10} + 2\sin\frac{2\pi}{5}$$
By considering $C + iS$, find a simplified expression for $C$ in terms of only integers and $\cot\frac{\pi}{10}$. [8]
\item Verify that $S = C - 2$ and, by considering the series in their original form, explain why this is so. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q9 [14]}}