| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2019 |
| 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 Further Maths question requiring recognition that C + iS forms a geometric series with complex terms, application of the sum to infinity formula, and algebraic manipulation using Euler's formula. While it involves multiple steps and complex number manipulation, the technique is directly taught in CP2 and follows a predictable pattern. The question guides students through parts (a) and (b), making it slightly above average difficulty but well within reach for prepared Further Maths students. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(C+iS = \cos\theta+i\sin\theta+\frac{1}{2}(\cos5\theta+i\sin5\theta)+\frac{1}{4}(\cos9\theta+i\sin9\theta)+\ldots\) | M1 | Combines two series by pairing multiples of \(\theta\) (at least up to \(5\theta\)) |
| \(= e^{i\theta}+\frac{1}{2}e^{5i\theta}+\frac{1}{4}e^{9i\theta}+\ldots\) | A1 | Converts to Euler form correctly (at least up to \(5\theta\)) |
| \(C+iS = \frac{e^{i\theta}}{1-\frac{1}{2}e^{4i\theta}}\) | M1 | Recognises convergent geometric series and uses sum to infinity of GP |
| \(= \frac{2e^{i\theta}}{2-e^{4i\theta}}\) * | A1* | Reaches printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(C+iS = \cos\theta+i\sin\theta+\frac{1}{2}(\cos\theta+i\sin\theta)^5+\frac{1}{4}(\cos\theta+i\sin\theta)^9+\ldots\) | A1 | Converts to power form correctly (at least up to \(5\theta\)) |
| \(C+iS = \frac{\cos\theta+i\sin\theta}{1-\frac{1}{2}(\cos\theta+i\sin\theta)^4} = \frac{e^{i\theta}}{1-\frac{1}{2}e^{4i\theta}}\) | M1 | Recognises convergent GP and uses sum to infinity |
| \(= \frac{2e^{i\theta}}{2-e^{4i\theta}}\) * | A1* | Reaches printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2e^{i\theta}}{2-e^{4i\theta}} \times \frac{2-e^{-4i\theta}}{2-e^{-4i\theta}}\) | M1 | Multiplies numerator and denominator by \(2-e^{-4i\theta}\) |
| \(\frac{4e^{i\theta}-2e^{-3i\theta}}{4-2e^{-4i\theta}-2e^{4i\theta}+1}\) | A1 | Correct fraction in terms of exponentials |
| \(\frac{4\cos\theta+4i\sin\theta-2\cos3\theta+2i\sin3\theta}{5-2\cos4\theta+2i\sin4\theta-2\cos4\theta-2i\sin4\theta}\) | dM1 | Converts back to trigonometric form; dependent on first M |
| \(S = \frac{4\sin\theta+2\sin3\theta}{5-4\cos4\theta}\) * | A1* | Reaches printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2e^{i\theta}}{2-e^{4i\theta}} = \frac{2(\cos\theta+i\sin\theta)}{2-(\cos4\theta+i\sin4\theta)} \times \frac{2-(\cos4\theta-i\sin4\theta)}{2-(\cos4\theta-i\sin4\theta)}\) | M1 | Converts to trig form; multiplies by complex conjugate of denominator |
| \(\frac{4\cos\theta+4i\sin\theta-2\cos\theta\cos4\theta-2\sin\theta\sin4\theta+2i\sin4\theta\cos\theta-2i\sin\theta\cos4\theta}{4+\cos^24\theta+\sin^24\theta-4\cos4\theta}\) | A1 | Correct fraction in terms of trig functions |
| \(\frac{4\cos\theta+4i\sin\theta-2\cos3\theta+2i\sin3\theta}{5-2\cos4\theta+2i\sin4\theta-2\cos4\theta-2i\sin4\theta}\) | dM1 | Uses correct addition formula to obtain \(\sin3\theta\) in numerator |
| \(S = \frac{4\sin\theta+2\sin3\theta}{5-4\cos4\theta}\) * | A1* | Reaches printed answer with no errors |
# Question 4(a) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C+iS = \cos\theta+i\sin\theta+\frac{1}{2}(\cos5\theta+i\sin5\theta)+\frac{1}{4}(\cos9\theta+i\sin9\theta)+\ldots$ | M1 | Combines two series by pairing multiples of $\theta$ (at least up to $5\theta$) |
| $= e^{i\theta}+\frac{1}{2}e^{5i\theta}+\frac{1}{4}e^{9i\theta}+\ldots$ | A1 | Converts to Euler form correctly (at least up to $5\theta$) |
| $C+iS = \frac{e^{i\theta}}{1-\frac{1}{2}e^{4i\theta}}$ | M1 | Recognises convergent geometric series and uses sum to infinity of GP |
| $= \frac{2e^{i\theta}}{2-e^{4i\theta}}$ * | A1* | Reaches printed answer with no errors |
# Question 4(a) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C+iS = \cos\theta+i\sin\theta+\frac{1}{2}(\cos\theta+i\sin\theta)^5+\frac{1}{4}(\cos\theta+i\sin\theta)^9+\ldots$ | A1 | Converts to power form correctly (at least up to $5\theta$) |
| $C+iS = \frac{\cos\theta+i\sin\theta}{1-\frac{1}{2}(\cos\theta+i\sin\theta)^4} = \frac{e^{i\theta}}{1-\frac{1}{2}e^{4i\theta}}$ | M1 | Recognises convergent GP and uses sum to infinity |
| $= \frac{2e^{i\theta}}{2-e^{4i\theta}}$ * | A1* | Reaches printed answer with no errors |
# Question 4(b) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2e^{i\theta}}{2-e^{4i\theta}} \times \frac{2-e^{-4i\theta}}{2-e^{-4i\theta}}$ | M1 | Multiplies numerator and denominator by $2-e^{-4i\theta}$ |
| $\frac{4e^{i\theta}-2e^{-3i\theta}}{4-2e^{-4i\theta}-2e^{4i\theta}+1}$ | A1 | Correct fraction in terms of exponentials |
| $\frac{4\cos\theta+4i\sin\theta-2\cos3\theta+2i\sin3\theta}{5-2\cos4\theta+2i\sin4\theta-2\cos4\theta-2i\sin4\theta}$ | dM1 | Converts back to trigonometric form; dependent on first M |
| $S = \frac{4\sin\theta+2\sin3\theta}{5-4\cos4\theta}$ * | A1* | Reaches printed answer with no errors |
# Question 4(b) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2e^{i\theta}}{2-e^{4i\theta}} = \frac{2(\cos\theta+i\sin\theta)}{2-(\cos4\theta+i\sin4\theta)} \times \frac{2-(\cos4\theta-i\sin4\theta)}{2-(\cos4\theta-i\sin4\theta)}$ | M1 | Converts to trig form; multiplies by complex conjugate of denominator |
| $\frac{4\cos\theta+4i\sin\theta-2\cos\theta\cos4\theta-2\sin\theta\sin4\theta+2i\sin4\theta\cos\theta-2i\sin\theta\cos4\theta}{4+\cos^24\theta+\sin^24\theta-4\cos4\theta}$ | A1 | Correct fraction in terms of trig functions |
| $\frac{4\cos\theta+4i\sin\theta-2\cos3\theta+2i\sin3\theta}{5-2\cos4\theta+2i\sin4\theta-2\cos4\theta-2i\sin4\theta}$ | dM1 | Uses correct addition formula to obtain $\sin3\theta$ in numerator |
| $S = \frac{4\sin\theta+2\sin3\theta}{5-4\cos4\theta}$ * | A1* | Reaches printed answer with no errors |
---\begin{enumerate}
\item The infinite series C and S are defined by
\end{enumerate}
$$\begin{aligned}
& \mathrm { C } = \cos \theta + \frac { 1 } { 2 } \cos 5 \theta + \frac { 1 } { 4 } \cos 9 \theta + \frac { 1 } { 8 } \cos 13 \theta + \ldots \\
& \mathrm { S } = \sin \theta + \frac { 1 } { 2 } \sin 5 \theta + \frac { 1 } { 4 } \sin 9 \theta + \frac { 1 } { 8 } \sin 13 \theta + \ldots
\end{aligned}$$
Given that the series C and S are both convergent,\\
(a) show that
$$\mathrm { C } + \mathrm { iS } = \frac { 2 \mathrm { e } ^ { \mathrm { i } \theta } } { 2 - \mathrm { e } ^ { 4 \mathrm { i } \theta } }$$
(b) Hence show that
$$\mathrm { S } = \frac { 4 \sin \theta + 2 \sin 3 \theta } { 5 - 4 \cos 4 \theta }$$
\hfill \mbox{\textit{Edexcel CP2 2019 Q4 [8]}}