| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Argument relationships and tangent identities |
| Difficulty | Standard +0.8 This is a multi-part Core Pure 2 question requiring geometric series summation, complex number manipulation with de Moivre's theorem, and algebraic proof involving separation of real/imaginary parts. While part (a) is routine recall, parts (b)(i) and (b)(ii) require non-trivial manipulation and insight about when complex expressions can be purely imaginary, placing it moderately above average difficulty. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{1}{1-z}\) | B1 | See scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 + z + z^2 + z^3 + ... = 1 + \left(\frac{1}{2}(\cos\theta + i\sin\theta)\right) + \left(\frac{1}{2}(\cos\theta + i\sin\theta)\right)^2 + ...\) \(= 1 + \frac{1}{2}(\cos\theta + i\sin\theta) + \frac{1}{4}(\cos 2\theta + i\sin 2\theta) + \frac{1}{8}(\cos 3\theta + i\sin 3\theta) + ...\) | M1 | Substitutes \(z = \frac{1}{2}(\cos\theta + i\sin\theta)\) into at least 3 terms and applies de Moivre's theorem |
| \(\dfrac{1}{1-z} = \dfrac{1}{1 - \frac{1}{2}(\cos\theta + i\sin\theta)} \times \dfrac{1 - \frac{1}{2}\cos\theta + \frac{1}{2}i\sin\theta}{1 - \frac{1}{2}\cos\theta + \frac{1}{2}i\sin\theta}\) | M1 | Substitutes \(z = \frac{1}{2}(\cos\theta + i\sin\theta)\) into part (a) and rationalises the denominator |
| \(\left\{\frac{1}{2}(\sin\theta) + \frac{1}{4}(\sin 2\theta) + \frac{1}{8}(\sin 3\theta) + ...\right\} = \dfrac{\frac{1}{2}\sin\theta}{\left(1-\frac{1}{2}\cos\theta\right)^2 + \left(\frac{1}{2}\sin\theta\right)^2}\) | M1 | Equates the imaginary terms |
| \(\left(1 - \frac{1}{2}\cos\theta\right)^2 + \left(\frac{1}{2}\sin\theta\right)^2 = 1 - \cos\theta + \frac{1}{4}\cos^2\theta + \frac{1}{4}\sin^2\theta = \frac{5}{4} - \cos\theta\) | M1 | Multiplies out denominator and simplifies using \(\cos^2\theta + \sin^2\theta = 1\) |
| \(\dfrac{1}{2}\sin\theta + \dfrac{1}{4}\sin 2\theta + \dfrac{1}{8}\sin 3\theta + ... = \dfrac{2\sin\theta}{5 - 4\cos\theta}\) \(\quad *\) | A1* | cso. Must have substituted \(z = \frac{1}{2}(\cos\theta + i\sin\theta)\) into 4 terms of the series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{1 - \frac{1}{2}\cos\theta}{\frac{5}{4} - \cos\theta} = 0 \Rightarrow \cos\theta = 2\) | M1 | Setting the real part of the series \(= 0\) and rearranging to find \(\cos\theta = ...\) |
| As \(-1 \leq \cos\theta \leq 1\), there is no solution to \(\cos\theta = 2\), so there will also be a real part, hence the sum cannot be purely imaginary. | A1 | See scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Imaginary part is \(\dfrac{4 - 2\cos\theta}{5 - 4\cos\theta} = \dfrac{1}{2} + \dfrac{3}{2(5-4\cos\theta)}\) | M1 | Rearranges imaginary part so that \(\cos\theta\) only appears once |
| \(-1 \leq \cos\theta \leq 1\) therefore \(\dfrac{1}{6} < \dfrac{3}{2(5-4\cos\theta)} < \dfrac{3}{2}\) so sum must contain real part | A1 | Uses \(-1 \leq \cos\theta \leq 1\) to show sum must always be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{1}{1-z} = ki \Rightarrow z = 1 + \dfrac{i}{k}\) | M1 | Sets sum as purely imaginary and rearranges to make \(z\) the subject |
| \(\text{mod } z > 1\), contradiction, hence cannot be purely imaginary | A1 | Shows a contradiction and draws an appropriate conclusion |
# Question 9:
## Part 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{1}{1-z}$ | B1 | See scheme |
**Total: (1 mark)**
---
## Part 9(b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 + z + z^2 + z^3 + ... = 1 + \left(\frac{1}{2}(\cos\theta + i\sin\theta)\right) + \left(\frac{1}{2}(\cos\theta + i\sin\theta)\right)^2 + ...$ $= 1 + \frac{1}{2}(\cos\theta + i\sin\theta) + \frac{1}{4}(\cos 2\theta + i\sin 2\theta) + \frac{1}{8}(\cos 3\theta + i\sin 3\theta) + ...$ | M1 | Substitutes $z = \frac{1}{2}(\cos\theta + i\sin\theta)$ into at least 3 terms and applies de Moivre's theorem |
| $\dfrac{1}{1-z} = \dfrac{1}{1 - \frac{1}{2}(\cos\theta + i\sin\theta)} \times \dfrac{1 - \frac{1}{2}\cos\theta + \frac{1}{2}i\sin\theta}{1 - \frac{1}{2}\cos\theta + \frac{1}{2}i\sin\theta}$ | M1 | Substitutes $z = \frac{1}{2}(\cos\theta + i\sin\theta)$ into part (a) and rationalises the denominator |
| $\left\{\frac{1}{2}(\sin\theta) + \frac{1}{4}(\sin 2\theta) + \frac{1}{8}(\sin 3\theta) + ...\right\} = \dfrac{\frac{1}{2}\sin\theta}{\left(1-\frac{1}{2}\cos\theta\right)^2 + \left(\frac{1}{2}\sin\theta\right)^2}$ | M1 | Equates the imaginary terms |
| $\left(1 - \frac{1}{2}\cos\theta\right)^2 + \left(\frac{1}{2}\sin\theta\right)^2 = 1 - \cos\theta + \frac{1}{4}\cos^2\theta + \frac{1}{4}\sin^2\theta = \frac{5}{4} - \cos\theta$ | M1 | Multiplies out denominator and simplifies using $\cos^2\theta + \sin^2\theta = 1$ |
| $\dfrac{1}{2}\sin\theta + \dfrac{1}{4}\sin 2\theta + \dfrac{1}{8}\sin 3\theta + ... = \dfrac{2\sin\theta}{5 - 4\cos\theta}$ $\quad *$ | A1* | cso. Must have substituted $z = \frac{1}{2}(\cos\theta + i\sin\theta)$ into 4 terms of the series |
**Total: (5 marks)**
---
## Part 9(b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{1 - \frac{1}{2}\cos\theta}{\frac{5}{4} - \cos\theta} = 0 \Rightarrow \cos\theta = 2$ | M1 | Setting the real part of the series $= 0$ and rearranging to find $\cos\theta = ...$ |
| As $-1 \leq \cos\theta \leq 1$, there is no solution to $\cos\theta = 2$, so there will also be a real part, hence the sum cannot be purely imaginary. | A1 | See scheme |
**Alternative 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Imaginary part is $\dfrac{4 - 2\cos\theta}{5 - 4\cos\theta} = \dfrac{1}{2} + \dfrac{3}{2(5-4\cos\theta)}$ | M1 | Rearranges imaginary part so that $\cos\theta$ only appears once |
| $-1 \leq \cos\theta \leq 1$ therefore $\dfrac{1}{6} < \dfrac{3}{2(5-4\cos\theta)} < \dfrac{3}{2}$ so sum must contain real part | A1 | Uses $-1 \leq \cos\theta \leq 1$ to show sum must always be positive |
**Alternative 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{1}{1-z} = ki \Rightarrow z = 1 + \dfrac{i}{k}$ | M1 | Sets sum as purely imaginary and rearranges to make $z$ the subject |
| $\text{mod } z > 1$, contradiction, hence cannot be purely imaginary | A1 | Shows a contradiction and draws an appropriate conclusion |
**Total: (2 marks)**
**Overall Total: (8 marks)**\begin{enumerate}
\item (a) Given that $| z | < 1$, write down the sum of the infinite series
\end{enumerate}
$$1 + z + z ^ { 2 } + z ^ { 3 } + \ldots$$
(b) Given that $z = \frac { 1 } { 2 } ( \cos \theta + \mathrm { i } \sin \theta )$,\\
(i) use the answer to part (a), and de Moivre's theorem or otherwise, to prove that
$$\frac { 1 } { 2 } \sin \theta + \frac { 1 } { 4 } \sin 2 \theta + \frac { 1 } { 8 } \sin 3 \theta + \ldots = \frac { 2 \sin \theta } { 5 - 4 \cos \theta }$$
(ii) show that the sum of the infinite series $1 + z + z ^ { 2 } + z ^ { 3 } + \ldots$ cannot be purely imaginary, giving a reason for your answer.
\hfill \mbox{\textit{Edexcel CP2 2021 Q9 [8]}}