| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Sum geometric series with complex terms |
| Difficulty | Challenging +1.3 This is a multi-part Further Maths question requiring manipulation of complex exponentials, binomial theorem with geometric series, and geometric applications. Part (a)(i) is routine verification, (a)(ii) requires recognizing the binomial expansion of (1+e^{j2θ})^n and applying de Moivre's theorem (moderately challenging), while part (b) involves standard rotations and distance calculations. The question demands solid technique across multiple areas but follows established patterns for FP2, making it above average but not exceptionally difficult for further maths students. |
| Spec | 4.02b Express complex numbers: cartesian and modulus-argument forms4.02k Argand diagrams: geometric interpretation4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 + e^{j2\theta} = 1 + \cos 2\theta + j\sin 2\theta\) | ||
| \(= 1 + (2\cos^2\theta - 1) + 2j\sin\theta\cos\theta\) | M1 | Using \(e^{2j\theta} = \cos 2\theta + j\sin 2\theta\) and double angle formulae; allow one error |
| \(= 2\cos^2\theta + 2j\sin\theta\cos\theta\) | ||
| \(= 2\cos\theta(\cos\theta + j\sin\theta)\) | A1(ag) | Completion www |
| OR \(1 + e^{j2\theta} = e^{j\theta}(e^{-j\theta} + e^{j\theta})\) | M1 | "Factorising" and complete replacement by trigonometric functions |
| \(= (\cos\theta + j\sin\theta) \times 2\cos\theta\) | A1(ag) | Completion www |
| OR \(1 + e^{j2\theta} = 1 + (\cos\theta + j\sin\theta)^2\) | ||
| \(= 1 + \cos^2\theta - \sin^2\theta + 2j\sin\theta\cos\theta\) | ||
| \(= 2\cos^2\theta + 2j\sin\theta\cos\theta\) | M1 | Using \(e^{j\theta} = \cos\theta + j\sin\theta\) and \(1 - \sin^2\theta = \cos^2\theta\) |
| \(= 2\cos\theta(\cos\theta + j\sin\theta)\) | A1(ag) | Completion www |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(C + jS = 1 + \binom{n}{1}e^{j2\theta} + \binom{n}{2}e^{j4\theta} + \ldots + e^{j2n\theta}\) | M1 | Forming \(C + jS\) |
| \(= (1 + e^{j2\theta})^n\) | M1, A1 | Recognising as binomial expansion |
| \(= 2^n\cos^n\theta(\cos\theta + j\sin\theta)^n\) | M1, A1 | Applying (i) and De Moivre o.e. (dependent on M1M1 above) |
| \(= 2^n\cos^n\theta(\cos n\theta + j\sin n\theta)\) | Need to see \(e^{jn\theta} = \cos n\theta + j\sin n\theta\) o.e. | |
| \(\Rightarrow C = 2^n\cos^n\theta\cos n\theta\) | A1(ag) | Completion www |
| and \(S = 2^n\cos^n\theta\sin n\theta\) | A1 | |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{j\frac{2\pi}{3}} = \cos\frac{2\pi}{3} + j\sin\frac{2\pi}{3} = -\frac{1}{2} + j\frac{\sqrt{3}}{2}\) | B1 | Must evaluate trigonometric functions |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Other two vertices are \((2+4j)e^{j\frac{2\pi}{3}}\) | M1 | Award for idea of rotation by \(\frac{2\pi}{3}\); e.g. use of \(\arctan 2 + \frac{2\pi}{3}\) (3.202 rad) (must be 2) |
| \(= (2+4j)\left(-\frac{1}{2}+j\frac{\sqrt{3}}{2}\right)\) | ||
| \(= (-1-2\sqrt{3}) + j(-2+\sqrt{3})\) | A1A1 | May be given as co-ordinates |
| and \((2+4j)e^{j\frac{4\pi}{3}} = (2+4j)e^{-j\frac{2\pi}{3}}\) | M1 | Award for idea of rotation by \(-\frac{2\pi}{3}\); e.g. use of \(\arctan 2 + \frac{4\pi}{3}\) (5.296 rad) (must be 2) |
| \(= (2+4j)\left(-\frac{1}{2}-j\frac{\sqrt{3}}{2}\right)\) | ||
| \(= (-1+2\sqrt{3}) + j(-2-\sqrt{3})\) | A1A1 | May be given as co-ordinates; if A0A0A0A0 award SC1 for awrt \(-4.46-0.27j\) and \(2.46-3.73j\) |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Length of \((2+4j) = \sqrt{20}\) | ||
| So length of side \(= 2\sqrt{20}\cos\frac{\pi}{6} = 2\sqrt{20} \times \frac{\sqrt{3}}{2}\) | M1 | Complete method; alternative: finding distance between \((2,4)\) and \((-1-2\sqrt{3}, -2+\sqrt{3})\) o.e. |
| \(= 2\sqrt{15}\) | A1(ag) | Completion www |
| [2] |
# Question 2:
## Part (a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 + e^{j2\theta} = 1 + \cos 2\theta + j\sin 2\theta$ | | |
| $= 1 + (2\cos^2\theta - 1) + 2j\sin\theta\cos\theta$ | M1 | Using $e^{2j\theta} = \cos 2\theta + j\sin 2\theta$ and double angle formulae; allow one error |
| $= 2\cos^2\theta + 2j\sin\theta\cos\theta$ | | |
| $= 2\cos\theta(\cos\theta + j\sin\theta)$ | A1(ag) | Completion www |
| **OR** $1 + e^{j2\theta} = e^{j\theta}(e^{-j\theta} + e^{j\theta})$ | M1 | "Factorising" and complete replacement by trigonometric functions |
| $= (\cos\theta + j\sin\theta) \times 2\cos\theta$ | A1(ag) | Completion www |
| **OR** $1 + e^{j2\theta} = 1 + (\cos\theta + j\sin\theta)^2$ | | |
| $= 1 + \cos^2\theta - \sin^2\theta + 2j\sin\theta\cos\theta$ | | |
| $= 2\cos^2\theta + 2j\sin\theta\cos\theta$ | M1 | Using $e^{j\theta} = \cos\theta + j\sin\theta$ and $1 - \sin^2\theta = \cos^2\theta$ |
| $= 2\cos\theta(\cos\theta + j\sin\theta)$ | A1(ag) | Completion www |
| **[2]** | | |
## Part (a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $C + jS = 1 + \binom{n}{1}e^{j2\theta} + \binom{n}{2}e^{j4\theta} + \ldots + e^{j2n\theta}$ | M1 | Forming $C + jS$ |
| $= (1 + e^{j2\theta})^n$ | M1, A1 | Recognising as binomial expansion |
| $= 2^n\cos^n\theta(\cos\theta + j\sin\theta)^n$ | M1, A1 | Applying (i) and De Moivre o.e. (dependent on M1M1 above) |
| $= 2^n\cos^n\theta(\cos n\theta + j\sin n\theta)$ | | Need to see $e^{jn\theta} = \cos n\theta + j\sin n\theta$ o.e. |
| $\Rightarrow C = 2^n\cos^n\theta\cos n\theta$ | A1(ag) | Completion www |
| and $S = 2^n\cos^n\theta\sin n\theta$ | A1 | |
| **[7]** | | |
## Part (b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{j\frac{2\pi}{3}} = \cos\frac{2\pi}{3} + j\sin\frac{2\pi}{3} = -\frac{1}{2} + j\frac{\sqrt{3}}{2}$ | B1 | Must evaluate trigonometric functions |
| **[1]** | | |
## Part (b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Other two vertices are $(2+4j)e^{j\frac{2\pi}{3}}$ | M1 | Award for idea of rotation by $\frac{2\pi}{3}$; e.g. use of $\arctan 2 + \frac{2\pi}{3}$ (3.202 rad) (must be 2) |
| $= (2+4j)\left(-\frac{1}{2}+j\frac{\sqrt{3}}{2}\right)$ | | |
| $= (-1-2\sqrt{3}) + j(-2+\sqrt{3})$ | A1A1 | May be given as co-ordinates |
| and $(2+4j)e^{j\frac{4\pi}{3}} = (2+4j)e^{-j\frac{2\pi}{3}}$ | M1 | Award for idea of rotation by $-\frac{2\pi}{3}$; e.g. use of $\arctan 2 + \frac{4\pi}{3}$ (5.296 rad) (must be 2) |
| $= (2+4j)\left(-\frac{1}{2}-j\frac{\sqrt{3}}{2}\right)$ | | |
| $= (-1+2\sqrt{3}) + j(-2-\sqrt{3})$ | A1A1 | May be given as co-ordinates; if A0A0A0A0 award SC1 for awrt $-4.46-0.27j$ and $2.46-3.73j$ |
| **[6]** | | |
## Part (b)(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Length of $(2+4j) = \sqrt{20}$ | | |
| So length of side $= 2\sqrt{20}\cos\frac{\pi}{6} = 2\sqrt{20} \times \frac{\sqrt{3}}{2}$ | M1 | Complete method; alternative: finding distance between $(2,4)$ and $(-1-2\sqrt{3}, -2+\sqrt{3})$ o.e. |
| $= 2\sqrt{15}$ | A1(ag) | Completion www |
| **[2]** | | |
---\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that
$$1 + \mathrm { e } ^ { \mathrm { j } 2 \theta } = 2 \cos \theta ( \cos \theta + \mathrm { j } \sin \theta )$$
\item The series $C$ and $S$ are defined as follows.
$$\begin{aligned}
& C = 1 + \binom { n } { 1 } \cos 2 \theta + \binom { n } { 2 } \cos 4 \theta + \ldots + \cos 2 n \theta \\
& S = \binom { n } { 1 } \sin 2 \theta + \binom { n } { 2 } \sin 4 \theta + \ldots + \sin 2 n \theta
\end{aligned}$$
By considering $C + \mathrm { j } S$, show that
$$C = 2 ^ { n } \cos ^ { n } \theta \cos n \theta$$
and find a corresponding expression for $S$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Express $\mathrm { e } ^ { \mathrm { j } 2 \pi / 3 }$ in the form $x + \mathrm { j } y$, where the real numbers $x$ and $y$ should be given exactly.
\item An equilateral triangle in the Argand diagram has its centre at the origin. One vertex of the triangle is at the point representing $2 + 4 \mathrm { j }$. Obtain the complex numbers representing the other two vertices, giving your answers in the form $x + \mathrm { j } y$, where the real numbers $x$ and $y$ should be given exactly.
\item Show that the length of a side of the triangle is $2 \sqrt { 15 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2013 Q2 [18]}}