| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Sum geometric series with complex terms |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring multiple sophisticated techniques: converting between exponential and Cartesian forms, manipulating complex exponentials, recognizing the binomial theorem applied to complex numbers, and summing series using De Moivre's theorem. Part (b)(ii) requires significant insight to connect C+jS with (1+e^{jθ})^n. However, it's well-structured with clear guidance through parts (i) and (ii), making it accessible to well-prepared FM students, though harder than typical A-level questions. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n4.02b Express complex numbers: cartesian and modulus-argument forms4.02c Complex notation: z, z*, Re(z), Im(z), |z|, arg(z)4.02d Exponential form: re^(i*theta)4.02e Arithmetic of complex numbers: add, subtract, multiply, divide |
| Answer | Marks | Guidance |
|---|---|---|
| \( | w | =3,\ \arg w = -\frac{\pi}{12}\) |
| \( | z | =2,\ \arg z = -\frac{\pi}{3}\) |
| \(\left | \frac{w}{z}\right | =\frac{3}{2},\ \arg\frac{w}{z}=\frac{\pi}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{w}{z} = \frac{3}{2}\left(\cos\frac{\pi}{4}+j\sin\frac{\pi}{4}\right)\) | M1 | |
| \(= \frac{3\sqrt{2}}{4} + \frac{3\sqrt{2}}{4}j\) | A1 | \(a=b=\frac{3\sqrt{2}}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-\frac{1}{2}j\theta}+e^{\frac{1}{2}j\theta} = 2\cos\frac{1}{2}\theta\) | M1 | Euler's formula |
| \(1+e^{j\theta} = e^{\frac{1}{2}j\theta}(e^{-\frac{1}{2}j\theta}+e^{\frac{1}{2}j\theta})\) | M1 | Factoring \(e^{\frac{1}{2}j\theta}\) |
| \(= 2e^{\frac{1}{2}j\theta}\cos\frac{1}{2}\theta\) | A1, A1 | Correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(C + jS = \sum_{r=0}^{n}\binom{n}{r}e^{jr\theta} = (1+e^{j\theta})^n\) | M1 | Binomial theorem |
| \(= \left(2e^{\frac{1}{2}j\theta}\cos\frac{1}{2}\theta\right)^n = 2^n e^{\frac{1}{2}jn\theta}\cos^n\frac{1}{2}\theta\) | M1, A1 | Using part (i) |
| \(C = 2^n\cos^n\frac{1}{2}\theta\cos\frac{1}{2}n\theta\) | A1 | Taking real part |
| \(S = 2^n\cos^n\frac{1}{2}\theta\sin\frac{1}{2}n\theta\) | A1 | Taking imaginary part |
| \(\frac{S}{C} = \frac{\sin\frac{1}{2}n\theta}{\cos\frac{1}{2}n\theta} = \tan\frac{1}{2}n\theta\) | M1, A1 | Dividing |
## Question 2:
**Part (a)(i):** Modulus and argument of $w$, $z$, $\frac{w}{z}$
| $|w|=3,\ \arg w = -\frac{\pi}{12}$ | B1 | |
| $|z|=2,\ \arg z = -\frac{\pi}{3}$ | B1, M1 | $|z|=\sqrt{1+3}=2$ |
| $\left|\frac{w}{z}\right|=\frac{3}{2},\ \arg\frac{w}{z}=\frac{\pi}{4}$ | A1, A1 | Dividing moduli, subtracting arguments |
**Part (a)(ii):** Write $\frac{w}{z}$ in form $a+bj$
| $\frac{w}{z} = \frac{3}{2}\left(\cos\frac{\pi}{4}+j\sin\frac{\pi}{4}\right)$ | M1 | |
| $= \frac{3\sqrt{2}}{4} + \frac{3\sqrt{2}}{4}j$ | A1 | $a=b=\frac{3\sqrt{2}}{4}$ |
**Part (b)(i):** Show $1+e^{j\theta}=2e^{\frac{1}{2}j\theta}\cos\frac{1}{2}\theta$
| $e^{-\frac{1}{2}j\theta}+e^{\frac{1}{2}j\theta} = 2\cos\frac{1}{2}\theta$ | M1 | Euler's formula |
| $1+e^{j\theta} = e^{\frac{1}{2}j\theta}(e^{-\frac{1}{2}j\theta}+e^{\frac{1}{2}j\theta})$ | M1 | Factoring $e^{\frac{1}{2}j\theta}$ |
| $= 2e^{\frac{1}{2}j\theta}\cos\frac{1}{2}\theta$ | A1, A1 | Correct conclusion |
**Part (b)(ii):** Find $C$ and $S$, show $\frac{S}{C}=\tan\frac{1}{2}n\theta$
| $C + jS = \sum_{r=0}^{n}\binom{n}{r}e^{jr\theta} = (1+e^{j\theta})^n$ | M1 | Binomial theorem |
| $= \left(2e^{\frac{1}{2}j\theta}\cos\frac{1}{2}\theta\right)^n = 2^n e^{\frac{1}{2}jn\theta}\cos^n\frac{1}{2}\theta$ | M1, A1 | Using part (i) |
| $C = 2^n\cos^n\frac{1}{2}\theta\cos\frac{1}{2}n\theta$ | A1 | Taking real part |
| $S = 2^n\cos^n\frac{1}{2}\theta\sin\frac{1}{2}n\theta$ | A1 | Taking imaginary part |
| $\frac{S}{C} = \frac{\sin\frac{1}{2}n\theta}{\cos\frac{1}{2}n\theta} = \tan\frac{1}{2}n\theta$ | M1, A1 | Dividing |
---2
\begin{enumerate}[label=(\alph*)]
\item You are given the complex numbers $w = 3 \mathrm { e } ^ { - \frac { 1 } { 12 } \pi \mathrm { j } }$ and $z = 1 - \sqrt { 3 } \mathrm { j }$.
\begin{enumerate}[label=(\roman*)]
\item Find the modulus and argument of each of the complex numbers $w , z$ and $\frac { w } { z }$.
\item Hence write $\frac { w } { z }$ in the form $a + b \mathrm { j }$, giving the exact values of $a$ and $b$.
\end{enumerate}\item In this part of the question, $n$ is a positive integer and $\theta$ is a real number with $0 < \theta < \frac { \pi } { n }$.
\begin{enumerate}[label=(\roman*)]
\item Express $\mathrm { e } ^ { - \frac { 1 } { 2 } \mathrm { j } \theta } + \mathrm { e } ^ { \frac { 1 } { 2 } \mathrm { j } \theta }$ in simplified trigonometric form, and hence, or otherwise, show that
$$1 + \mathrm { e } ^ { \mathrm { j } \theta } = 2 \mathrm { e } ^ { \frac { 1 } { 2 } \mathrm { j } \theta } \cos \frac { 1 } { 2 } \theta$$
Series $C$ and $S$ are defined by
$$\begin{aligned}
& C = 1 + \binom { n } { 1 } \cos \theta + \binom { n } { 2 } \cos 2 \theta + \binom { n } { 3 } \cos 3 \theta + \ldots + \binom { n } { n } \cos n \theta \\
& S = \binom { n } { 1 } \sin \theta + \binom { n } { 2 } \sin 2 \theta + \binom { n } { 3 } \sin 3 \theta + \ldots + \binom { n } { n } \sin n \theta
\end{aligned}$$
\item Find $C$ and $S$, and show that $\frac { S } { C } = \tan \frac { 1 } { 2 } n \theta$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2007 Q2 [18]}}