| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Sum geometric series with complex terms |
| Difficulty | Standard +0.3 This is a standard Further Maths question combining two routine techniques: (a) uses the geometric series formula with complex exponentials (a well-known method for summing trigonometric series), and (b) involves standard polar form conversions and nth roots of complex numbers. Both parts are textbook exercises requiring methodical application of learned techniques rather than problem-solving insight. Slightly above average difficulty due to being Further Maths content, but straightforward for the target cohort. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(C+jS = 1+ae^{j\theta}+a^2e^{2j\theta}+\ldots\) | M1 | Forming \(C+jS\) as a series of powers. Powers must be correct. \(\ldots a^2(\cos 2\theta + j\sin 2\theta)\) insufficient |
| This is a geometric series with \(r=ae^{j\theta}\) | M1 | Identifying G.P. and attempting sum. Dependent on first M1 |
| Sum to infinity \(= \frac{1}{1-ae^{j\theta}}\) | A1 | |
| \(= \frac{1}{1-ae^{j\theta}}\times\frac{1-ae^{-j\theta}}{1-ae^{-j\theta}}\) | M1* | Multiplying numerator and denominator by \(1-ae^{-j\theta}\) o.e. |
| \(= \frac{1-ae^{-j\theta}}{1-ae^{j\theta}-ae^{-j\theta}+a^2}\) | M1 | Multiplying out denominator. Dependent on M1*. Use of FOIL with powers combined correctly (allow one slip) |
| \(= \frac{1-a(\cos\theta - j\sin\theta)}{1-2a\cos\theta+a^2}\) | M1 | Introducing trig functions. Dependent on M1*. Condone e.g. \(e^{-j\theta}=\cos\theta+j\sin\theta\) |
| \(= \frac{1-a\cos\theta}{1-2a\cos\theta+a^2} + \frac{aj\sin\theta}{1-2a\cos\theta+a^2}\) | ||
| \(\Rightarrow C = \frac{1-a\cos\theta}{1-2a\cos\theta+a^2}\) | E1 | Answer given. www which leads to \(C\) |
| and \(S = \frac{a\sin\theta}{1-2a\cos\theta+a^2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Modulus \(= 2\) | B1 | |
| Argument \(= \frac{2\pi}{3}\) | B1 | |
| \(\Rightarrow -1+j\sqrt{3} = 2e^{j\frac{2\pi}{3}}\) | ||
| \(\Rightarrow\) fourth roots have \(r = \sqrt[4]{2}\) | B1 | Allow 1.19 or better |
| and \(\theta = \frac{\pi}{6}\) | ||
| \(\theta = \frac{\pi}{6} + \frac{2k\pi}{4}\) scores M1 | M1 | \(\div\) arg \(z\) by 4 and adding \(\frac{\pi}{2}\). \(\theta = \frac{\pi}{6}+\frac{2k\pi}{4}\) scores M1; \(k=0,1,2,3\) (or \(-2,-1,0,1\)) A1 |
| \(\Rightarrow\) roots are \(\sqrt[4]{2}e^{j\frac{\pi}{6}},\,\sqrt[4]{2}e^{j\frac{2\pi}{3}},\,\sqrt[4]{2}e^{j\frac{7\pi}{6}},\,\sqrt[4]{2}e^{j\frac{5\pi}{3}}\) | A1 | All arguments correct |
| Product of \(4^{\text{th}}\) roots \(= 2e^{j(1+4+7+10)\frac{\pi}{6}}\) | M1 | Attempting to find product |
| \(= 2e^{j\frac{5\pi}{3}}\) | A1 | Or \(-\frac{\pi}{3}\) o.e. |
| Argand diagram: position of \(z\) in 2nd quadrant; roots forming square; position of product | G1, G1ft, G1ft | In 2nd quadrant; ignore marked angles; correct or their \(-z\); must consider both modulus and argument |
# Question 2:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $C+jS = 1+ae^{j\theta}+a^2e^{2j\theta}+\ldots$ | M1 | Forming $C+jS$ as a series of powers. Powers must be correct. $\ldots a^2(\cos 2\theta + j\sin 2\theta)$ insufficient |
| This is a geometric series with $r=ae^{j\theta}$ | M1 | Identifying G.P. and attempting sum. Dependent on first M1 |
| Sum to infinity $= \frac{1}{1-ae^{j\theta}}$ | A1 | |
| $= \frac{1}{1-ae^{j\theta}}\times\frac{1-ae^{-j\theta}}{1-ae^{-j\theta}}$ | M1* | Multiplying numerator and denominator by $1-ae^{-j\theta}$ o.e. |
| $= \frac{1-ae^{-j\theta}}{1-ae^{j\theta}-ae^{-j\theta}+a^2}$ | M1 | Multiplying out denominator. Dependent on M1*. Use of FOIL with powers combined correctly (allow one slip) |
| $= \frac{1-a(\cos\theta - j\sin\theta)}{1-2a\cos\theta+a^2}$ | M1 | Introducing trig functions. Dependent on M1*. Condone e.g. $e^{-j\theta}=\cos\theta+j\sin\theta$ |
| $= \frac{1-a\cos\theta}{1-2a\cos\theta+a^2} + \frac{aj\sin\theta}{1-2a\cos\theta+a^2}$ | | |
| $\Rightarrow C = \frac{1-a\cos\theta}{1-2a\cos\theta+a^2}$ | E1 | Answer given. www which leads to $C$ |
| and $S = \frac{a\sin\theta}{1-2a\cos\theta+a^2}$ | A1 | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Modulus $= 2$ | B1 | |
| Argument $= \frac{2\pi}{3}$ | B1 | |
| $\Rightarrow -1+j\sqrt{3} = 2e^{j\frac{2\pi}{3}}$ | | |
| $\Rightarrow$ fourth roots have $r = \sqrt[4]{2}$ | B1 | Allow 1.19 or better |
| and $\theta = \frac{\pi}{6}$ | | |
| $\theta = \frac{\pi}{6} + \frac{2k\pi}{4}$ scores M1 | M1 | $\div$ arg $z$ by 4 and adding $\frac{\pi}{2}$. $\theta = \frac{\pi}{6}+\frac{2k\pi}{4}$ scores M1; $k=0,1,2,3$ (or $-2,-1,0,1$) A1 |
| $\Rightarrow$ roots are $\sqrt[4]{2}e^{j\frac{\pi}{6}},\,\sqrt[4]{2}e^{j\frac{2\pi}{3}},\,\sqrt[4]{2}e^{j\frac{7\pi}{6}},\,\sqrt[4]{2}e^{j\frac{5\pi}{3}}$ | A1 | All arguments correct |
| Product of $4^{\text{th}}$ roots $= 2e^{j(1+4+7+10)\frac{\pi}{6}}$ | M1 | Attempting to find product |
| $= 2e^{j\frac{5\pi}{3}}$ | A1 | Or $-\frac{\pi}{3}$ o.e. |
| Argand diagram: position of $z$ in 2nd quadrant; roots forming square; position of product | G1, G1ft, G1ft | In 2nd quadrant; ignore marked angles; correct or their $-z$; must consider both modulus and argument |
---2
\begin{enumerate}[label=(\alph*)]
\item The infinite series $C$ and $S$ are defined as follows.
$$\begin{aligned}
& C = 1 + a \cos \theta + a ^ { 2 } \cos 2 \theta + \ldots \\
& S = a \sin \theta + a ^ { 2 } \sin 2 \theta + a ^ { 3 } \sin 3 \theta + \ldots
\end{aligned}$$
where $a$ is a real number and $| a | < 1$.\\
By considering $C + \mathrm { j } S$, show that $C = \frac { 1 - a \cos \theta } { 1 + a ^ { 2 } - 2 a \cos \theta }$ and find a corresponding expression for $S$.
\item Express the complex number $z = - 1 + \mathrm { j } \sqrt { 3 }$ in the form $r \mathrm { e } ^ { \mathrm { j } \theta }$.
Find the 4th roots of $z$ in the form $r \mathrm { e } ^ { \mathrm { j } \theta }$.\\
Show $z$ and its 4th roots in an Argand diagram.\\
Find the product of the 4th roots and mark this as a point on your Argand diagram.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2012 Q2 [18]}}