| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Standard +0.8 This is a substantial Further Maths question requiring multiple applications of De Moivre's theorem, binomial expansion, and manipulation of complex exponentials. Part (a)(ii) requires careful algebraic manipulation to derive a trigonometric identity, and part (b) involves finding complex roots and working with exponential form. While the techniques are standard for FP2, the multi-step nature and requirement to synthesize several concepts makes this moderately challenging, though not requiring novel insight. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02k Argand diagrams: geometric interpretation4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(z^n + \frac{1}{z^n} = 2 \cos n\theta\) | B1 [2] | Mark final answer |
| \(z^n - \frac{1}{z^n} = 2j \sin n\theta\) | B1 | Mark final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(z + \frac{1}{z}\right)^4 = z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4} = z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4} + 4\left(z^2 + \frac{1}{z^2}\right) + 6\) | M1 | Expanding by Binomial or complete equivalent |
| \(\Rightarrow (2\cos \theta)^4 = 2\cos 4\theta + 8\cos 2\theta + 6\) | M1 A1 | Introducing cosines of multiple angles; RHS correct; Condone lost 2s; Both As depend on both Ms |
| \(\Rightarrow \cos^4 \theta = \frac{3}{8} + \frac{1}{2}\cos 2\theta + \frac{1}{8}\cos 4\theta\) | A1ft [4] | Dividing both sides by 16. F.t. line above; \(A = \frac{3}{8}\), \(B = \frac{1}{2}\), \(C = \frac{1}{8}\); Give SC2 for fully correct answer found "otherwise" |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos^6 \theta = \frac{3}{8} + \frac{1}{2}\left(2\cos^2 \theta - 1\right) + \frac{1}{8}\cos 4\theta\) | M1 | Using (ii), obtaining \(\cos 4\theta\) and expressing \(\cos 2\theta\) in terms of \(\cos \theta\); Condone \(\cos 2\theta = \pm 1 \pm 2\cos^2 \theta\) |
| \(\Rightarrow \cos^6 \theta = \cos^2 \theta - \frac{1}{4} + \frac{1}{8}\cos 4\theta\) | A1 | |
| \(\Rightarrow \cos 4\theta = 8\cos^6 \theta - 8\cos^2 \theta + 1\) | A1 [2] | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = 4e^{j\frac{\pi}{6}}\) and \(w^2 = z; \Rightarrow w^2 = r^2 e^{j2\theta}\) | B1 | Condone \(r = \pm 2\) |
| \(\Rightarrow r^2 = 4 \Rightarrow r = 2\) | B1B1 | |
| and \(\theta = \frac{\pi}{6}, \frac{7\pi}{6}\) | Or \(-\frac{5\pi}{6}\); Award B2 for \(\pi\left(k + \frac{1}{6}\right)\) | |
| [A diagram showing two roots with approximately equal moduli and approximately correct argument] | B1 | Roots with approx. equal moduli and approx. correct argument |
| [Dependent on first B1] | B1 | Dependent on first B1; \(z\) in correct position |
| B1 [5] | Modulus and argument bigger |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = 4e^{j\frac{\pi}{3}} \Rightarrow z^n = 4^n e^{j\frac{n\pi}{3}}\) so if \(\frac{n\pi}{3} = n \Rightarrow n = 3\) | B1 | Ignore other larger values |
| Imaginary if \(\frac{n\pi}{3} = \frac{\pi}{2} + k\pi \Rightarrow n = \frac{3}{2} + 3k\) | M1 | \(\cos \frac{n\pi}{3} = 0\) or \(\frac{n\pi}{3} = \frac{\pi}{2}\)...; An argument which covers the positive and negative im. axis |
| which is not an integer for any \(k\) | A1(ag) | |
| \(w_1 = 2e^{j\frac{\pi}{6}} \Rightarrow w_1^3 = 8e^{j\frac{\pi}{2}} = 8j\) | M1 | Attempting their \(w^3\) in any form |
| \(w_2 = 2e^{j\frac{7\pi}{6}} \Rightarrow w_2^3 = 8e^{j\frac{7\pi}{2}} = -8j\) | A1 | 8j, -8j [5] |
### (a)(i)
$z^n + \frac{1}{z^n} = 2 \cos n\theta$ | B1 [2] | Mark final answer
$z^n - \frac{1}{z^n} = 2j \sin n\theta$ | B1 | Mark final answer
### (a)(ii)
$\left(z + \frac{1}{z}\right)^4 = z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4} = z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4} + 4\left(z^2 + \frac{1}{z^2}\right) + 6$ | M1 | Expanding by Binomial or complete equivalent
$\Rightarrow (2\cos \theta)^4 = 2\cos 4\theta + 8\cos 2\theta + 6$ | M1 A1 | Introducing cosines of multiple angles; RHS correct; Condone lost 2s; Both As depend on both Ms
$\Rightarrow \cos^4 \theta = \frac{3}{8} + \frac{1}{2}\cos 2\theta + \frac{1}{8}\cos 4\theta$ | A1ft [4] | Dividing both sides by 16. F.t. line above; $A = \frac{3}{8}$, $B = \frac{1}{2}$, $C = \frac{1}{8}$; Give SC2 for fully correct answer found "otherwise"
### (a)(iii)
$\cos^6 \theta = \frac{3}{8} + \frac{1}{2}\left(2\cos^2 \theta - 1\right) + \frac{1}{8}\cos 4\theta$ | M1 | Using (ii), obtaining $\cos 4\theta$ and expressing $\cos 2\theta$ in terms of $\cos \theta$; Condone $\cos 2\theta = \pm 1 \pm 2\cos^2 \theta$
$\Rightarrow \cos^6 \theta = \cos^2 \theta - \frac{1}{4} + \frac{1}{8}\cos 4\theta$ | A1 |
$\Rightarrow \cos 4\theta = 8\cos^6 \theta - 8\cos^2 \theta + 1$ | A1 [2] | c.a.o.
### (b)(i)
$z = 4e^{j\frac{\pi}{6}}$ and $w^2 = z; \Rightarrow w^2 = r^2 e^{j2\theta}$ | B1 | Condone $r = \pm 2$
$\Rightarrow r^2 = 4 \Rightarrow r = 2$ | B1B1 |
and $\theta = \frac{\pi}{6}, \frac{7\pi}{6}$ | | Or $-\frac{5\pi}{6}$; Award B2 for $\pi\left(k + \frac{1}{6}\right)$
[A diagram showing two roots with approximately equal moduli and approximately correct argument] | B1 | Roots with approx. equal moduli and approx. correct argument
[Dependent on first B1] | B1 | Dependent on first B1; $z$ in correct position
| B1 [5] | Modulus and argument bigger
### (b)(ii)
$z = 4e^{j\frac{\pi}{3}} \Rightarrow z^n = 4^n e^{j\frac{n\pi}{3}}$ so if $\frac{n\pi}{3} = n \Rightarrow n = 3$ | B1 | Ignore other larger values
Imaginary if $\frac{n\pi}{3} = \frac{\pi}{2} + k\pi \Rightarrow n = \frac{3}{2} + 3k$ | M1 | $\cos \frac{n\pi}{3} = 0$ or $\frac{n\pi}{3} = \frac{\pi}{2}$...; An argument which covers the positive and negative im. axis
which is not an integer for any $k$ | A1(ag) |
$w_1 = 2e^{j\frac{\pi}{6}} \Rightarrow w_1^3 = 8e^{j\frac{\pi}{2}} = 8j$ | M1 | Attempting their $w^3$ in any form
$w_2 = 2e^{j\frac{7\pi}{6}} \Rightarrow w_2^3 = 8e^{j\frac{7\pi}{2}} = -8j$ | A1 | 8j, -8j [5]
---2
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $z = \cos \theta + \mathrm { j } \sin \theta$, express $z ^ { n } + \frac { 1 } { z ^ { n } }$ and $z ^ { n } - \frac { 1 } { z ^ { n } }$ in simplified trigonometric form.
\item Beginning with an expression for $\left( z + \frac { 1 } { z } \right) ^ { 4 }$, find the constants $A , B , C$ in the identity
$$\cos ^ { 4 } \theta \equiv A + B \cos 2 \theta + C \cos 4 \theta$$
\item Use the identity in part (ii) to obtain an expression for $\cos 4 \theta$ as a polynomial in $\cos \theta$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that $z = 4 \mathrm { e } ^ { \mathrm { j } \pi / 3 }$ and that $w ^ { 2 } = z$, write down the possible values of $w$ in the form $r \mathrm { e } ^ { \mathrm { j } \theta }$, where $r > 0$. Show $z$ and the possible values of $w$ in an Argand diagram.
\item Find the least positive integer $n$ for which $z ^ { n }$ is real.
Show that there is no positive integer $n$ for which $z ^ { n }$ is imaginary.\\
For each possible value of $w$, find the value of $w ^ { 3 }$ in the form $a + \mathrm { j } b$ where $a$ and $b$ are real.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2012 Q2 [18]}}