| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Standard +0.3 This is a standard FP2 complex numbers question covering De Moivre's theorem and exponential form. Part (a) involves routine application of De Moivre's theorem and binomial expansion—techniques that are core FP2 syllabus material. Part (b) requires finding roots of complex numbers in exponential form using standard formulas, which is straightforward once the method is known. While it requires multiple steps and careful bookkeeping with angles, it demands no novel insight and follows predictable patterns that students practice extensively. Slightly easier than average A-level due to being methodical rather than conceptually challenging. |
| Spec | 4.02d Exponential form: re^(i*theta)4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks |
|---|---|
| B1 | |
| B1 | |
| Total | 2 |
| Answer | Marks |
|---|---|
| M1 | Expanding \((z + z^{-1})^6\) |
| M1 | Using \(z^n + z^{-n} = 2\cos n\theta\) with \(n = 2, 4\) or \(6\). Allow M1 if 2 omitted, etc. |
| A1 (ag) | |
| Total | 3 |
| Answer | Marks |
|---|---|
| B1 | |
| M1 | Using (i) as in part (ii) |
| A1 | Correct expression in any form |
| M1 | Attempting to add or subtract |
| A1 | |
| B1 | This used |
| M1 | Obtaining an expression for \(\cos^6\theta\) |
| A1 | Correct expression in any form |
| M1A1 | Attempting to add or subtract |
| Total | 5 |
| Answer | Marks |
|---|---|
| M1 | Correctly manipulating modulus and argument |
| A1 | \(\sqrt{8}, \frac{7\pi}{6}\) or \(\frac{5\pi}{6}\) Condone \(r(c + js)\) |
| M1 | Correctly manipulating modulus and argument |
| A1 | \(2, \frac{13\pi}{9}\) or \(\frac{5\pi}{9}\). Condone \(r(c + js)\) |
| Total | 6 |
| Answer | Marks |
|---|---|
| M1 | Correctly manipulating modulus and argument |
| A1 | Accept any equivalent form |
| A1 | |
| Total | 3 |
## (a)(i)
$z^6 + z^{-6} = 2\cos 6\theta$
$z^6 - z^{-6} = 2j\sin 6\theta$
| | B1 | |
| | B1 | |
| **Total** | **2** | |
## (a)(ii)
$(z + z^{-1})^6 = z^6 + 6z^4 + 15z^2 + 20 + 15z^{-2} + 6z^{-4} + z^{-6}$
$= (z^6 + z^{-6}) + 6(z^4 + z^{-4}) + 15(z^2 + z^{-2}) + 20$
$\Rightarrow 64\cos^6\theta = 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20$
$\Rightarrow \cos^6\theta = \frac{1}{32}\cos 6\theta + \frac{3}{16}\cos 4\theta + \frac{15}{32}\cos 2\theta + \frac{5}{16}$
$\Rightarrow \cos^6\theta = \frac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)$
| | M1 | Expanding $(z + z^{-1})^6$ |
| | M1 | Using $z^n + z^{-n} = 2\cos n\theta$ with $n = 2, 4$ or $6$. Allow M1 if 2 omitted, etc. |
| | A1 (ag) | |
| **Total** | **3** | |
## (a)(iii)
$(z - z^{-1})^6 = z^6 + z^{-6} - 6(z^4 + z^{-4}) + 15(z^2 + z^{-2}) - 20$
$\Rightarrow -64\sin^6\theta = 2\cos 6\theta - 12\cos 4\theta + 30\cos 2\theta - 20$
$\Rightarrow -\sin^6\theta = \frac{1}{32}\cos 6\theta - \frac{3}{16}\cos 4\theta + \frac{15}{32}\cos 2\theta - \frac{5}{16}$
$\Rightarrow \cos^6\theta - \sin^6\theta = \frac{1}{16}\cos 6\theta + \frac{3}{8}\cos 4\theta$
---
**OR** $\cos^6\theta = \frac{1}{2}(\cos 2\theta + 1)$
$16\cos^6\theta = 2\cos 4\theta + 8\cos 2\theta + 6$
$\Rightarrow \cos^6\theta = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}$
$\cos^6\theta - \sin^6\theta = 2\cos^6\theta - 3\cos^2\theta + 3\cos^2\theta - 1$
$= \frac{1}{6}\cos 6\theta + \frac{3}{8}\cos 4\theta$
| | B1 | |
| | M1 | Using (i) as in part (ii) |
| | A1 | Correct expression in any form |
| | M1 | Attempting to add or subtract |
| | A1 | |
| | --- | --- |
| | B1 | This used |
| | M1 | Obtaining an expression for $\cos^6\theta$ |
| | A1 | Correct expression in any form |
| | M1A1 | Attempting to add or subtract |
| **Total** | **5** | |
## (b)(i)
$z_1^3 = 8e^{\frac{j\pi}{6}} \Rightarrow z_1 = 2\sqrt[4]{2}e^{j(\frac{\pi}{6} + \frac{4\pi}{3})}$
$= 2\sqrt{2}e^{\frac{j7\pi}{6}}$
$z_2^3 = 8e^{\frac{j(x + 4\pi)}{3}} \Rightarrow z_2 = 2e^{\frac{j13\pi}{9}}$
$= 2e^{\frac{j13\pi}{9}}$
| | M1 | Correctly manipulating modulus and argument |
| | A1 | $\sqrt{8}, \frac{7\pi}{6}$ or $\frac{5\pi}{6}$ Condone $r(c + js)$ |
| | M1 | Correctly manipulating modulus and argument |
| | A1 | $2, \frac{13\pi}{9}$ or $\frac{5\pi}{9}$. Condone $r(c + js)$ |
| **Total** | **6** | |
## (b)(ii)
$z_1 z_2 = 2\sqrt{2}e^{\frac{j7\pi}{6}} \times 2e^{\frac{j13\pi}{9}}$
$= 4\sqrt{2}e^{j(\frac{7\pi + 13\pi}{6\quad 9})}$
$= 4\sqrt{2}e^{\frac{j11\pi}{6}}$
$= 4\sqrt{2}e^{\frac{j11\pi}{6}}$
Lies in second quadrant
| | M1 | Correctly manipulating modulus and argument |
| | A1 | Accept any equivalent form |
| | A1 | |
| **Total** | **3** | |
---\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $z = \cos \theta + j \sin \theta$, express $z^n + z^{-n}$ and $z^n - z^{-n}$ in simplified trigonometrical form. [2]
\item By considering $(z + z^{-1})^6$, show that
$$\cos^6 \theta = \frac{1}{32}(\cos 6\theta + 6 \cos 4\theta + 15 \cos 2\theta + 10).$$ [3]
\item Obtain an expression for $\cos^6 \theta - \sin^6 \theta$ in terms of $\cos 2\theta$ and $\cos 6\theta$. [5]
\end{enumerate}
\item The complex number $w$ is $8e^{i\pi/3}$. You are given that $z_1$ is a square root of $w$ and that $z_2$ is a cube root of $w$. The points representing $z_1$ and $z_2$ in the Argand diagram both lie in the third quadrant.
\begin{enumerate}[label=(\roman*)]
\item Find $z_1$ and $z_2$ in the form $re^{i\theta}$. Draw an Argand diagram showing $w$, $z_1$ and $z_2$. [6]
\item Find the product $z_1z_2$, and determine the quadrant of the Argand diagram in which it lies. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2011 Q2 [19]}}