| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| 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 standard Further Maths FP2 question combining two classic techniques: deriving trig identities via de Moivre (requiring binomial expansion and equating imaginary parts) and finding nth roots with geometric interpretation. Part (a) is routine but algebraically intensive; parts (b)(i)-(iv) require multiple steps including converting to polar form, finding roots, geometric reasoning about midpoints, and computing powers. The question is moderately challenging for Further Maths students but follows well-established patterns without requiring novel insight. |
| Spec | 4.02d Exponential form: re^(i*theta)4.02k Argand diagrams: geometric interpretation4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| M1: Convert to modular form: \( | -2 + 2i | = 2\sqrt{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| M1: Calculate modulus of cube roots: \( | z_k | = 2^{1/2}\) or \(\sqrt[3]{2\sqrt{2}} = \sqrt{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| A1: \( | w | = 2\cos\frac{\pi}{12}\) or \( |
| Answer | Marks | Guidance |
|---|---|---|
| M1: Calculate \( | w^6 | = 2^6\cos^6\frac{\pi}{12} = 8\) (using \( |
**(a) Show $\sin 5\theta = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta$** [5]
M1: Use de Moivre's theorem: $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$
M1: Expand left side using binomial theorem
M1: Separate into real and imaginary parts
M1: Identify imaginary part equals $\sin 5\theta$
A1: Correct simplification to required form
**(b)(i) Find cube roots of $-2 + 2i$** [6]
M1: Convert to modular form: $|-2 + 2i| = 2\sqrt{2}$
M1: Find argument: $\arg(-2 + 2i) = \frac{3\pi}{4}$
M1: Express as $2\sqrt{2} e^{i(3\pi/4 + 2\pi k)}$ for $k = 0, 1, 2$
M1: Calculate modulus of cube roots: $|z_k| = 2^{1/2}$ or $\sqrt[3]{2\sqrt{2}} = \sqrt{2}$
M1: Calculate arguments: $\theta_k = \frac{\pi}{4}, \frac{11\pi}{12}, \frac{19\pi}{12}$ (or equivalent forms in range $[-\pi, \pi]$)
A1: All three roots in form $re^{i\theta}$ or $r(\cos\theta + i\sin\theta)$: $\sqrt{2}e^{i\pi/4}$, $\sqrt{2}e^{i11\pi/12}$, $\sqrt{2}e^{-i5\pi/12}$
**(b)(ii) Draw Argand diagram** [2]
B1: Three points correctly plotted at equal angular intervals of $\frac{2\pi}{3}$
B1: Point M (midpoint of A and B) correctly marked
**(b)(iii) Find modulus and argument of $w$** [2]
A1: $|w| = 2\cos\frac{\pi}{12}$ or $|w| = \frac{\sqrt{6}+\sqrt{2}}{2}$
A1: $\arg(w) = \frac{7\pi}{12}$
**(b)(iv) Find $w^6$ in form $a + bi$** [3]
M1: Calculate $|w^6| = 2^6\cos^6\frac{\pi}{12} = 8$ (using $|w|^6$)
M1: Calculate argument: $6 \times \frac{7\pi}{12} = \frac{7\pi}{2} = \frac{3\pi}{2} \pmod{2\pi}$
A1: $w^6 = -8i$2
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to show that $\sin 5 \theta = 5 \sin \theta - 20 \sin ^ { 3 } \theta + 16 \sin ^ { 5 } \theta$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the cube roots of $- 2 + 2 \mathrm { j }$ in the form $r \mathrm { e } ^ { \mathrm { j } \theta }$ where $r > 0$ and $- \pi < \theta \leqslant \pi$.
These cube roots are represented by points $\mathrm { A } , \mathrm { B }$ and C in the Argand diagram, with A in the first quadrant and ABC going anticlockwise. The midpoint of AB is M , and M represents the complex number $w$.
\item Draw an Argand diagram, showing the points $\mathrm { A } , \mathrm { B } , \mathrm { C }$ and M .
\item Find the modulus and argument of $w$.
\item Find $w ^ { 6 }$ in the form $a + b \mathrm { j }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2007 Q2 [18]}}