| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Roots of unity with derived equations |
| Difficulty | Challenging +1.2 This is a structured multi-part FP2 question on roots of unity with systematic algebraic manipulations. Part (a) is routine, parts (b)(i)-(iv) are guided 'show that' proofs requiring standard exponential form manipulations, and part (c) applies the setup to find specific roots. While lengthy (typical of FP2), each step is scaffolded with clear targets, requiring competent technique but minimal novel insight—moderately above average difficulty. |
| Spec | 4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(z = e^{\frac{2k\pi i}{6}}, k = 0, \pm 1, \pm 2, 3\) | M1, A2,1,0 | OE; M1A1 only if: (1) range for k is incorrect eg 0,1,2,3,4,5; (2) i is missing (3 marks) |
| (b)(i) \(\frac{w^2-1}{w} - w - \frac{1}{w} = 2i\sin\theta\) | M1A1 | AG (2 marks) |
| (ii) \(\frac{w}{w^2-1} = \frac{1}{2i\sin\theta} = \frac{i}{-2\sin\theta}\) | M1, A1 | AG (2 marks) |
| (iii) \(\frac{2i}{w^2-1} = \frac{-2i\nu^{-1}i}{2\sin\theta} = \frac{1}{\sin\theta}(\cos\theta - \sin\theta) = \cot\theta - i\) | M1, A1, A1 | AG (3 marks) |
| (iv) \(z = \frac{2i}{w^2-1}\) Or \(z + 2i = \frac{2i}{w^2-1} + 2i\); \(z + 2i = zw^2\) | M1, A1 | ie any correct method; AG (2 marks) |
| (c)(i) No coefficient of \(z^6\) | E1 | (1 mark) |
| (ii) \((w^2)^6 = 1\), \(w^2 = e^{\frac{k\pi i}{3}}\); \(z = \cot\frac{k\pi}{6} - i\), \(k = \pm 1, \pm 2, 3\) | B1, M1, A2,1,0 | Alternatively: \(z + 2i = e^{\frac{k\pi i}{3}}z\); \(z = \frac{2i}{e^{\frac{k\pi i}{3}} - 1}\); roots A2,1,0. (NB roots are \(\pm\sqrt{3} - i; \pm\frac{1}{\sqrt{3}} - i; -i\)) (4 marks) |
**(a)** $z = e^{\frac{2k\pi i}{6}}, k = 0, \pm 1, \pm 2, 3$ | M1, A2,1,0 | OE; M1A1 only if: (1) range for k is incorrect eg 0,1,2,3,4,5; (2) i is missing (3 marks)
**(b)(i)** $\frac{w^2-1}{w} - w - \frac{1}{w} = 2i\sin\theta$ | M1A1 | AG (2 marks)
**(ii)** $\frac{w}{w^2-1} = \frac{1}{2i\sin\theta} = \frac{i}{-2\sin\theta}$ | M1, A1 | AG (2 marks)
**(iii)** $\frac{2i}{w^2-1} = \frac{-2i\nu^{-1}i}{2\sin\theta} = \frac{1}{\sin\theta}(\cos\theta - \sin\theta) = \cot\theta - i$ | M1, A1, A1 | AG (3 marks)
**(iv)** $z = \frac{2i}{w^2-1}$ Or $z + 2i = \frac{2i}{w^2-1} + 2i$; $z + 2i = zw^2$ | M1, A1 | ie any correct method; AG (2 marks)
**(c)(i)** No coefficient of $z^6$ | E1 | (1 mark)
**(ii)** $(w^2)^6 = 1$, $w^2 = e^{\frac{k\pi i}{3}}$; $z = \cot\frac{k\pi}{6} - i$, $k = \pm 1, \pm 2, 3$ | B1, M1, A2,1,0 | **Alternatively:** $z + 2i = e^{\frac{k\pi i}{3}}z$; $z = \frac{2i}{e^{\frac{k\pi i}{3}} - 1}$; roots A2,1,0. (NB roots are $\pm\sqrt{3} - i; \pm\frac{1}{\sqrt{3}} - i; -i$) (4 marks)
**Total: 17 marks**
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# TOTAL: 75 marks7
\begin{enumerate}[label=(\alph*)]
\item Find the six roots of the equation $z ^ { 6 } = 1$, giving your answers in the form $\mathrm { e } ^ { \mathrm { i } \phi }$, where $- \pi < \phi \leqslant \pi$.
\item It is given that $w = \mathrm { e } ^ { \mathrm { i } \theta }$, where $\theta \neq n \pi$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac { w ^ { 2 } - 1 } { w } = 2 \mathrm { i } \sin \theta$.
\item Show that $\frac { w } { w ^ { 2 } - 1 } = - \frac { \mathrm { i } } { 2 \sin \theta }$.
\item Show that $\frac { 2 \mathrm { i } } { w ^ { 2 } - 1 } = \cot \theta - \mathrm { i }$.
\item Given that $z = \cot \theta - \mathrm { i }$, show that $z + 2 \mathrm { i } = z w ^ { 2 }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Explain why the equation
$$( z + 2 \mathrm { i } ) ^ { 6 } = z ^ { 6 }$$
has five roots.
\item Find the five roots of the equation
$$( z + 2 \mathrm { i } ) ^ { 6 } = z ^ { 6 }$$
giving your answers in the form $a + \mathrm { i } b$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2006 Q7 [17]}}