| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Roots of unity with trigonometric identities |
| Difficulty | Standard +0.8 This is a structured Further Maths question on roots of unity requiring multiple techniques: verifying roots, using geometric series sum, exploiting conjugate pairs to extract real parts, and manipulating complex exponentials. While guided through parts (a)-(c), it demands solid understanding of De Moivre's theorem and algebraic manipulation beyond standard A-level, placing it moderately above average difficulty. |
| Spec | 4.02n 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 |
|---|---|---|
| \(\left(e^{\frac{2\pi i}{7}}\right)^7 = e^{2\pi i} = 1\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Roots are \(\omega^0, \omega^1, \omega^3, \omega^4, \omega^6\) | M1A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Sum of roots considered = 0 | M1, A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\omega^2 + \omega^5 = e^{\frac{4\pi i}{7}} + e^{\frac{10\pi i}{7}}\) | M1 | |
| \(= e^{\frac{4\pi i}{7}} + e^{\frac{-4\pi i}{7}}\) | A1 | |
| \(= 2\cos\frac{4\pi}{7}\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\omega + \omega^6 = 2\cos\frac{2\pi}{7}\) ; \(\omega^3 + \omega^4 = 2\cos\frac{6\pi}{7}\) | B1, B1 | |
| Using part (b) Result | M1, A1 | 4 marks |
### Part (a)(i)
$\left(e^{\frac{2\pi i}{7}}\right)^7 = e^{2\pi i} = 1$ | B1 | 1 mark | Or $z^7 = e^{2\pi i}$, $z = e^{\frac{2\pi ki}{7}}$, $k = 1$
### Part (a)(ii)
Roots are $\omega^0, \omega^1, \omega^3, \omega^4, \omega^6$ | M1A1 | 2 marks | OE; M1A0 for incomplete set. SC B1 for a set of correct roots in terms of $e^{i\theta}$
### Part (b)
Sum of roots considered = 0 | M1, A1 | 2 marks | $\left\{\text{or } \sum_{r=0}^6 \omega^6 = \frac{\omega^7-1}{\omega-1} = 0\right\}$
### Part (c)(i)
$\omega^2 + \omega^5 = e^{\frac{4\pi i}{7}} + e^{\frac{10\pi i}{7}}$ | M1 |
$= e^{\frac{4\pi i}{7}} + e^{\frac{-4\pi i}{7}}$ | A1 |
$= 2\cos\frac{4\pi}{7}$ | A1 | 3 marks | AG
### Part (c)(ii)
$\omega + \omega^6 = 2\cos\frac{2\pi}{7}$ ; $\omega^3 + \omega^4 = 2\cos\frac{6\pi}{7}$ | B1, B1 |
Using part (b) Result | M1, A1 | 4 marks | AG
**TOTAL: 75 marks**8 (a) (i) Show that $\omega = \mathrm { e } ^ { \frac { 2 \pi \mathrm { i } } { 7 } }$ is a root of the equation $z ^ { 7 } = 1$.\\
(ii) Write down the five other non-real roots in terms of $\omega$.\\
(b) Show that
$$1 + \omega + \omega ^ { 2 } + \omega ^ { 3 } + \omega ^ { 4 } + \omega ^ { 5 } + \omega ^ { 6 } = 0$$
(c) Show that:\\
(i) $\quad \omega ^ { 2 } + \omega ^ { 5 } = 2 \cos \frac { 4 \pi } { 7 }$;\\
(ii) $\cos \frac { 2 \pi } { 7 } + \cos \frac { 4 \pi } { 7 } + \cos \frac { 6 \pi } { 7 } = - \frac { 1 } { 2 }$.
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\hfill \mbox{\textit{AQA FP2 2010 Q8 [12]}}