AQA FP2 2010 January — Question 3 14 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2010
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeComplex roots with real coefficients
DifficultyStandard +0.8 This is a substantial FP2 question requiring complex conjugate roots, Vieta's formulas, conversion to exponential form, De Moivre's theorem, and algebraic manipulation of powers. While the techniques are standard for Further Maths, the multi-part structure with the final algebraic proof involving three roots and simplification to the given form requires careful execution across multiple steps, placing it moderately above average difficulty.
Spec4.01a Mathematical induction: construct proofs4.02d Exponential form: re^(i*theta)4.02g Conjugate pairs: real coefficient polynomials4.02i Quadratic equations: with complex roots4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae

3 The cubic equation $$2 z ^ { 3 } + p z ^ { 2 } + q z + 16 = 0$$ where \(p\) and \(q\) are real, has roots \(\alpha , \beta\) and \(\gamma\).
It is given that \(\alpha = 2 + 2 \sqrt { 3 } \mathrm { i }\).
    1. Write down another root, \(\beta\), of the equation.
    2. Find the third root, \(\gamma\).
    3. Find the values of \(p\) and \(q\).
    1. Express \(\alpha\) in the form \(r \mathrm { e } ^ { \mathrm { i } \theta }\), where \(r > 0\) and \(- \pi < \theta \leqslant \pi\).
    2. Show that $$( 2 + 2 \sqrt { 3 } \mathrm { i } ) ^ { n } = 4 ^ { n } \left( \cos \frac { n \pi } { 3 } + \mathrm { i } \sin \frac { n \pi } { 3 } \right)$$
    3. Show that $$\alpha ^ { n } + \beta ^ { n } + \gamma ^ { n } = 2 ^ { 2 n + 1 } \cos \frac { n \pi } { 3 } + \left( - \frac { 1 } { 2 } \right) ^ { n }$$ where \(n\) is an integer.

Part (a)(i)
AnswerMarks Guidance
\(\beta = 2 - 2\sqrt{3}i\)B1 1 mark
Part (a)(ii)
AnswerMarks Guidance
\(\alpha\beta = -8\)M1 Allow for +8 but not ±16
\(\alpha\beta = 16\)B1
\(\gamma = -\frac{1}{2}\)A1 3 marks total
Part (a)(iii)
AnswerMarks Guidance
Either \(-\frac{p}{2} = \alpha + \beta + \gamma\) or \(\frac{q}{2} = \alpha\beta + \beta\gamma + \gamma\alpha\)M1 SC if failure to divide by 2 throughout, allow M1A1 for either p or q correct ft
\(p = -7, q = 28\)A1F, A1F 3 marks
Alternative to (a)(i) and (a)(iii):
AnswerMarks
\((z^2 - 4z + 16)(az + b)\)M1
\(\alpha\beta = 16\)B1
\(a = 2, b = +1, \gamma = -\frac{1}{2}\)A1
Equating coefficientsM1
\(p = -7\)A1F
\(q = 28\)A1F
Part (b)(i)
AnswerMarks Guidance
\(r = 4, \theta = \frac{\pi}{3}\)B1, B1 2 marks
Part (b)(ii)
AnswerMarks Guidance
\((2 + 2\sqrt{3}i) = \left(4e^{i\frac{\pi}{3}}\right)\)M1
\(= 4^n\left(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3}\right)\)A1 2 marks
Part (b)(iii)
AnswerMarks Guidance
\((2 - 2\sqrt{3}i)^n = 4^n\left(\cos\frac{n\pi}{3} - i\sin\frac{n\pi}{3}\right)\)B1
\(\alpha^n + \beta^n + \gamma^n = 4^n\left(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3}\right) + 4^n\left(\cos\frac{n\pi}{3} - i\sin\frac{n\pi}{3}\right) + \left(-\frac{1}{2}\right)^n\)M1
\(= 2^{2n+1}\cos\frac{n\pi}{3} + \left(-\frac{1}{2}\right)^n\)A1 3 marks
### Part (a)(i)
$\beta = 2 - 2\sqrt{3}i$ | B1 | 1 mark

### Part (a)(ii)
$\alpha\beta = -8$ | M1 | Allow for +8 but not ±16
$\alpha\beta = 16$ | B1 |
$\gamma = -\frac{1}{2}$ | A1 | 3 marks total

### Part (a)(iii)
Either $-\frac{p}{2} = \alpha + \beta + \gamma$ or $\frac{q}{2} = \alpha\beta + \beta\gamma + \gamma\alpha$ | M1 | SC if failure to divide by 2 throughout, allow M1A1 for either p or q correct ft
$p = -7, q = 28$ | A1F, A1F | 3 marks | ft incorrect $\gamma$

**Alternative to (a)(i) and (a)(iii):**
$(z^2 - 4z + 16)(az + b)$ | M1 |
$\alpha\beta = 16$ | B1 |
$a = 2, b = +1, \gamma = -\frac{1}{2}$ | A1 |
Equating coefficients | M1 |
$p = -7$ | A1F |
$q = 28$ | A1F |

### Part (b)(i)
$r = 4, \theta = \frac{\pi}{3}$ | B1, B1 | 2 marks

### Part (b)(ii)
$(2 + 2\sqrt{3}i) = \left(4e^{i\frac{\pi}{3}}\right)$ | M1 |
$= 4^n\left(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3}\right)$ | A1 | 2 marks | AG

### Part (b)(iii)
$(2 - 2\sqrt{3}i)^n = 4^n\left(\cos\frac{n\pi}{3} - i\sin\frac{n\pi}{3}\right)$ | B1 |
$\alpha^n + \beta^n + \gamma^n = 4^n\left(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3}\right) + 4^n\left(\cos\frac{n\pi}{3} - i\sin\frac{n\pi}{3}\right) + \left(-\frac{1}{2}\right)^n$ | M1 |
$= 2^{2n+1}\cos\frac{n\pi}{3} + \left(-\frac{1}{2}\right)^n$ | A1 | 3 marks | AG
3 The cubic equation

$$2 z ^ { 3 } + p z ^ { 2 } + q z + 16 = 0$$

where $p$ and $q$ are real, has roots $\alpha , \beta$ and $\gamma$.\\
It is given that $\alpha = 2 + 2 \sqrt { 3 } \mathrm { i }$.\\
(a) (i) Write down another root, $\beta$, of the equation.\\
(ii) Find the third root, $\gamma$.\\
(iii) Find the values of $p$ and $q$.\\
(b) (i) Express $\alpha$ in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r > 0$ and $- \pi < \theta \leqslant \pi$.\\
(ii) Show that

$$( 2 + 2 \sqrt { 3 } \mathrm { i } ) ^ { n } = 4 ^ { n } \left( \cos \frac { n \pi } { 3 } + \mathrm { i } \sin \frac { n \pi } { 3 } \right)$$

(iii) Show that

$$\alpha ^ { n } + \beta ^ { n } + \gamma ^ { n } = 2 ^ { 2 n + 1 } \cos \frac { n \pi } { 3 } + \left( - \frac { 1 } { 2 } \right) ^ { n }$$

where $n$ is an integer.

\hfill \mbox{\textit{AQA FP2 2010 Q3 [14]}}