AQA FP2 2007 June — Question 8 13 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2007
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
Typenth roots via factorization
DifficultyChallenging +1.2 This is a structured Further Maths question on complex nth roots with clear scaffolding. Part (a)(i) is a straightforward quadratic in z³, (a)(ii) requires converting to modulus-argument form and finding cube roots (standard FP2 technique), part (b) is algebraic verification, and part (c) applies the result to factorize. While it involves multiple steps and Further Maths content, the heavy scaffolding and standard techniques make it more accessible than typical FP2 questions requiring independent insight.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02d Exponential form: re^(i*theta)4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers4.05c Partial fractions: extended to quadratic denominators
8
    1. Given that \(z ^ { 6 } - 4 z ^ { 3 } + 8 = 0\), show that \(z ^ { 3 } = 2 \pm 2 \mathrm { i }\).
    2. Hence solve the equation $$z ^ { 6 } - 4 z ^ { 3 } + 8 = 0$$ giving your answers in the form \(r \mathrm { e } ^ { \mathrm { i } \theta }\), where \(r > 0\) and \(- \pi < \theta \leqslant \pi\).
  2. Show that, for any real values of \(k\) and \(\theta\), $$\left( z - k \mathrm { e } ^ { \mathrm { i } \theta } \right) \left( z - k \mathrm { e } ^ { - \mathrm { i } \theta } \right) = z ^ { 2 } - 2 k z \cos \theta + k ^ { 2 }$$
  3. Express \(z ^ { 6 } - 4 z ^ { 3 } + 8\) as the product of three quadratic factors with real coefficients.
Question 8:
Part (a)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(z^3 = \dfrac{4 \pm \sqrt{16-32}}{2}\)M1
\(= 2 \pm 2i\)A1 AG
Part (a)(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(2+2i = 2\sqrt{2}e^{\frac{\pi i}{4}}\), \(\quad 2-2i = 2\sqrt{2}e^{\frac{-\pi i}{4}}\)M1, A1A1 M1 for either result or for one of \(r=2\sqrt{2}\), \(\theta = \pm\dfrac{\pi}{4}\)
\(z = \sqrt{2}e^{\frac{\pi i}{12}+\frac{2k\pi i}{3}}\) or \(\sqrt{2}e^{\frac{-\pi i}{12}+\frac{2k\pi i}{3}}\)M1 M1 for either; allow A1 for any 3 correct; ft errors in \(\pm\dfrac{\pi}{4}\)
\(z = \sqrt{2}\,e^{\frac{\pm\pi i}{12}},\ \sqrt{2}\,e^{\frac{\pm 3\pi i}{4}},\ \sqrt{2}\,e^{\frac{\pm 7\pi i}{12}}\)A2,1,0 F Total: 6 marks
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Multiplication of bracketsM1
Use of \(e^{i\theta} + e^{-i\theta} = 2\cos\theta\)A1 AG
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\left(z - \sqrt{2}e^{\frac{\pi i}{12}}\right)\!\left(z - \sqrt{2}e^{\frac{-\pi i}{12}}\right) = z^2 - 2\sqrt{2}\cos\dfrac{\pi}{12}\,z + 2\)M1A1F PI
Product is \(\left(z^2 - 2\sqrt{2}\cos\dfrac{7\pi}{12}\,z+2\right)\left(z^2 - 2\sqrt{2}\cos\dfrac{3\pi}{4}\,z+2\right)\)A1F \(\left(\text{or } z^2+2z+2\right)\); Total: 3 marks 
## Question 8:

### Part (a)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $z^3 = \dfrac{4 \pm \sqrt{16-32}}{2}$ | M1 | |
| $= 2 \pm 2i$ | A1 | AG |

---

### Part (a)(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $2+2i = 2\sqrt{2}e^{\frac{\pi i}{4}}$, $\quad 2-2i = 2\sqrt{2}e^{\frac{-\pi i}{4}}$ | M1, A1A1 | M1 for either result or for one of $r=2\sqrt{2}$, $\theta = \pm\dfrac{\pi}{4}$ |
| $z = \sqrt{2}e^{\frac{\pi i}{12}+\frac{2k\pi i}{3}}$ or $\sqrt{2}e^{\frac{-\pi i}{12}+\frac{2k\pi i}{3}}$ | M1 | M1 for either; allow A1 for any 3 correct; ft errors in $\pm\dfrac{\pi}{4}$ |
| $z = \sqrt{2}\,e^{\frac{\pm\pi i}{12}},\ \sqrt{2}\,e^{\frac{\pm 3\pi i}{4}},\ \sqrt{2}\,e^{\frac{\pm 7\pi i}{12}}$ | A2,1,0 F | Total: 6 marks |

---

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Multiplication of brackets | M1 | |
| Use of $e^{i\theta} + e^{-i\theta} = 2\cos\theta$ | A1 | AG |

---

### Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(z - \sqrt{2}e^{\frac{\pi i}{12}}\right)\!\left(z - \sqrt{2}e^{\frac{-\pi i}{12}}\right) = z^2 - 2\sqrt{2}\cos\dfrac{\pi}{12}\,z + 2$ | M1A1F | PI |
| Product is $\left(z^2 - 2\sqrt{2}\cos\dfrac{7\pi}{12}\,z+2\right)\left(z^2 - 2\sqrt{2}\cos\dfrac{3\pi}{4}\,z+2\right)$ | A1F | $\left(\text{or } z^2+2z+2\right)$; Total: 3 marks |
8
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $z ^ { 6 } - 4 z ^ { 3 } + 8 = 0$, show that $z ^ { 3 } = 2 \pm 2 \mathrm { i }$.
\item Hence solve the equation

$$z ^ { 6 } - 4 z ^ { 3 } + 8 = 0$$

giving your answers in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r > 0$ and $- \pi < \theta \leqslant \pi$.
\end{enumerate}\item Show that, for any real values of $k$ and $\theta$,

$$\left( z - k \mathrm { e } ^ { \mathrm { i } \theta } \right) \left( z - k \mathrm { e } ^ { - \mathrm { i } \theta } \right) = z ^ { 2 } - 2 k z \cos \theta + k ^ { 2 }$$
\item Express $z ^ { 6 } - 4 z ^ { 3 } + 8$ as the product of three quadratic factors with real coefficients.
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2007 Q8 [13]}}
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