| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Modulus and argument calculations |
| Difficulty | Standard +0.3 This is a standard FP2 complex numbers question testing routine techniques: finding modulus (trivial calculation), sketching a circle locus (standard), finding argument from geometry (straightforward trigonometry), and solving z³=w using De Moivre's theorem (textbook procedure). While it requires multiple steps and FP2 content, each part follows well-practiced methods with no novel insight needed, placing it slightly above average difficulty. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02k Argand diagrams: geometric interpretation4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \((-4\sqrt{3})^2 + 4^2 = (48 + 16)\) | M1 | PI by correct answer |
| (Modulus) = 8 | A1 | 2 marks |
| (b)(i) circle | M1 | condone freehand circle |
| centre at \(-4\sqrt{3} + 4i\) | A1 | |
| circle touching negative real axis and not meeting imaginary axis | A1 | 3 marks |
| (b)(ii) Right angled triangle hyp = 8 & radius = 4 | M1 | |
| & \(\alpha = \frac{\pi}{6}\) as in diagram | [diagram shown: right triangle with hypotenuse 8 on real axis, height 4] | |
| May consider the triangle with one side on real axis but only earns M1 when angle doubled to \(\frac{2\pi}{3}\) | ||
| \(\arg w = \frac{2\pi}{3}\) | A1 | 2 marks |
| (c) \(r = (8)^{\frac{1}{3}} (= 2)\) | B1F | \(r = \)(modulus from (a))\(^{\frac{1}{3}}\) |
| \(\arg(-4\sqrt{3}+4i) = \frac{5\pi}{6}\) | B1 | |
| Use of de Moivre "their" arg/3 | M1 | 3 correct values of \(\theta\) mod \(2\pi\) |
| \(\theta = \frac{5\pi}{18}, \frac{17\pi}{18}, \frac{-7\pi}{18}\) | A1 | eg third angle is \(\frac{29\pi}{18}\) |
| Roots are \(2e^{i\frac{5\pi}{18}}, 2e^{i\frac{17\pi}{18}}, 2e^{(-\frac{7\pi}{18}i)}\) | A1 | 5 marks |
**(a)** $(-4\sqrt{3})^2 + 4^2 = (48 + 16)$ | M1 | PI by correct answer
(Modulus) = 8 | A1 | **2 marks**
**(b)(i)** circle | M1 | condone freehand circle
centre at $-4\sqrt{3} + 4i$ | A1 |
circle touching negative real axis and not meeting imaginary axis | A1 | **3 marks** | [diagram shown: circle in second quadrant touching negative real axis at $-4\sqrt{3}$, with radius 4]
**(b)(ii)** Right angled triangle hyp = 8 & radius = 4 | M1 |
& $\alpha = \frac{\pi}{6}$ as in diagram | | [diagram shown: right triangle with hypotenuse 8 on real axis, height 4]
| | | May consider the triangle with one side on real axis but only earns M1 when angle doubled to $\frac{2\pi}{3}$
$\arg w = \frac{2\pi}{3}$ | A1 | **2 marks** | must be exact but allow $\frac{4\pi}{6}$ etc
**(c)** $r = (8)^{\frac{1}{3}} (= 2)$ | B1F | $r = $(modulus from (a))$^{\frac{1}{3}}$
$\arg(-4\sqrt{3}+4i) = \frac{5\pi}{6}$ | B1 |
Use of de Moivre "their" arg/3 | M1 | 3 correct values of $\theta$ mod $2\pi$
$\theta = \frac{5\pi}{18}, \frac{17\pi}{18}, \frac{-7\pi}{18}$ | A1 | eg third angle is $\frac{29\pi}{18}$
Roots are $2e^{i\frac{5\pi}{18}}, 2e^{i\frac{17\pi}{18}}, 2e^{(-\frac{7\pi}{18}i)}$ | A1 | **5 marks** | must be in exactly this form for final mark; final root may be written as $2e^{-i\frac{7\pi}{18}}$ etc
**Total: 12 marks**
**Notes:**
- **(a)** NMS (Modulus) = 8 earns M1(implied) A1
- **(b)(i)** The two A1 marks are independent; first A1 PI by $-4\sqrt{3}$ marked on Re($z$) axis & 4 marked on Im($z$) axis; condone centre stated as $(-4\sqrt{3}, 4)$ for first A1 but withhold **first A1** if point of contact labelled as anything other than $-4\sqrt{3}$; second A1 is awarded if clear intention to touch the negative real axis but radius = 4 need not be marked
- **(b)(ii)** Condone arg $w = \ldots \frac{2\pi}{3}$
- **(c)** Example: $r = 2$; $\theta = \frac{2k\pi}{3} + \frac{5\pi}{18}$; $k = 0, 1, -1$ scores B1F, B1, M1, A1, A0
---\begin{enumerate}[label=(\alph*)]
\item Find the modulus of the complex number $-4\sqrt{3} + 4\mathrm{i}$, giving your answer as an integer. [2 marks]
\item The locus of points, $L$, satisfies the equation $|z + 4\sqrt{3} - 4\mathrm{i}| = 4$.
\begin{enumerate}[label=(\roman*)]
\item Sketch the locus $L$ on the Argand diagram below. [3 marks]
\item The complex number $w$ lies on $L$ so that $-\pi < \arg w \leq \pi$.
Find the least possible value of $\arg w$, giving your answer in terms of $\pi$. [2 marks]
\end{enumerate}
\item Solve the equation $z^3 = -4\sqrt{3} + 4\mathrm{i}$, giving your answers in the form $re^{\mathrm{i}\theta}$, where $r > 0$ and $-\pi < \theta \leq \pi$. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2016 Q5 [12]}}