| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Modulus-argument form conversion |
| Difficulty | Moderate -0.3 This is a standard modulus-argument form conversion question with a common error (quadrant identification) that students are guided through step-by-step. Parts (a)-(c) are routine Further Maths AS content requiring basic trigonometric values and understanding of argument in different quadrants. Part (d) is trivial recall. The scaffolding makes this easier than a typical Further Maths question, placing it slightly below average overall difficulty. |
| Spec | 4.02b Express complex numbers: cartesian and modulus-argument forms4.02f Convert between forms: cartesian and modulus-argument |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right) = 2\left(\frac{1}{2} + i\left(-\frac{\sqrt{3}}{2}\right)\right) = 1 - i\sqrt{3}\) | B1 | Accept \(-(-1 + i\sqrt{3})\); condone no conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| There are two solutions to \(\tan\theta = -\sqrt{3}\) in the interval \(-\pi < \theta \leq \pi\); Abdoallah has chosen the wrong one; \(-1 + i\sqrt{3}\) is in the 2nd quadrant so \(\theta\) should be obtuse | B1 | Accept any indication that there is another solution to \(\tan\theta = -\sqrt{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = -\frac{\pi}{3} + \pi = \frac{2\pi}{3}\); \(-1 + i\sqrt{3} = 2\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right)\) | B1 | Obtains correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(-1 - i\sqrt{3}\) | B1 | Writes correct answer in any form |
## Question 8:
### Part 8(a):
$2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right) = 2\left(\frac{1}{2} + i\left(-\frac{\sqrt{3}}{2}\right)\right) = 1 - i\sqrt{3}$ | B1 | Accept $-(-1 + i\sqrt{3})$; condone no conclusion
### Part 8(b):
There are two solutions to $\tan\theta = -\sqrt{3}$ in the interval $-\pi < \theta \leq \pi$; Abdoallah has chosen the wrong one; $-1 + i\sqrt{3}$ is in the 2nd quadrant so $\theta$ should be obtuse | B1 | Accept any indication that there is another solution to $\tan\theta = -\sqrt{3}$
### Part 8(c):
$\theta = -\frac{\pi}{3} + \pi = \frac{2\pi}{3}$; $-1 + i\sqrt{3} = 2\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right)$ | B1 | Obtains correct answer
### Part 8(d):
$-1 - i\sqrt{3}$ | B1 | Writes correct answer in any form
---8 Abdoallah wants to write the complex number $- 1 + \mathrm { i } \sqrt { 3 }$ in the form $r ( \cos \theta + \mathrm { i } \sin \theta )$ where $r \geq 0$ and $- \pi < \theta \leq \pi$
Here is his method:
$$\begin{array} { r l r l }
r & = \sqrt { ( - 1 ) ^ { 2 } + ( \sqrt { 3 } ) ^ { 2 } } & & \tan \theta = \frac { \sqrt { 3 } } { - 1 } \\
& = \sqrt { 1 + 3 } & & \Rightarrow \\
& = \sqrt { 4 } & & \tan \theta = - \sqrt { 3 } \\
& = 2 & & \theta = \tan ^ { - 1 } ( - \sqrt { 3 } ) \\
& & \theta = - \frac { \pi } { 3 } \\
& - 1 + i \sqrt { 3 } = 2 \left( \cos \left( - \frac { \pi } { 3 } \right) + i \sin \left( - \frac { \pi } { 3 } \right) \right)
\end{array}$$
There is an error in Abdoallah's method.
8
\begin{enumerate}[label=(\alph*)]
\item Show that Abdoallah's answer is wrong by writing
$$2 \left( \cos \left( - \frac { \pi } { 3 } \right) + i \sin \left( - \frac { \pi } { 3 } \right) \right)$$
in the form $x + \mathrm { i } y$\\
Simplify your answer.\\
8
\item Explain the error in Abdoallah's method.\\
8
\item Express $- 1 + \mathrm { i } \sqrt { 3 }$ in the form $r ( \cos \theta + \mathrm { i } \sin \theta )$\\
8
\item Write down the complex conjugate of $- 1 + i \sqrt { 3 }$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2023 Q8 [4]}}