| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Direct nth roots: general complex RHS |
| Difficulty | Standard +0.3 This is a standard Further Maths question on converting complex numbers to modulus-argument form and finding cube roots. Part (a) requires routine calculation of modulus and argument. Part (b) applies De Moivre's theorem systematically to find three cube roots. While it's Further Maths content (inherently harder), it follows a well-practiced algorithmic procedure with no novel insight required, making it slightly easier than average overall. |
| Spec | 4.02b Express complex numbers: cartesian and modulus-argument forms4.02d Exponential form: re^(i*theta)4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = \sqrt{(-4)^2 + (-4\sqrt{3})^2} = \ldots\) | M1 | Correct attempt at modulus; implied by correct modulus if no method seen |
| \(\tan\theta = \frac{-4\sqrt{3}}{-4} \Rightarrow \theta = \tan^{-1}(\sqrt{3}) \pm \pi\) | M1 | Attempt to find \(\theta\) in correct quadrant; accept \(\tan^{-1}(\sqrt{3})\pm\pi\) or \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\pm\pi\); may be implied by sight of \(-\frac{2}{3}\pi, \frac{4}{3}\pi, -\frac{5}{6}\pi, \frac{7}{6}\pi\) |
| \(8\left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right)\) | A1 | cao, no other solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = re^{i\theta} \Rightarrow (re^{i\theta})^3 = -4-4\sqrt{3} \Rightarrow r^3(e^{3i\theta}) = 8e^{-i\frac{2\pi}{3}}\) | ||
| \(\Rightarrow r = \sqrt[3]{8} = 2\) | M1 | Applies De Moivre's Theorem and finds value for \(r\), i.e. \((\text{their } 8)^{\frac{1}{3}}\) |
| \(3\theta = -\frac{2\pi}{3}(+2k\pi) \Rightarrow \theta = -\frac{2\pi}{9} + \left(\frac{2k\pi}{3}\right)\) | M1 | Proceeds to find at least one value for \(\theta\), i.e. their argument\(/3\) |
| So \(z = 2e^{-\frac{8\pi}{9}i},\ 2e^{-\frac{2\pi}{9}i},\ 2e^{\frac{4\pi}{9}i}\) | A1ft, A1 | A1ft: at least two roots correct for their \(r\) and \(\theta\); A1: all three correct roots and no others (accept e.g. \(2e^{i-\frac{8\pi}{9}}\) as slip in notation) |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = \sqrt{(-4)^2 + (-4\sqrt{3})^2} = \ldots$ | M1 | Correct attempt at modulus; implied by correct modulus if no method seen |
| $\tan\theta = \frac{-4\sqrt{3}}{-4} \Rightarrow \theta = \tan^{-1}(\sqrt{3}) \pm \pi$ | M1 | Attempt to find $\theta$ in correct quadrant; accept $\tan^{-1}(\sqrt{3})\pm\pi$ or $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\pm\pi$; may be implied by sight of $-\frac{2}{3}\pi, \frac{4}{3}\pi, -\frac{5}{6}\pi, \frac{7}{6}\pi$ |
| $8\left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right)$ | A1 | cao, no other solution |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = re^{i\theta} \Rightarrow (re^{i\theta})^3 = -4-4\sqrt{3} \Rightarrow r^3(e^{3i\theta}) = 8e^{-i\frac{2\pi}{3}}$ | | |
| $\Rightarrow r = \sqrt[3]{8} = 2$ | M1 | Applies De Moivre's Theorem and finds value for $r$, i.e. $(\text{their } 8)^{\frac{1}{3}}$ |
| $3\theta = -\frac{2\pi}{3}(+2k\pi) \Rightarrow \theta = -\frac{2\pi}{9} + \left(\frac{2k\pi}{3}\right)$ | M1 | Proceeds to find at least one value for $\theta$, i.e. their argument$/3$ |
| So $z = 2e^{-\frac{8\pi}{9}i},\ 2e^{-\frac{2\pi}{9}i},\ 2e^{\frac{4\pi}{9}i}$ | A1ft, A1 | A1ft: at least two roots correct for their $r$ and $\theta$; A1: all three correct roots and no others (accept e.g. $2e^{i-\frac{8\pi}{9}}$ as slip in notation) |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.\\
(a) Express the complex number
\end{enumerate}
$$- 4 - 4 \sqrt { 3 } i$$
in the form $r ( \cos \theta + \mathrm { i } \sin \theta )$, where $r > 0$ and $- \pi < \theta \leqslant \pi$\\
(b) Solve the equation
$$z ^ { 3 } + 4 + 4 \sqrt { 3 } i = 0$$
giving your answers in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r > 0$ and $- \pi < \theta \leqslant \pi$\\
\hfill \mbox{\textit{Edexcel F2 2022 Q1 [7]}}