| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Complex number arithmetic and simplification |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard complex number techniques: finding modulus/argument (routine calculation), applying de Moivre's theorem to find z^4 (direct application of formula), and finding square roots using half-angle formulas. All parts follow textbook procedures with no novel problem-solving required, though it's slightly above average difficulty due to being Further Maths content. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02h Square roots: of complex numbers4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(z = 6 - 6\sqrt{3}i \Rightarrow | z | = \sqrt{6^2 + (6\sqrt{3})^2} = 12\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Attempts \(\pm\arctan\left(\pm\frac{6\sqrt{3}}{6}\right)\) or e.g. \(\pm\tan^{-1}\left(\pm\frac{1}{\sqrt{3}}\right)\) | M1 | If arctan not seen allow e.g. \(\tan\alpha = \frac{6\sqrt{3}}{6} \Rightarrow \alpha = \frac{\pi}{3}\). If using sin or cos the hypotenuse must be 12 |
| \(\arg z = -\frac{\pi}{3}\) | A1* | Correct proof with no incorrect work. LHS required. Allow \(\theta =\) if consistent |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(z = \text{"12"}\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)\) or \(\text{"12"}e^{-\frac{\pi}{3}i}\) | M1 | Correct trig or exp form with their 12. Could be implied by \(z^4\) in trig or exp form |
| \(z^4 = 20736\left(\cos\left(-\frac{4\pi}{3}\right) + i\sin\left(-\frac{4\pi}{3}\right)\right)\) or \(20736e^{-\frac{4\pi}{3}i}\) | A1 | \(12^4\) evaluated and arg. of \(-\frac{4\pi}{3}\). Final answer: \(-10368 + 10368\sqrt{3}i\) or \(20736\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\) provided evidence of de Moivre |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(w = z^{\frac{1}{2}} = (\pm)\sqrt{\text{"12"}}\left(\cos\left(\frac{-\frac{\pi}{3}}{2}\right) + i\sin\left(\frac{-\frac{\pi}{3}}{2}\right)\right)\) or \((\pm)\text{"2}\sqrt{3}\text{"}e^{-\frac{\pi}{6}i}\) | M1 | Correct use of de Moivre's with \(-\frac{\pi}{3}\) and their 12 to attempt one square root. M0 if \(z^4\) used for \(z\) |
| \(w = 3 - \sqrt{3}i,\ -3 + \sqrt{3}i\) | A1ft, A1 | A1ft: One correct exact root. A1: Both exact roots (no others) correct in \(a+ib\) form |
# Question 2:
## Part 2(a)(i):
| Working | Mark | Notes |
|---------|------|-------|
| $z = 6 - 6\sqrt{3}i \Rightarrow |z| = \sqrt{6^2 + (6\sqrt{3})^2} = 12$ | B1 | +12 only. Accept if just stated |
## Part 2(a)(ii):
| Working | Mark | Notes |
|---------|------|-------|
| Attempts $\pm\arctan\left(\pm\frac{6\sqrt{3}}{6}\right)$ or e.g. $\pm\tan^{-1}\left(\pm\frac{1}{\sqrt{3}}\right)$ | M1 | If arctan not seen allow e.g. $\tan\alpha = \frac{6\sqrt{3}}{6} \Rightarrow \alpha = \frac{\pi}{3}$. If using sin or cos the hypotenuse must be 12 |
| $\arg z = -\frac{\pi}{3}$ | A1* | Correct proof with no incorrect work. LHS required. Allow $\theta =$ if consistent |
## Part 2(b):
| Working | Mark | Notes |
|---------|------|-------|
| $z = \text{"12"}\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)$ or $\text{"12"}e^{-\frac{\pi}{3}i}$ | M1 | Correct trig or exp form with their 12. Could be implied by $z^4$ in trig or exp form |
| $z^4 = 20736\left(\cos\left(-\frac{4\pi}{3}\right) + i\sin\left(-\frac{4\pi}{3}\right)\right)$ or $20736e^{-\frac{4\pi}{3}i}$ | A1 | $12^4$ evaluated and arg. of $-\frac{4\pi}{3}$. Final answer: $-10368 + 10368\sqrt{3}i$ or $20736\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)$ provided evidence of de Moivre |
## Part 2(c):
| Working | Mark | Notes |
|---------|------|-------|
| $w = z^{\frac{1}{2}} = (\pm)\sqrt{\text{"12"}}\left(\cos\left(\frac{-\frac{\pi}{3}}{2}\right) + i\sin\left(\frac{-\frac{\pi}{3}}{2}\right)\right)$ or $(\pm)\text{"2}\sqrt{3}\text{"}e^{-\frac{\pi}{6}i}$ | M1 | Correct use of de Moivre's with $-\frac{\pi}{3}$ and their 12 to attempt one square root. M0 if $z^4$ used for $z$ |
| $w = 3 - \sqrt{3}i,\ -3 + \sqrt{3}i$ | A1ft, A1 | A1ft: One correct exact root. A1: Both exact roots (no others) correct in $a+ib$ form |
---2.
$$z = 6 - 6 \sqrt { 3 } i$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Determine the modulus of $z$
\item Show that the argument of $z$ is $- \frac { \pi } { 3 }$
Using de Moivre's theorem, and making your method clear,
\end{enumerate}\item determine, in simplest form, $z ^ { 4 }$
\item Determine the values of $w$ such that $w ^ { 2 } = z$, giving your answers in the form $a + \mathrm { i } b$, where $a$ and $b$ are real numbers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2024 Q2 [8]}}