| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Complex number arithmetic and simplification |
| Difficulty | Standard +0.3 This is a standard Further Maths FP2 question testing routine application of modulus-argument form and de Moivre's theorem. Part (a) requires straightforward calculation of r and θ, part (b) is direct application of de Moivre's theorem for a power, and part (c) involves finding fourth roots using the standard formula. While it's Further Maths content (inherently harder), the question follows a completely standard template with no novel problem-solving required, making it slightly easier than average overall. |
| Spec | 4.02b Express complex numbers: cartesian and modulus-argument forms4.02c Complex notation: z, z*, Re(z), Im(z), |z|, arg(z)4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Modulus \(= 16\) | B1 | |
| Argument \(= \arctan(-\sqrt{3}) = \frac{2\pi}{3}\) | M1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z^3 = 16^3\!\left(\cos\!\frac{2\pi}{3} + i\sin\!\frac{2\pi}{3}\right)^3 = 16^3(\cos 2\pi + i\sin 2\pi) = 4096\) or \(16^3\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(w = 16^{\frac{1}{4}}\!\left(\cos\!\frac{2\pi}{3} + i\sin\!\frac{2\pi}{3}\right)^{\frac{1}{4}} = 2\!\left(\cos\!\frac{\pi}{6} + i\sin\!\frac{\pi}{6}\right)\ \left(= \sqrt{3}+i\right)\) | M1 A1ft | |
| OR \(-1+\sqrt{3}i\) OR \(-\sqrt{3}-i\) OR \(1-\sqrt{3}i\) | M1 A2(1,0) | (5 marks) |
## Question 4:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Modulus $= 16$ | B1 | |
| Argument $= \arctan(-\sqrt{3}) = \frac{2\pi}{3}$ | M1 A1 | (3 marks) |
**Part (b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z^3 = 16^3\!\left(\cos\!\frac{2\pi}{3} + i\sin\!\frac{2\pi}{3}\right)^3 = 16^3(\cos 2\pi + i\sin 2\pi) = 4096$ or $16^3$ | M1 A1 | (2 marks) |
**Part (c)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $w = 16^{\frac{1}{4}}\!\left(\cos\!\frac{2\pi}{3} + i\sin\!\frac{2\pi}{3}\right)^{\frac{1}{4}} = 2\!\left(\cos\!\frac{\pi}{6} + i\sin\!\frac{\pi}{6}\right)\ \left(= \sqrt{3}+i\right)$ | M1 A1ft | |
| OR $-1+\sqrt{3}i$ OR $-\sqrt{3}-i$ OR $1-\sqrt{3}i$ | M1 A2(1,0) | (5 marks) |4.
$$z = - 8 + ( 8 \sqrt { } 3 ) i$$
\begin{enumerate}[label=(\alph*)]
\item Find the modulus of $z$ and the argument of $z$.
Using de Moivre's theorem,
\item find $z ^ { 3 }$,
\item find the values of $w$ such that $w ^ { 4 } = z$, giving your answers in the form $a + \mathrm { i } b$, where $a , b \in \mathbb { R }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2010 Q4 [10]}}