| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Modulus and argument calculations |
| Difficulty | Moderate -0.8 This is a straightforward application of standard complex number formulas: finding modulus using √(a²+b²), argument from arctan(b/a) with a recognizable special angle (tan⁻¹(1/√3) = π/6), and using the properties |w/z| = |w|/|z| and arg(w/z) = arg(w) - arg(z). All steps are routine recall with no problem-solving required, making it easier than average even for Further Maths. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(r = \sqrt{5^2 \times 3 + 5^2} = 10\) | B1 (1) | No working needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\arg z = \arctan\left(-\dfrac{5}{5\sqrt{3}}\right) = -\dfrac{\pi}{6}\) | M1 | For \(\arg z = \arctan\left(\pm\dfrac{5}{5\sqrt{3}}\right)\), \(\tan(\arg z) = \pm\dfrac{5}{5\sqrt{3}}\), or use \( |
| \(= -\dfrac{\pi}{6}\) \(\left(\text{or } -\dfrac{\pi}{6} \pm 2n\pi\right)\) | A1 (2) | Must be 4th quadrant |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left\ | \dfrac{w}{z}\right\ | = \dfrac{2}{10} = \dfrac{1}{5}\) or \(0.2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\arg\left(\dfrac{w}{z}\right) = \dfrac{\pi}{4} - \left(-\dfrac{\pi}{6}\right) = \dfrac{5\pi}{12}\) | M1 | For \(\arg\left(\dfrac{w}{z}\right) = \dfrac{\pi}{4} - \arg z\) using their \(\arg z\) |
| \(= \dfrac{5\pi}{12}\) \(\left(\text{or } \dfrac{5\pi}{12} \pm 2n\pi\right)\) | A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Find \(\dfrac{w}{z} = \dfrac{(\sqrt{6}-\sqrt{2})+(\sqrt{6}+\sqrt{2})\mathrm{i}}{20}\) | ||
| \(\tan\left(\arg\dfrac{w}{z}\right) = \dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\) | M1 | From their \(\dfrac{w}{z}\) |
| \(\arg\left(\dfrac{w}{z}\right) = \dfrac{5\pi}{12}\) | A1 cao |
## Question 2:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $r = \sqrt{5^2 \times 3 + 5^2} = 10$ | B1 (1) | No working needed |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\arg z = \arctan\left(-\dfrac{5}{5\sqrt{3}}\right) = -\dfrac{\pi}{6}$ | M1 | For $\arg z = \arctan\left(\pm\dfrac{5}{5\sqrt{3}}\right)$, $\tan(\arg z) = \pm\dfrac{5}{5\sqrt{3}}$, or use $|z|$ with sin or cos used correctly |
| $= -\dfrac{\pi}{6}$ $\left(\text{or } -\dfrac{\pi}{6} \pm 2n\pi\right)$ | A1 (2) | Must be 4th quadrant |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left\|\dfrac{w}{z}\right\| = \dfrac{2}{10} = \dfrac{1}{5}$ or $0.2$ | B1 (1) | |
### Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\arg\left(\dfrac{w}{z}\right) = \dfrac{\pi}{4} - \left(-\dfrac{\pi}{6}\right) = \dfrac{5\pi}{12}$ | M1 | For $\arg\left(\dfrac{w}{z}\right) = \dfrac{\pi}{4} - \arg z$ using their $\arg z$ |
| $= \dfrac{5\pi}{12}$ $\left(\text{or } \dfrac{5\pi}{12} \pm 2n\pi\right)$ | A1 (2) | |
*Alternative for (d):*
| Answer | Mark | Guidance |
|--------|------|----------|
| Find $\dfrac{w}{z} = \dfrac{(\sqrt{6}-\sqrt{2})+(\sqrt{6}+\sqrt{2})\mathrm{i}}{20}$ | | |
| $\tan\left(\arg\dfrac{w}{z}\right) = \dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}$ | M1 | From their $\dfrac{w}{z}$ |
| $\arg\left(\dfrac{w}{z}\right) = \dfrac{5\pi}{12}$ | A1 cao | |
*Note: Work for (c) and (d) seen together — give B and A marks only if modulus and argument are clearly identified. $\dfrac{1}{5}\left(\cos\dfrac{5\pi}{12} + \mathrm{i}\sin\dfrac{5\pi}{12}\right)$ alone scores B0M1A0*
---2.
$$z = 5 \sqrt { } 3 - 5 i$$
Find
\begin{enumerate}[label=(\alph*)]
\item $| z |$,
\item $\arg ( z )$, in terms of $\pi$.
$$w = 2 \left( \cos \frac { \pi } { 4 } + i \sin \frac { \pi } { 4 } \right)$$
Find
\item $\left| \frac { w } { z } \right|$,
\item $\quad \arg \left( \frac { w } { z } \right)$, in terms of $\pi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2013 Q2 [6]}}