FP2 June 2013 Q2
2. \[z = 5\sqrt{3} - 5\mathrm{i}\]
Find
\[w = 2\left(\cos\frac{\pi}{4} + \mathrm{i}\sin\frac{\pi}{4}\right)\]
Find
| Scheme | Marks |
|---|---|
| \(z = 5\sqrt{3} - 5\mathrm{i} = r(\cos\theta + \mathrm{i}\sin\theta)\) | |
| \(r = \sqrt{\left(5^2 \times 3 + 5^2\right)} = 10\) | B1 |
| (1) |
Notes
B1 for \(|z| = 10\) no working needed
| Scheme | Marks |
|---|---|
| \(\arg z = \arctan\left(-\dfrac{5}{5\sqrt{3}}\right) = -\dfrac{\pi}{6}\) \(\left(\text{or } -\dfrac{\pi}{6} \pm 2n\pi\right)\) | M1A1 |
| (2) |
Notes
M1 for \(\arg z = \arctan\left(\pm\dfrac{5}{5\sqrt{3}}\right)\), \(\tan(\arg z) = \pm\dfrac{5}{5\sqrt{3}}\), \(\arg z = \arctan\left(\pm\dfrac{5\sqrt{3}}{5}\right)\) or \(\tan(\arg z) = \pm\dfrac{5\sqrt{3}}{5}\) OR use their \(|z|\) with sin or cos used correctly
A1 for \(= -\dfrac{\pi}{6}\) \(\left(\text{or } -\dfrac{\pi}{6} \pm 2n\pi\right)\) (must be 4th quadrant)
| Scheme | Marks |
|---|---|
| \(\left|\dfrac{w}{z}\right| = \dfrac{2}{10} = \dfrac{1}{5}\) or 0.2 | B1 |
| (1) |
Notes
B1 for \(\left|\dfrac{w}{z}\right| = \dfrac{2}{10}\) or \(\dfrac{1}{5}\) or 0.2
| Scheme | Marks |
|---|---|
| \(\arg\left(\dfrac{w}{z}\right) = \dfrac{\pi}{4} - \left(-\dfrac{\pi}{6}\right),\ = \dfrac{5\pi}{12}\) \(\left(\text{or } \dfrac{5\pi}{12} \pm 2n\pi\right)\) | M1,A1 |
| (2) | |
| (6 marks) |
Notes
M1 for \(\arg\left(\dfrac{w}{z}\right) = \dfrac{\pi}{4} - \arg z\) using their \(\arg z\)
A1 for \(\dfrac{5\pi}{12}\) \(\left(\text{or } \dfrac{5\pi}{12} \pm 2n\pi\right)\)
Alternative for (d)
| Scheme | Marks |
|---|---|
| Find \(\dfrac{w}{z} = \dfrac{\left(\sqrt{6} - \sqrt{2}\right) + \left(\sqrt{6} + \sqrt{2}\right)\mathrm{i}}{20}\) | |
| \(\tan\left(\arg\dfrac{w}{z}\right) = \dfrac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}}\) | M1 from their \(\dfrac{w}{z}\) |
| \(\arg\left(\dfrac{w}{z}\right) = \dfrac{5\pi}{12}\) | A1 cao |
Work for (c) and (d) may be seen together – give B and A marks only if modulus and argument are clearly identified
ie \(\dfrac{1}{5}\left(\cos\dfrac{5\pi}{12} + \mathrm{i}\sin\dfrac{5\pi}{12}\right)\) alone scores B0M1A0