FP2 June 2015 Q2
2. \[z = -2 + \left(2\sqrt{3}\right)\mathrm{i}\]
Using de Moivre’s theorem,
| Scheme | Marks |
|---|---|
| \(|z| = 4\) | B1 |
| \(\arg z = \arctan\left(\dfrac{-2\sqrt{3}}{2}\right) = \arctan\left(-\sqrt{3}\right) = \dfrac{2\pi}{3}\) or \(120^\circ\) | M1A1 |
| (3) |
Notes
B1: Correct modulus seen Must be 4
M1: Attempt arg using arctan, nos either way up. Must include minus sign or other consideration of quadrant. (\(\arg = \dfrac{\pi}{3}\) scores M0)
A1: \(\dfrac{2\pi}{3}\) or \(120^\circ\) Correct answer only seen, award M1A1
| Scheme | Marks |
|---|---|
| \(z^6 = \left(4\left(\cos\dfrac{2\pi}{3} + \mathrm{i}\sin\dfrac{2\pi}{3}\right)\right)^6 = 4^6\left(\cos 4\pi + \mathrm{i}\sin 4\pi\right)\) or \(z^6 = \left(4\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}\right)^6\) | M1 |
| \(= 4096\) or \(4^6\) or \(2^{12}\) (a) and (b) can be marked together | A1 cso |
| (2) |
Notes
M1: apply de Moivre
A1cso: 4096 or \(4^6\) Must have been obtained with the correct argument for \(z\)
| Scheme | Marks |
|---|---|
| \(z^{\frac{3}{4}} = 4^{\frac{3}{4}}\left(\cos\dfrac{2\pi}{3} + \mathrm{i}\sin\dfrac{2\pi}{3}\right)^{\frac{3}{4}} = 4^{\frac{3}{4}}\left(\cos\dfrac{\pi}{2} + \mathrm{i}\sin\dfrac{\pi}{2}\right)\) | |
| \(w = \mathrm{i}2\sqrt{2}\) oe or any other correct root | B1 |
| \(4^{\frac{3}{4}}\left(\cos\left(\dfrac{2\pi}{3} + 2n\pi\right) + \mathrm{i}\sin\left(\dfrac{2\pi}{3} + 2n\pi\right)\right)^{\frac{3}{4}}\) | M1 |
| \((n = 0\) see above\()\) \(n = 1 \quad w = 2\sqrt{2}\) oe \(n = 2 \quad w = -\mathrm{i}2\sqrt{2}\) oe \(n = 3 \quad w = -2\sqrt{2}\) oe | A1A1 |
| (4) | |
| (9 marks) |
Notes
B1: For \(w = \mathrm{i}2\sqrt{2}\) or any single correct root (0 or 0i may be included in all roots) in any Form including polar
M1: Applying de Moivre and use a correct method to attempt 2 or 3 further roots
A1A1: For the other roots (3 correct scores A1A1; 2 correct scores A1)
Accept eg \(2\sqrt{2},\ \sqrt{8},\ 2.83,\ 64^{\frac{1}{4}},\ 4^{\frac{3}{4}},\ 4096^{\frac{1}{8}}\) Decimals must be 3 sf min.
ALT 1 for (c):
| Scheme | Marks |
|---|---|
| \(z^3 = 64 = w^4 \Rightarrow w = (\pm)2\sqrt{2}\) (\(\pm\) not needed) | B1 |
| Use rotational symmetry to find other 2/3 roots | M1 |
| Remaining roots as above | A1A1 |
ALT 2:
\(z^4 = 64 \quad z^2 = \pm 8\)
\(z = \pm 2\sqrt{2} \quad z = \pm\sqrt{-8} = \pm\mathrm{i}2\sqrt{2}\)
B1 any one correct, M1 attempt remaining 2/3 roots; A1A1 as above