| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Direct nth roots: general complex RHS |
| Difficulty | Standard +0.3 This is a standard Further Maths question on finding nth roots of complex numbers. Part (a) requires routine conversion to polar form (finding modulus and argument), and part (b) applies the standard formula for fourth roots by dividing the argument by 4 and adding multiples of 2π/4. While this is Further Maths content, it's a textbook exercise requiring only direct application of learned procedures with no novel insight or problem-solving. |
| Spec | 4.02b Express complex numbers: cartesian and modulus-argument forms4.02r nth roots: of complex numbers |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\ | 18\sqrt{3}-18i\ | = 18\sqrt{(3+1)} = 36\) |
| \(\tan\theta = \frac{-18}{18\sqrt{3}}\), \(\theta = -\frac{\pi}{6}\) | M1 | Attempt argument using \(\tan\theta = \frac{\pm18}{18\sqrt{3}}\) or other valid method. Can be implied by \(\theta = \pm\frac{\pi}{6}\) |
| \(18\sqrt{3}-18i = 36\left(\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right)\) | A1cao (3) | Correct answer in required form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z^4 = 36\left(\cos\left(2k\pi-\frac{\pi}{6}\right)+i\sin\left(2k\pi-\frac{\pi}{6}\right)\right)\) | M1 | Valid method for generating at least 2 roots; rotation through \(\frac{\pi}{2}\) accepted |
| \(z = \sqrt{6}\left(\cos\left(\frac{12k\pi-\pi}{24}\right)+i\sin\left(\frac{12k\pi-\pi}{24}\right)\right)\) | M1 | Apply de Moivre or use the rotation method |
| \(k=0\): \(z_0 = \sqrt{6}\left(\cos\left(\frac{-\pi}{24}\right)+i\sin\left(\frac{-\pi}{24}\right)\right) = \sqrt{6}\,e^{i\left(-\frac{\pi}{24}\right)}\) | B1 | Any one correct root |
| \(k=1\): \(z_1 = \sqrt{6}\left(\cos\left(\frac{11\pi}{24}\right)+i\sin\left(\frac{11\pi}{24}\right)\right) = \sqrt{6}\,e^{i\frac{11\pi}{24}}\) | A1ft | Second root in required form |
| \(k=2\): \(z_2 = \sqrt{6}\left(\cos\left(\frac{23\pi}{24}\right)+i\sin\left(\frac{23\pi}{24}\right)\right) = \sqrt{6}\,e^{i\frac{23\pi}{24}}\) | ||
| \(k=-1\): \(z_3 = \sqrt{6}\left(\cos\left(-\frac{13\pi}{24}\right)+i\sin\left(-\frac{13\pi}{24}\right)\right) = \sqrt{6}\,e^{i\left(-\frac{13\pi}{24}\right)}\) | A1ft (5) | All 4 roots in required form. Follow through on \(\sqrt[4]{36}\) but 36 not acceptable. Arguments in degrees: M1M1B1A0A0. Incorrect argument: B0A1ftA1ft available. Answers in \(r(\cos\theta+i\sin\theta)\) form — deduct final A marks |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\|18\sqrt{3}-18i\| = 18\sqrt{(3+1)} = 36$ | B1 | Correct modulus |
| $\tan\theta = \frac{-18}{18\sqrt{3}}$, $\theta = -\frac{\pi}{6}$ | M1 | Attempt argument using $\tan\theta = \frac{\pm18}{18\sqrt{3}}$ or other valid method. Can be implied by $\theta = \pm\frac{\pi}{6}$ |
| $18\sqrt{3}-18i = 36\left(\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right)$ | A1cao (3) | Correct answer in required form |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z^4 = 36\left(\cos\left(2k\pi-\frac{\pi}{6}\right)+i\sin\left(2k\pi-\frac{\pi}{6}\right)\right)$ | M1 | Valid method for generating at least 2 roots; rotation through $\frac{\pi}{2}$ accepted |
| $z = \sqrt{6}\left(\cos\left(\frac{12k\pi-\pi}{24}\right)+i\sin\left(\frac{12k\pi-\pi}{24}\right)\right)$ | M1 | Apply de Moivre or use the rotation method |
| $k=0$: $z_0 = \sqrt{6}\left(\cos\left(\frac{-\pi}{24}\right)+i\sin\left(\frac{-\pi}{24}\right)\right) = \sqrt{6}\,e^{i\left(-\frac{\pi}{24}\right)}$ | B1 | Any one correct root |
| $k=1$: $z_1 = \sqrt{6}\left(\cos\left(\frac{11\pi}{24}\right)+i\sin\left(\frac{11\pi}{24}\right)\right) = \sqrt{6}\,e^{i\frac{11\pi}{24}}$ | A1ft | Second root in required form |
| $k=2$: $z_2 = \sqrt{6}\left(\cos\left(\frac{23\pi}{24}\right)+i\sin\left(\frac{23\pi}{24}\right)\right) = \sqrt{6}\,e^{i\frac{23\pi}{24}}$ | | |
| $k=-1$: $z_3 = \sqrt{6}\left(\cos\left(-\frac{13\pi}{24}\right)+i\sin\left(-\frac{13\pi}{24}\right)\right) = \sqrt{6}\,e^{i\left(-\frac{13\pi}{24}\right)}$ | A1ft (5) | All 4 roots in required form. Follow through on $\sqrt[4]{36}$ but 36 not acceptable. Arguments in degrees: M1M1B1A0A0. Incorrect argument: B0A1ftA1ft available. Answers in $r(\cos\theta+i\sin\theta)$ form — deduct final A marks |
**Total: [8]**4. (a) Express the complex number $18 \sqrt { 3 } - 18 \mathrm { i }$ in the form
$$r ( \cos \theta + \mathrm { i } \sin \theta ) \quad - \pi < \theta \leqslant \pi$$
(b) Solve the equation
$$z ^ { 4 } = 18 \sqrt { 3 } - 18 i$$
giving your answers in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$ where $- \pi < \theta \leqslant \pi$\\
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\hfill \mbox{\textit{Edexcel F2 2020 Q4 [8]}}