| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Direct nth roots: roots with geometric or algebraic follow-up |
| Difficulty | Standard +0.3 This is a standard Further Maths question on finding nth roots of complex numbers using De Moivre's theorem. Part (i) requires routine conversion to polar form, part (ii) applies the standard cube root formula, and part (iii) involves sketching and using symmetry. While it's Further Maths content (inherently harder), it's a textbook application of well-practiced techniques with no novel insight required, making it slightly easier than average overall. |
| Spec | 4.02b Express complex numbers: cartesian and modulus-argument forms4.02d Exponential form: re^(i*theta)4.02k Argand diagrams: geometric interpretation4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(4((√3) - i) = 8e^{-\frac{i\pi}{6}}\) | B1 | For \(r = 8\) |
| B1 | For \(\theta = -\frac{\pi}{6}\) | |
| 2 | ||
| (ii) One cube root is \(2e^{\frac{i\pi}{8}}\) | B1√ | For modulus and argument both correct |
| Others are found be multiplying by \(e^{\frac{2\pi i}{3}}\) | M1 | For multiplication by either cube root of 1 (or equivalent use of symmetry) |
| Giving \(2e^{\frac{i\pi}{8}}\) and \(2e^{-\frac{i5\pi}{8}}\) | A1 A1 | For either one of these roots; For both correct |
| 4 | ||
| (iii) [Correct diagram showing three roots equally spaced] | B1√ | For correct diagram from their (ii) |
| The roots have equal modulus and args differing by \(\frac{2\pi}{3}\), so adding them geometrically makes a closed equilateral triangle; i.e. sum is zero | M1 | For geometrical interpretation of addition |
| A1 | For a correct proof (or via components, etc) | |
| 3 | ||
| 9 |
**(i)** $4((√3) - i) = 8e^{-\frac{i\pi}{6}}$ | B1 | For $r = 8$
| B1 | For $\theta = -\frac{\pi}{6}$
| **2** |
**(ii)** One cube root is $2e^{\frac{i\pi}{8}}$ | B1√ | For modulus and argument both correct
Others are found be multiplying by $e^{\frac{2\pi i}{3}}$ | M1 | For multiplication by either cube root of 1 (or equivalent use of symmetry)
Giving $2e^{\frac{i\pi}{8}}$ and $2e^{-\frac{i5\pi}{8}}$ | A1 A1 | For either one of these roots; For both correct
| **4** |
**(iii)** [Correct diagram showing three roots equally spaced] | B1√ | For correct diagram from their (ii)
The roots have equal modulus and args differing by $\frac{2\pi}{3}$, so adding them geometrically makes a closed equilateral triangle; i.e. sum is zero | M1 | For geometrical interpretation of addition
| A1 | For a correct proof (or via components, etc)
| **3** |
| **9** |4 In this question, give your answers exactly in polar form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r > 0$ and $- \pi < \theta \leqslant \pi$.\\
(i) Express $4 ( ( \sqrt { } 3 ) - \mathrm { i } )$ in polar form.\\
(ii) Find the cube roots of $4 ( ( \sqrt { } 3 ) - \mathrm { i } )$ in polar form.\\
(iii) Sketch an Argand diagram showing the positions of the cube roots found in part (ii). Hence, or otherwise, prove that the sum of these cube roots is zero.
\hfill \mbox{\textit{OCR FP3 Q4 [9]}}