| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| 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 fourth roots of a complex number. It requires converting to polar form, applying De Moivre's theorem, and sketching roots—all routine techniques for FP3 students with no novel problem-solving required. Slightly above average difficulty due to being Further Maths content, but mechanically straightforward. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z^4 = 4\!\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = 4\operatorname{cis}\frac{1}{3}\pi\) | B1 | For \(\arg(z^4)=\frac{1}{3}\pi\) soi |
| Dividing \(\arg(z^4)\) by 4 | M1 | |
| \(z = \sqrt{2}\,\operatorname{cis}\!\left(k\frac{\pi}{12}\right),\; k=1,7,13,19\) | A1 | For any 2 correct values of \(k\) |
| A1 | For all 4 values of \(k\) and no extras. Ignore values outside range | |
| Modulus of all stated roots \(= \sqrt{2}\) | B1 | Don't accept \(1.41...\) or \(\sqrt[4]{4}\). For second A1 must be in correct form |
| SR For \(\arg(z^4)=\frac{1}{6}\pi\) award B0 M1 A1 FT for all \(\operatorname{cis}\!\left(k\frac{\pi}{24}\right), k=1,13,25,37\), A0 B0/B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Roots forming a square, centre \(O\), on equal-scale axes | B1 | Must be roots distinct from \(z^4\). Penalise once use of points not lines |
| \(z^4\) and only one root in first quadrant with arguments in ratio approximately 3:1 | B1 | |
| \(\ | z^4\ | : |
# Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z^4 = 4\!\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = 4\operatorname{cis}\frac{1}{3}\pi$ | B1 | For $\arg(z^4)=\frac{1}{3}\pi$ soi |
| Dividing $\arg(z^4)$ by 4 | M1 | |
| $z = \sqrt{2}\,\operatorname{cis}\!\left(k\frac{\pi}{12}\right),\; k=1,7,13,19$ | A1 | For any 2 correct values of $k$ |
| | A1 | For all 4 values of $k$ and no extras. Ignore values outside range |
| Modulus of all stated roots $= \sqrt{2}$ | B1 | Don't accept $1.41...$ or $\sqrt[4]{4}$. For second A1 must be in correct form |
| **SR** For $\arg(z^4)=\frac{1}{6}\pi$ award B0 M1 A1 FT for all $\operatorname{cis}\!\left(k\frac{\pi}{24}\right), k=1,13,25,37$, A0 B0/B1 | | |
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# Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Roots forming a square, centre $O$, on equal-scale axes | B1 | Must be roots distinct from $z^4$. Penalise once use of points not lines |
| $z^4$ and only one root in first quadrant with arguments in ratio approximately 3:1 | B1 | |
| $\|z^4\|:|z| \approx 4:\sqrt{2}$ (allow $(2,4):1$) | B1 | For all four roots |
---2 (i) Solve the equation $z ^ { 4 } = 2 ( 1 + \mathrm { i } \sqrt { 3 } )$, giving the roots exactly in the form $r ( \cos \theta + \mathrm { i } \sin \theta )$, where $r > 0$ and $0 \leqslant \theta < 2 \pi$.\\
(ii) Sketch an Argand diagram to show the lines from the origin to the point representing $2 ( 1 + i \sqrt { 3 } )$ and from the origin to the points which represent the roots of the equation in part (i).
\hfill \mbox{\textit{OCR FP3 2012 Q2 [8]}}