CAIE Further Paper 2 2022 June — Question 1 5 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2022
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDirect nth roots: general complex RHS
DifficultyStandard +0.3 This is a standard Further Maths question requiring conversion to polar form, application of De Moivre's theorem for cube roots, and careful angle arithmetic. While it involves multiple steps (finding modulus, argument, then three roots), the procedure is algorithmic and well-practiced in Further Maths courses with no novel insight required.
Spec4.02d Exponential form: re^(i*theta)4.02r nth roots: of complex numbers

1 Find the roots of the equation \(z ^ { 3 } = 7 \sqrt { 3 } - 7 \mathrm { i }\), giving your answers in the form \(r \mathrm { e } ^ { \mathrm { i } \theta }\), where \(r > 0\) and \(- \pi \leqslant \theta < \pi\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(z^3 = 7\sqrt{3} - 7i = 14e^{-i\frac{\pi}{6}}\)B1 Converts \(7\sqrt{3} - 7i\) to exponential form
\(z_1 = 14^{\frac{1}{3}}e^{-i\frac{\pi}{18}}\)M1 A1 Finds one root
\(z_2 = 14^{\frac{1}{3}}e^{i\frac{11\pi}{18}}\), \(z_3 = 14^{\frac{1}{3}}e^{-i\frac{13\pi}{18}}\)A1FT A1FT Finds other two roots. FT on modulus. Others given scores A1 A0
Total: 5
**Question 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $z^3 = 7\sqrt{3} - 7i = 14e^{-i\frac{\pi}{6}}$ | B1 | Converts $7\sqrt{3} - 7i$ to exponential form |
| $z_1 = 14^{\frac{1}{3}}e^{-i\frac{\pi}{18}}$ | M1 A1 | Finds one root |
| $z_2 = 14^{\frac{1}{3}}e^{i\frac{11\pi}{18}}$, $z_3 = 14^{\frac{1}{3}}e^{-i\frac{13\pi}{18}}$ | A1FT A1FT | Finds other two roots. FT on modulus. Others given scores A1 A0 |
| **Total: 5** | | |

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1 Find the roots of the equation $z ^ { 3 } = 7 \sqrt { 3 } - 7 \mathrm { i }$, giving your answers in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r > 0$ and $- \pi \leqslant \theta < \pi$.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q1 [5]}}