Edexcel FP2 2017 June — Question 3 6 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2017
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDirect nth roots: general complex RHS
DifficultyStandard +0.3 This is a straightforward Further Maths FP2 question requiring students to find cube roots of a complex number. The process is mechanical: convert -32-32i√3 to modulus-argument form (r=64, θ=-2π/3), then apply the nth root formula with k=0,1,2. While it's a Further Maths topic, it's a standard textbook exercise with no novel insight required, making it slightly easier than average overall.
Spec4.02r nth roots: of complex numbers

3. Solve the equation $$z ^ { 3 } + 32 + 32 i \sqrt { 3 } = 0$$ giving your answers in the form \(r \mathrm { e } ^ { \mathrm { i } \theta }\) where \(r > 0\) and \(- \pi < \theta \leqslant \pi\)

Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(z^3 + 32 + 32i\sqrt{3} = 0\)
\(\arg(z^3) = \frac{4\pi}{3}\) or \(-\frac{2\pi}{3}\)M1A1 M1: Uses tan to find \(\arg z^3\); \(\arctan\sqrt{3}\), \(\arctan\frac{1}{\sqrt{3}}\), \(\frac{\pi}{3}\) or \(\frac{\pi}{6}\) seen. Allow equivalent angles. A1: Either value shown
\(z = r = 4\)
\(3\theta = \frac{4\pi}{3},\ -\frac{2\pi}{3},\ -\frac{8\pi}{3}\)
\(\theta = \frac{4\pi}{9},\ -\frac{2\pi}{9},\ -\frac{8\pi}{9}\)M1 Divides by 3 to obtain at least 2 values of \(\theta\) differing by \(\frac{2\pi}{3}\) or \(\frac{4\pi}{3}\)
\(\theta = \frac{4\pi}{9},\ -\frac{2\pi}{9}\) or \(\frac{16\pi}{9},\ -\frac{8\pi}{9}\) or \(\frac{10\pi}{9}\)A1 At least 2 correct and distinct values from list
\(z = 4e^{\frac{4\pi}{9}i},\ 4e^{-\frac{2\pi}{9}i},\ 4e^{-\frac{8\pi}{9}i}\) or \(4e^{i\theta}\) where \(\theta = \ldots\)A1 (6) All correct and in either of the forms shown. Ignore extra answers outside range
# Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z^3 + 32 + 32i\sqrt{3} = 0$ | | |
| $\arg(z^3) = \frac{4\pi}{3}$ or $-\frac{2\pi}{3}$ | M1A1 | M1: Uses tan to find $\arg z^3$; $\arctan\sqrt{3}$, $\arctan\frac{1}{\sqrt{3}}$, $\frac{\pi}{3}$ or $\frac{\pi}{6}$ seen. Allow equivalent angles. A1: Either value shown |
| $|z| = r = 4$ | B1 | Correct $r$ seen anywhere |
| $3\theta = \frac{4\pi}{3},\ -\frac{2\pi}{3},\ -\frac{8\pi}{3}$ | | |
| $\theta = \frac{4\pi}{9},\ -\frac{2\pi}{9},\ -\frac{8\pi}{9}$ | M1 | Divides by 3 to obtain at least 2 values of $\theta$ differing by $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$ |
| $\theta = \frac{4\pi}{9},\ -\frac{2\pi}{9}$ or $\frac{16\pi}{9},\ -\frac{8\pi}{9}$ or $\frac{10\pi}{9}$ | A1 | At least 2 correct and distinct values from list |
| $z = 4e^{\frac{4\pi}{9}i},\ 4e^{-\frac{2\pi}{9}i},\ 4e^{-\frac{8\pi}{9}i}$ or $4e^{i\theta}$ where $\theta = \ldots$ | A1 **(6)** | All correct and in either of the forms shown. Ignore extra answers outside range |

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3. Solve the equation

$$z ^ { 3 } + 32 + 32 i \sqrt { 3 } = 0$$

giving your answers in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$ where $r > 0$ and $- \pi < \theta \leqslant \pi$\\

\hfill \mbox{\textit{Edexcel FP2 2017 Q3 [6]}}