Question 2 - A-Level Maths

Edexcel FP2 2009 June — Question 2 6 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2009
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex numbers 2
Type	Direct nth roots: general complex RHS
Difficulty	Standard +0.8 This is a standard FP2 cube roots question requiring conversion to modulus-argument form, application of de Moivre's theorem, and finding three roots. While methodical, it involves multiple steps (finding r and θ, dividing argument by 3, adding 2π/3 for each root) and careful angle arithmetic to ensure -π < θ ≤ π, making it moderately challenging but still a textbook exercise.
Spec	4.02r nth roots: of complex numbers

Solve the equation $$z ^ { 3 } = 4 \sqrt { } 2 - 4 \sqrt { } 2 i$$ giving your answers in the form $r ( \cos \theta + \mathrm { i } \sin \theta )$, where $- \pi < \theta \leqslant \pi$.

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Question 2:

Part (a)

Answer	Marks	Guidance
$r = \sqrt{(4\sqrt{2})^2+(-4\sqrt{2})^2} = \sqrt{32+32} = \sqrt{64} = 8$; $\theta = -\tan^{-1}\!\left(\frac{4\sqrt{2}}{4\sqrt{2}}\right) = -\frac{\pi}{4}$	M1	Valid attempt to find modulus and argument of $4\sqrt{2}-4\sqrt{2}\,\mathrm{i}$
$z^3 = 8\!\left(\cos\!\left(-\frac{\pi}{4}\right)+\mathrm{i}\sin\!\left(-\frac{\pi}{4}\right)\right)$
$z = (8)^{\frac{1}{3}}\!\left(\cos\!\left(\frac{-\pi/4}{3}\right)+\mathrm{i}\sin\!\left(\frac{-\pi/4}{3}\right)\right)$	M1	Taking cube root of modulus and dividing argument by 3
$z = 2\!\left(\cos\!\left(-\frac{\pi}{12}\right)+\mathrm{i}\sin\!\left(-\frac{\pi}{12}\right)\right)$	A1	$2\!\left(\cos\!\left(-\frac{\pi}{12}\right)+\mathrm{i}\sin\!\left(-\frac{\pi}{12}\right)\right)$
Also $z^3 = 8\!\left(\cos\!\left(\frac{7\pi}{4}\right)+\mathrm{i}\sin\!\left(\frac{7\pi}{4}\right)\right)$ or $z^3 = 8\!\left(\cos\!\left(-\frac{9\pi}{4}\right)+\mathrm{i}\sin\!\left(-\frac{9\pi}{4}\right)\right)$	M1	Adding or subtracting $2\pi$ to argument of $z^3$ to find other roots
$z = 2\!\left(\cos\frac{7\pi}{12}+\mathrm{i}\sin\frac{7\pi}{12}\right)$	A1	Any one of the final two roots
$z = 2\!\left(\cos\!\left(-\frac{3\pi}{4}\right)+\mathrm{i}\sin\!\left(-\frac{3\pi}{4}\right)\right)$	A1	Both of the final two roots (6 marks)

# Question 2:

## Part (a)
| $r = \sqrt{(4\sqrt{2})^2+(-4\sqrt{2})^2} = \sqrt{32+32} = \sqrt{64} = 8$; $\theta = -\tan^{-1}\!\left(\frac{4\sqrt{2}}{4\sqrt{2}}\right) = -\frac{\pi}{4}$ | M1 | Valid attempt to find modulus and argument of $4\sqrt{2}-4\sqrt{2}\,\mathrm{i}$ |
|---|---|---|
| $z^3 = 8\!\left(\cos\!\left(-\frac{\pi}{4}\right)+\mathrm{i}\sin\!\left(-\frac{\pi}{4}\right)\right)$ | | |
| $z = (8)^{\frac{1}{3}}\!\left(\cos\!\left(\frac{-\pi/4}{3}\right)+\mathrm{i}\sin\!\left(\frac{-\pi/4}{3}\right)\right)$ | M1 | Taking cube root of modulus and dividing argument by 3 |
| $z = 2\!\left(\cos\!\left(-\frac{\pi}{12}\right)+\mathrm{i}\sin\!\left(-\frac{\pi}{12}\right)\right)$ | A1 | $2\!\left(\cos\!\left(-\frac{\pi}{12}\right)+\mathrm{i}\sin\!\left(-\frac{\pi}{12}\right)\right)$ |
| Also $z^3 = 8\!\left(\cos\!\left(\frac{7\pi}{4}\right)+\mathrm{i}\sin\!\left(\frac{7\pi}{4}\right)\right)$ or $z^3 = 8\!\left(\cos\!\left(-\frac{9\pi}{4}\right)+\mathrm{i}\sin\!\left(-\frac{9\pi}{4}\right)\right)$ | M1 | Adding or subtracting $2\pi$ to argument of $z^3$ to find other roots |
| $z = 2\!\left(\cos\frac{7\pi}{12}+\mathrm{i}\sin\frac{7\pi}{12}\right)$ | A1 | Any one of the final two roots |
| $z = 2\!\left(\cos\!\left(-\frac{3\pi}{4}\right)+\mathrm{i}\sin\!\left(-\frac{3\pi}{4}\right)\right)$ | A1 | Both of the final two roots (6 marks) |

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Show LaTeX source

Solve the equation

$$z ^ { 3 } = 4 \sqrt { } 2 - 4 \sqrt { } 2 i$$

giving your answers in the form $r ( \cos \theta + \mathrm { i } \sin \theta )$, where $- \pi < \theta \leqslant \pi$.\\

\hfill \mbox{\textit{Edexcel FP2 2009 Q2 [6]}}

This paper (8 questions)

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Q1 6 Q2 6 Q3 8 Q4 8 Q5 10 Q6 10 Q7 12 Q8 15

Answer	Marks	Guidance
\(r = \sqrt{(4\sqrt{2})^2+(-4\sqrt{2})^2} = \sqrt{32+32} = \sqrt{64} = 8\); \(\theta = -\tan^{-1}\!\left(\frac{4\sqrt{2}}{4\sqrt{2}}\right) = -\frac{\pi}{4}\)	M1	Valid attempt to find modulus and argument of \(4\sqrt{2}-4\sqrt{2}\,\mathrm{i}\)
\(z^3 = 8\!\left(\cos\!\left(-\frac{\pi}{4}\right)+\mathrm{i}\sin\!\left(-\frac{\pi}{4}\right)\right)\)
\(z = (8)^{\frac{1}{3}}\!\left(\cos\!\left(\frac{-\pi/4}{3}\right)+\mathrm{i}\sin\!\left(\frac{-\pi/4}{3}\right)\right)\)	M1	Taking cube root of modulus and dividing argument by 3
\(z = 2\!\left(\cos\!\left(-\frac{\pi}{12}\right)+\mathrm{i}\sin\!\left(-\frac{\pi}{12}\right)\right)\)	A1	\(2\!\left(\cos\!\left(-\frac{\pi}{12}\right)+\mathrm{i}\sin\!\left(-\frac{\pi}{12}\right)\right)\)
Also \(z^3 = 8\!\left(\cos\!\left(\frac{7\pi}{4}\right)+\mathrm{i}\sin\!\left(\frac{7\pi}{4}\right)\right)\) or \(z^3 = 8\!\left(\cos\!\left(-\frac{9\pi}{4}\right)+\mathrm{i}\sin\!\left(-\frac{9\pi}{4}\right)\right)\)	M1	Adding or subtracting \(2\pi\) to argument of \(z^3\) to find other roots
\(z = 2\!\left(\cos\frac{7\pi}{12}+\mathrm{i}\sin\frac{7\pi}{12}\right)\)	A1	Any one of the final two roots
\(z = 2\!\left(\cos\!\left(-\frac{3\pi}{4}\right)+\mathrm{i}\sin\!\left(-\frac{3\pi}{4}\right)\right)\)	A1	Both of the final two roots (6 marks)