FP2 June 2012 Q3

EdexcelOld spec8 marksComplex Numbers De Moivre's Theorem

(a) Express the complex number \(-2 + \left(2\sqrt{3}\right)\mathrm{i}\) in the form \(r(\cos\theta + \mathrm{i}\sin\theta)\), \(-\pi < \theta \leqslant \pi\). (3)

(b) Solve the equation \[z^4 = -2 + \left(2\sqrt{3}\right)\mathrm{i}\] giving the roots in the form \(r(\cos\theta + \mathrm{i}\sin\theta)\), \(-\pi < \theta \leqslant \pi\). (5)

Mark scheme (a)

Scheme	Marks
\(r = \sqrt{(-2)^2 + (2\sqrt{3})^2} = 4\)	B1
\(\tan\theta = -\sqrt{3}\) (Also allow M mark for \(\tan\theta = \sqrt{3}\)) M mark can be implied by \(\theta = \pm\dfrac{2\pi}{3}\) or \(\theta = \pm\dfrac{\pi}{3}\)	M1
\(\theta = \dfrac{2\pi}{3}\)	A1
	(3)

Notes

M1 Accept \(\pm\sqrt{3}\) or \(\pm\dfrac{1}{\sqrt{3}}\)

A1 Accept awrt 2.1. A0 if in degrees.

Mark scheme (b)

Scheme	Marks
Finding the 4^th root of their \(r\): \(r = 4^{\frac{1}{4}}\ (= \sqrt{2})\)	M1
For one root, dividing their \(\theta\) by 4: \(\theta = \dfrac{2\pi}{3} \div 4 = \dfrac{\pi}{6}\)	M1
For another root, add or subtract a multiple of \(2\pi\) to their \(\theta\) and divide by 4 in correct order.	M1
\(\sqrt{2}(\cos\theta + \mathrm{i}\sin\theta)\), where \(\theta = -\dfrac{5\pi}{6},\ -\dfrac{\pi}{3},\ \dfrac{\pi}{6},\ \dfrac{2\pi}{3}\)	A1 A1
	(5)
	(8 marks)

Notes

2^nd M1 for awrt 0.52

1^st A1 for two correct values

2^nd A1 for all correct values values in correct form and no more