FP2 June 2013 (R) Q6

EdexcelOld spec11 marksDe Moivre's Theorem

6. The complex number \(z = \mathrm{e}^{\mathrm{i}\theta}\), where \(\theta\) is real.

(a) Use de Moivre’s theorem to show that \[z^n + \frac{1}{z^n} = 2\cos n\theta\] where \(n\) is a positive integer. (2)

(b) Show that \[\cos^5\theta = \frac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos\theta)\] (5)

(c) Hence find all the solutions of \[\cos 5\theta + 5\cos 3\theta + 12\cos\theta = 0\] in the interval \(0 \leqslant \theta < 2\pi\) (4)

Mark scheme (a)

Scheme	Marks
\(z^n + z^{-n} = \mathrm{e}^{\mathrm{i}n\theta} + \mathrm{e}^{-\mathrm{i}n\theta}\) \(= \cos n\theta + \mathrm{i}\sin n\theta + \cos n\theta - \mathrm{i}\sin n\theta\)
\(= 2\cos n\theta\) *	M1A1
	(2)

Notes

M1 for using de Moivre's theorem to show that either \(z^n = \cos n\theta + \mathrm{i}\sin n\theta\) or \(z^{-n} = \cos n\theta - \mathrm{i}\sin n\theta\)

A1 for completing to the given result \(z^n + z^{-n} = 2\cos n\theta\) * (corrected from the printed mark scheme, which prints \(z^n + z^n\))

Mark scheme (b)

Scheme	Marks
\(\left(z + z^{-1}\right)^5 = 32\cos^5\theta\)	B1
\(\left(z + z^{-1}\right)^5 = z^5 + 5z^3 + 10z + 10z^{-1} + 5z^{-3} + z^{-5}\)	M1A1
\(32\cos^5\theta = \left(z^5 + z^{-5}\right) + 5\left(z^3 + z^{-3}\right) + 10\left(z + z^{-1}\right)\)
\(= 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta\)	M1
\(\cos^5\theta = \dfrac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos\theta)\) *	A1
	(5)

Notes

B1 for using the result in (a) to obtain \(\left(z + z^{-1}\right)^5 = 32\cos^5\theta\) Need not be shown explicitly.

M1 for attempting to expand \(\left(z + z^{-1}\right)^5\) by binomial, Pascal's triangle or multiplying out the brackets. If \({}^nC_r\) is used do not award marks until changed to numbers

A1 for a correct expansion \(\left(z + z^{-1}\right)^5 = z^5 + 5z^3 + 10z + 10z^{-1} + 5z^{-3} + z^{-5}\)

M1 for replacing \(\left(z^5 + z^{-5}\right), \left(z^3 + z^{-3}\right), \left(z + z^{-1}\right)\) with \(2\cos 5\theta, 2\cos 3\theta, 2\cos\theta\) and equating their revised expression to their result for \(\left(z + z^{-1}\right)^5 = 32\cos^5\theta\)

A1cso for \(\cos^5\theta = \dfrac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos\theta)\) *

Mark scheme (c)

Scheme	Marks
\(\cos 5\theta + 5\cos 3\theta + 10\cos\theta = -2\cos\theta\)	M1
\(16\cos^5\theta = -2\cos\theta\)	A1
\(2\cos\theta\left(8\cos^4\theta + 1\right) = 0\)
\(8\cos^4\theta + 1 = 0\) no solution	B1
\(\cos\theta = 0\)
\(\theta = \dfrac{\pi}{2},\ \dfrac{3\pi}{2}\)	A1
	(4)
	(11 marks)

Notes

M1 for attempting re-arrange the equation with one side matching the bracket in the result in (b) Question states "hence", so no other method is allowed.

A1 for using the result in (b) to obtain \(16\cos^5\theta = -2\cos\theta\) oe

B1 for stating that there is no solution for \(8\cos^4\theta + 1 = 0\) oe eg \(8\cos^4\theta + 1 \neq 0\), \(8\cos^4\theta + 1 > 0\) or "ignore" but \(\cos\theta = \sqrt[4]{-\dfrac{1}{8}}\) without comment gets B0

A1 for \(\theta = \dfrac{\pi}{2}\) and \(\dfrac{3\pi}{2}\) and no more in the range. Must be in radians, can be in decimals (1.57..., 4.71... 3 sf or better)