A2 June 2023 Paper 2 Q15

AQACurrent spec10 marksDe Moivre's Theorem Series

(a) Given that \(z = \cos\theta + \mathrm{i}\sin\theta\), use de Moivre’s theorem to show that\[z^n - z^{-n} = 2\mathrm{i}\sin n\theta\] [2 marks]

(b) The series \(S\) is defined as\[S = \sin\theta + \sin 3\theta + \ldots + \sin(2n - 1)\theta\]

Use part (a) to express \(S\) in the form

\[S = \frac{1}{2\mathrm{i}}(G_1) - \frac{1}{2\mathrm{i}}(G_2)\]

where each of \(G_1\) and \(G_2\) is a geometric series. [3 marks]

Mark scheme (a)

Scheme	Marks	AO
Uses de Moivre’s theorem.	M1	1.1a
Completes a reasoned argument to obtain the required result. Must see \(\cos(-n\theta) + \mathrm{i}\sin(-n\theta)\) and \(z^n - z^{-n}\)	R1	2.1
	(2)

Typical solution

By de Moivre’s theorem,

\[z^n = \cos n\theta + \mathrm{i}\sin n\theta\]\[\begin{aligned} z^{-n} &= \cos(-n\theta) + \mathrm{i}\sin(-n\theta) \\ &= \cos(n\theta) - \mathrm{i}\sin(n\theta) \end{aligned}\]\[z^n - z^{-n} = 2\mathrm{i}\sin n\theta\]

Mark scheme (b)

Scheme	Marks	AO
Uses part (a) to express at least three terms of \(S\) in terms of \(z\)	M1	3.1a
Expresses \(S\) or \(2\mathrm{i}S\) as the difference of two series.	M1	1.1a
Completes a reasoned argument to obtain the required result.	R1	2.1
	(3)

Typical solution

\[\begin{aligned} 2\mathrm{i}S &= 2\mathrm{i}\sin\theta + 2\mathrm{i}\sin 3\theta + \ldots + 2\mathrm{i}\sin(2n - 1)\theta \\ &= z - z^{-1} + z^3 - z^{-3} + \ldots + z^{2n-1} - z^{-(2n-1)} \\ &= z + z^3 + \ldots + z^{2n-1} - \left(z^{-1} + z^{-3} + \ldots + z^{-(2n-1)}\right) \end{aligned}\]\[S = \frac{1}{2\mathrm{i}}\left(z + z^3 + \ldots + z^{2n-1}\right) - \frac{1}{2\mathrm{i}}\left(z^{-1} + z^{-3} + \ldots + z^{-(2n-1)}\right)\]

Mark scheme (c)

Scheme	Marks	AO
Obtains expressions for the sums of their geometric series.	M1	3.1a
Obtains fully correct expressions for the sums of \(G_1\) and \(G_2\)	A1	1.1b
Rearranges to obtain \(z - z^{-1}\) in the denominator of any fraction.	B1	3.1a
Obtains \(z^{2n} + z^{-2n} - 2\) in the numerator of their single fraction.	B1	3.1a
Completes a reasoned argument to obtain the required result.	R1	2.1
	(5)
	(10 marks)

Typical solution

\[2\mathrm{i}S = \frac{z\left(1 - z^{2n}\right)}{\left(1 - z^2\right)} - \frac{z^{-1}\left(1 - z^{-2n}\right)}{\left(1 - z^{-2}\right)}\]\[\begin{aligned} &= \frac{z^{2n} - 1}{z - z^{-1}} - \frac{1 - z^{-2n}}{z - z^{-1}} = \frac{z^{2n} + z^{-2n} - 2}{z - z^{-1}} \\ &= \frac{\left(z^n - z^{-n}\right)^2}{2\mathrm{i}\sin\theta} = \frac{(2\mathrm{i}\sin n\theta)^2}{2\mathrm{i}\sin\theta} = -\frac{2\sin^2 n\theta}{\mathrm{i}\sin\theta} \end{aligned}\]\[-2S = -\frac{2\sin^2 n\theta}{\sin\theta}\]\[S = \frac{\sin^2 n\theta}{\sin\theta}\]