A2 June 2024 Paper 1 Q13

OCR MEICurrent spec10 marksComplex Numbers De Moivre's Theorem

13 The complex number \(z\) is defined as \(z = \frac{1}{3}\mathrm{e}^{\mathrm{i}\theta}\) where \(0 \lt \theta \lt \frac{1}{2}\pi\).

On an Argand diagram, the point O represents the complex number 0, and the points \(\mathrm{P}_1, \mathrm{P}_2, \mathrm{P}_3, \ldots\) represent the complex numbers \(z, z^2, z^3, \ldots\) respectively.

(a) Write down each of the following.

(i) The ratio of the lengths \(\mathrm{OP}_{n+1} : \mathrm{OP}_n\) [1]

(ii) The angle \(\mathrm{P}_{n+1}\mathrm{OP}_n\) [1]

(b)

(i) Show that \((3 - \mathrm{e}^{\mathrm{i}\theta})(3 - \mathrm{e}^{-\mathrm{i}\theta}) = a + b\cos\theta\), where \(a\) and \(b\) are integers to be determined. [2]

(ii) By considering the sum to infinity of the series \(z + z^2 + z^3 + \ldots\), show that
\(\frac{1}{3}\sin\theta + \frac{1}{9}\sin 2\theta + \frac{1}{27}\sin 3\theta + \ldots = \dfrac{3\sin\theta}{10 - 6\cos\theta}\). [6]

Scheme	Marks	AO
(i) \(1 : 3\)	B1	1.1
	[1]
(ii) \(\theta\)	B1	1.1
	[1]

Mark scheme (b)

Scheme	Marks	AO
(i) \((3 - \mathrm{e}^{\mathrm{i}\theta})(3 - \mathrm{e}^{-\mathrm{i}\theta}) = 9 - 3\mathrm{e}^{\mathrm{i}\theta} - 3\mathrm{e}^{-\mathrm{i}\theta} + 1\) \(= 10 - 3(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta})\)	M1	2.1
\(= 10 - 6\cos\theta\)	A1	2.1
	[2]
(ii) \([z + z^2 + \ldots =]\ \frac{1}{3}\mathrm{e}^{\mathrm{i}\theta} + \frac{1}{9}\mathrm{e}^{2\mathrm{i}\theta} + \ldots\)	M1	2.1
\(= \dfrac{\frac{1}{3}\mathrm{e}^{\mathrm{i}\theta}}{1 - \frac{1}{3}\mathrm{e}^{\mathrm{i}\theta}}\ \left[= \dfrac{\mathrm{e}^{\mathrm{i}\theta}}{3 - \mathrm{e}^{\mathrm{i}\theta}}\right]\)	A1	2.1
\(= \dfrac{\mathrm{e}^{\mathrm{i}\theta}(3 - \mathrm{e}^{-\mathrm{i}\theta})}{(3 - \mathrm{e}^{\mathrm{i}\theta})(3 - \mathrm{e}^{-\mathrm{i}\theta})}\)	M1*	3.1a
\(= \dfrac{3\mathrm{e}^{\mathrm{i}\theta} - 1}{10 - 6\cos\theta}\)	A1	2.1
\(= \dfrac{3(\cos\theta + \mathrm{i}\sin\theta) - 1}{10 - 6\cos\theta}\)	M1dep	2.1
\(\left[\frac{1}{3}\sin\theta + \frac{1}{9}\sin 2\theta \ldots =\right] \dfrac{3\sin\theta}{10 - 6\cos\theta}\)	A1	2.2a
	[6]

Notes

(b)(i)

M1: Expanding correctly to give at least three terms. Condone \(e^0 = 1\).

A1: www. Condone only incorrect values quoted for \(a\) and \(b\). Intermediate step not required here.

(b)(ii)

M1: At least two terms of series in exponential form soi by correct GP formula or \(\frac{z}{1 - z}\) seen. Condone modulus-argument form.

A1: Using sum to infinity formula correctly

M1*: Multiplying their numerator and denominator by a multiple of \(3 - \mathrm{e}^{-\mathrm{i}\theta}\). Must be a clear attempt at a sum to infinity.

A1: oe

M1dep: \(\mathrm{e}^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta\) used when denominator has been simplified to a real expression. No errors allowed.

A1: AG www