A2 October 2021 Paper 2 Q9
9.
(b) Given that \(z = \dfrac{1}{2}(\cos\theta + \mathrm{i}\sin\theta)\),
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{1 - z}\) | B1 | 2.2a |
| (1) |
Notes
(a)
B1: See scheme
| Scheme | Marks | AO |
|---|---|---|
\(1 + z + z^2 + z^3 + \ldots\) \(= 1 + \left(\dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\right) + \left(\dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\right)^2 + \left(\dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\right)^3 + \ldots\) \(= 1 + \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right) + \dfrac{1}{4}\left(\cos 2\theta + \mathrm{i}\sin 2\theta\right) + \dfrac{1}{8}\left(\cos 3\theta + \mathrm{i}\sin 3\theta\right) + \ldots\) | M1 | 3.1a |
\(\dfrac{1}{1 - z} = \dfrac{1}{1 - \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)} \times \dfrac{1 - \frac{1}{2}\cos\theta + \frac{1}{2}\mathrm{i}\sin\theta}{1 - \frac{1}{2}\cos\theta + \frac{1}{2}\mathrm{i}\sin\theta}\) or \(\dfrac{1}{1 - z} = \dfrac{2}{2 - (\cos\theta + \mathrm{i}\sin\theta)} \times \dfrac{2 - (\cos\theta - \mathrm{i}\sin\theta)}{2 - (\cos\theta - \mathrm{i}\sin\theta)}\) | M1 | 3.1a |
\(\left\{\dfrac{1}{2}(\sin\theta) + \dfrac{1}{4}(\sin 2\theta) + \dfrac{1}{8}(\sin 3\theta) + \ldots\right\} = \dfrac{\frac{1}{2}\sin\theta}{\left(1 - \frac{1}{2}\cos\theta\right)^2 + \left(\frac{1}{2}\sin\theta\right)^2}\) or \(\left\{\dfrac{1}{2}(\sin\theta) + \dfrac{1}{4}(\sin 2\theta) + \dfrac{1}{8}(\sin 3\theta) + \ldots\right\} = \dfrac{2\sin\theta}{(2 - \cos\theta)^2 + (\sin\theta)^2}\) | M1 | 2.1 |
\(\left(1 - \dfrac{1}{2}\cos\theta\right)^2 + \left(\dfrac{1}{2}\sin\theta\right)^2 = 1 - \cos\theta + \dfrac{1}{4}\cos^2\theta + \dfrac{1}{4}\sin^2\theta\) \(= \dfrac{5}{4} - \cos\theta\) or \((2 - \cos\theta)^2 + (\sin\theta)^2 = 4 - 4\cos\theta + \cos^2\theta + \sin^2\theta\) \(= 5 - 4\cos\theta\) | M1 | 1.1b |
| \(\dfrac{1}{2}\sin\theta + \dfrac{1}{4}\sin 2\theta + \dfrac{1}{8}\sin 3\theta + \ldots = \dfrac{\frac{1}{2}\sin\theta}{\frac{5}{4} - \cos\theta} = \dfrac{2\sin\theta}{5 - 4\cos\theta}\) * | A1* | 1.1b |
| (5) |
Alternative
| Scheme | Marks | AO |
|---|---|---|
\(1 + z + z^2 + z^3 + \ldots\) \(= 1 + \left(\dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\right) + \left(\dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\right)^2 + \left(\dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\right)^3 + \ldots\) \(= 1 + \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right) + \dfrac{1}{4}\left(\cos 2\theta + \mathrm{i}\sin 2\theta\right) + \dfrac{1}{8}\left(\cos 3\theta + \mathrm{i}\sin 3\theta\right) + \ldots\) | M1 | 3.1a |
| \(\dfrac{1}{1 - z} = \dfrac{1}{1 - \frac{1}{2}\mathrm{e}^{\mathrm{i}\theta}} \times \dfrac{1 - \frac{1}{2}\mathrm{e}^{-\mathrm{i}\theta}}{1 - \frac{1}{2}\mathrm{e}^{-\mathrm{i}\theta}}\) | M1 | 3.1a |
| \(\dfrac{1 - \frac{1}{2}\mathrm{e}^{-\mathrm{i}\theta}}{1 - \frac{1}{2}\mathrm{e}^{\mathrm{i}\theta} - \frac{1}{2}\mathrm{e}^{-\mathrm{i}\theta} + \frac{1}{4}} = \dfrac{4 - 2\mathrm{e}^{-\mathrm{i}\theta}}{5 - 2\left(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta}\right)} = \dfrac{4 - 2(\cos\theta - \mathrm{i}\sin\theta)}{5 - 2(2\cos\theta)}\) | M1 | 2.1 |
| Select the imaginary part \(\dfrac{2\sin\theta}{5 - 4\cos\theta}\) | M1 | 1.1b |
| \(\dfrac{1}{2}\sin\theta + \dfrac{1}{4}\sin 2\theta + \dfrac{1}{8}\sin 3\theta + \ldots = \dfrac{2\sin\theta}{5 - 4\cos\theta}\) * | A1* | 1.1b |
| (5) |
Notes
(b)(i)
M1: Substitutes \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into at least 3 terms of the series and applies de Moivre’s theorem.
M1: Substitutes \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into their answer to part (a) and rationalises the denominator.
M1: Equates the imaginary terms.
M1: Multiplies out the denominator and simplifies by using the identity \(\cos^2\theta + \sin^2\theta = 1\)
A1*: cso. Achieves the printed answer having substituted \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into 4 terms of the series.
Alternative
M1: Substitutes \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into at least 3 terms of the series and applies de Moivre’s theorem.
M1: Substitutes \(z = \dfrac{1}{2}\mathrm{e}^{\mathrm{i}\theta}\) into their answer to part (a) and rationalises the denominator.
M1: Uses \(\mathrm{e}^{-\mathrm{i}\theta} = \cos\theta - \mathrm{i}\sin\theta\) and \(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} = 2\cos\theta\) to express in terms of \(\sin\theta\) and \(\cos\theta\)
M1: Select the imaginary terms.
A1*: cso Achieves the printed answer having substituted \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into 4 terms of the series.
(corrected from the printed mark scheme: in the alternative, the denominator \(1 - \frac{1}{2}\mathrm{e}^{\mathrm{i}\theta} - \frac{1}{2}\mathrm{e}^{-\mathrm{i}\theta} + \frac{1}{4}\) is printed with \(\frac{1}{4}\mathrm{e}^{\mathrm{i}\theta}\) and \(\frac{1}{4}\mathrm{e}^{-\mathrm{i}\theta}\))
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1 - \frac{1}{2}\cos\theta}{\frac{5}{4} - \cos\theta} = 0 \Rightarrow \cos\theta = 2\) | M1 | 3.1a |
| As \((-1 \leqslant)\cos\theta \leqslant 1\) therefore there is no solution to \(\cos\theta = 2\) so there will also be a real part, hence the sum cannot be purely imaginary. | A1 | 2.4 |
| (2) |
Alternative 1
| Scheme | Marks | AO |
|---|---|---|
| Real part is \(\dfrac{4 - 2\cos\theta}{5 - 4\cos\theta} = \dfrac{1}{2} + \dfrac{3}{2(5 - 4\cos\theta)}\) | M1 | 3.1a |
| \(-1 \leqslant \cos\theta \leqslant 1\) therefore \(\dfrac{1}{6} \leqslant \dfrac{3}{2(5 - 4\cos\theta)} \leqslant \dfrac{3}{2}\) so sum must contain real part | A1 | 2.4 |
| (2) |
Alternative 2
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{1 - z} = k\mathrm{i} \Rightarrow z = 1 + \dfrac{\mathrm{i}}{k}\) | M1 | 3.1a |
| mod \(z \gt 1\) contradiction hence cannot be purely imaginary | A1 | 2.4 |
| (2) | ||
| (8 marks) |
Notes
(b)(ii)
M1: Setting the real part of the series \(= 0\) and rearranges to find \(\cos\theta = \ldots\)
A1: See scheme
Alternative 1
M1: Rearranges the real part so that \(\cos\theta\) only appears once
A1: Uses \(-1 \leqslant \cos\theta \leqslant 1\) to show that the sum must always be positive so must contain a real part
Alternative 2
M1: Sets sum as purely imaginary and rearranges to make \(z\) the subject
A1: Shows a contradiction and draws an appropriate conclusion
(corrected from the printed mark scheme: in Alternative 1 the expression \(\dfrac{4 - 2\cos\theta}{5 - 4\cos\theta}\) is the real part but is printed as “Imaginary part”, and the note says “Rearranges imaginary part”; the bounds \(\dfrac{1}{6}\) and \(\dfrac{3}{2}\) are reached at \(\cos\theta = -1\) and \(\cos\theta = 1\), so they are printed as strict inequalities but should be \(\leqslant\))