| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | November |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Sum geometric series with complex terms |
| Difficulty | Challenging +1.8 This is a sophisticated multi-part Further Maths question requiring: (a) standard geometric series recall, (b) complex number manipulation with De Moivre's theorem and algebraic manipulation to derive a non-trivial trigonometric identity, (c) connecting the algebraic result to Riemann sums for integration bounds, and (d) adapting the method independently. The conceptual leap from complex series to integration bounds and the extended algebraic manipulation place this well above average difficulty, though the individual techniques are within Further Maths syllabus. |
| Spec | 1.08g Integration as limit of sum: Riemann sums4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.06b Method of differences: telescoping series4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{z^n - 1}{z - 1}\) | B1 | |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{z^n - 1}{z - 1} = \frac{\cos n\theta - 1 + \mathrm{i}\sin n\theta}{\cos\theta - 1 + \mathrm{i}\sin\theta}\) | B1 | |
| \(\frac{(\cos n\theta - 1 + \mathrm{i}\sin n\theta)(\cos\theta - 1 - \mathrm{i}\sin\theta)}{(\cos\theta - 1 + \mathrm{i}\sin\theta)(\cos\theta - 1 - \mathrm{i}\sin\theta)}\) | M1 | Multiplies numerator and denominator by complex conjugate |
| \(\mathrm{Re}\!\left(\frac{z^n-1}{z-1}\right) = \frac{\cos n\theta\cos\theta + \sin n\theta\sin\theta - \cos n\theta - \cos\theta + 1}{(\cos\theta - 1)^2 + \sin^2\theta}\) | M1 | Takes real part; \(\cos(n-1)\theta = \cos n\theta\cos\theta + \sin n\theta\sin\theta\) |
| \(= \frac{\cos n\theta\cos\theta + \sin n\theta\sin\theta - \cos n\theta}{2(1-\cos\theta)} + \frac{1}{2}\) | A1 | |
| \(= \frac{\cos n\theta(\cos\theta - 1) + \sin n\theta\sin\theta}{2(1-\cos\theta)} + \frac{1}{2}\) | M1 | Factorises |
| \(= \frac{1}{2}\!\left(1 - \cos n\theta + \frac{\sin n\theta\sin\theta}{1 - \cos\theta}\right)\) | M1 A1 | Divides through by denominator. AG |
| Alternative method: | ||
| \(\frac{z^n - 1}{z-1} = \frac{e^{\mathrm{i}n\theta}-1}{e^{\mathrm{i}\theta}-1}\) | B1 | |
| \(\frac{e^{\mathrm{i}(n-\frac{1}{2})\theta} - e^{-\mathrm{i}\frac{1}{2}\theta}}{e^{\mathrm{i}\frac{1}{2}\theta} - e^{-\mathrm{i}\frac{1}{2}\theta}} = \frac{\cos(n-\frac{1}{2})\theta + \mathrm{i}\sin(n-\frac{1}{2})\theta - \cos\frac{1}{2}\theta + \mathrm{i}\sin(\frac{1}{2}\theta)}{2\mathrm{i}\sin\frac{1}{2}\theta}\) | M1 | |
| \(\mathrm{Re}\!\left(\frac{z^n-1}{z-1}\right) = \frac{\sin(n-\frac{1}{2})\theta + \sin\frac{1}{2}\theta}{2\sin\frac{1}{2}\theta}\) | M1 | Takes real part |
| \(= \frac{\sin(n-\frac{1}{2})\theta}{2\sin\frac{1}{2}\theta} + \frac{1}{2}\) | A1 | |
| \(= \frac{\sin n\theta\cos\frac{1}{2}\theta - \cos n\theta\sin\frac{1}{2}\theta}{2\sin\frac{1}{2}\theta} + \frac{1}{2}\) | M1 | Uses compound angle identity |
| \(= \frac{\sin n\theta\cos\frac{1}{2}\theta}{2\sin\frac{1}{2}\theta} - \frac{1}{2}\cos n\theta + \frac{1}{2}\) | M1 | Divides through by denominator |
| \(\frac{\sin n\theta\sin\theta}{4\sin^2\frac{1}{2}\theta} - \frac{1}{2}\cos n\theta + \frac{1}{2} = \frac{\sin n\theta\sin\theta}{2(1-\cos\theta)} - \frac{1}{2}\cos n\theta + \frac{1}{2}\) | A1 | AG. \(\sin\theta = 2\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta\) and \(2\sin^2\frac{1}{2}\theta = 1 - \cos\theta\) |
| 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 \cos x\, dx < \frac{1}{n}\cos\frac{1}{n} + \frac{1}{n}\cos\frac{2}{n} + \ldots + \frac{1}{n}\cos\frac{n-1}{n}\) | M1 A1 | Forms sum of areas of rectangles given in diagram. A0 if comparison with integral missing or unclear |
| \(= \frac{1}{n}\!\left(1 + \cos\frac{1}{n} + \cos\frac{2}{n} + \ldots + \cos\frac{n-1}{n}\right) = \frac{1}{2n}\!\left(1 - \cos\frac{n}{n} + \frac{\sin\frac{n}{n}\sin\frac{1}{n}}{1 - \cos\frac{1}{n}}\right)\) | M1 A1 | Applies result from part (b) with \(\theta = \frac{1}{n}\). AG |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 \cos x\, dx > \frac{1}{n}\cos\frac{1}{n} + \frac{1}{n}\cos\frac{2}{n} + \ldots + \frac{1}{n}\cos\frac{n}{n}\) | M1 A1 | Forms sum of areas of rectangles. A0 if comparison with integral missing or unclear |
| \(= \frac{1}{2n}\!\left(1 - \cos 1 + \frac{\sin 1\sin\frac{1}{n}}{1-\cos\frac{1}{n}}\right) + \frac{1}{n}\cos 1 - \frac{1}{n} = \frac{1}{2n}\!\left(\cos 1 - 1 + \frac{\sin 1\sin\frac{1}{n}}{1-\cos\frac{1}{n}}\right)\) | A1 | |
| 3 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{z^n - 1}{z - 1}$ | B1 | |
| | **1** | |
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{z^n - 1}{z - 1} = \frac{\cos n\theta - 1 + \mathrm{i}\sin n\theta}{\cos\theta - 1 + \mathrm{i}\sin\theta}$ | B1 | |
| $\frac{(\cos n\theta - 1 + \mathrm{i}\sin n\theta)(\cos\theta - 1 - \mathrm{i}\sin\theta)}{(\cos\theta - 1 + \mathrm{i}\sin\theta)(\cos\theta - 1 - \mathrm{i}\sin\theta)}$ | M1 | Multiplies numerator and denominator by complex conjugate |
| $\mathrm{Re}\!\left(\frac{z^n-1}{z-1}\right) = \frac{\cos n\theta\cos\theta + \sin n\theta\sin\theta - \cos n\theta - \cos\theta + 1}{(\cos\theta - 1)^2 + \sin^2\theta}$ | M1 | Takes real part; $\cos(n-1)\theta = \cos n\theta\cos\theta + \sin n\theta\sin\theta$ |
| $= \frac{\cos n\theta\cos\theta + \sin n\theta\sin\theta - \cos n\theta}{2(1-\cos\theta)} + \frac{1}{2}$ | A1 | |
| $= \frac{\cos n\theta(\cos\theta - 1) + \sin n\theta\sin\theta}{2(1-\cos\theta)} + \frac{1}{2}$ | M1 | Factorises |
| $= \frac{1}{2}\!\left(1 - \cos n\theta + \frac{\sin n\theta\sin\theta}{1 - \cos\theta}\right)$ | M1 A1 | Divides through by denominator. AG |
| **Alternative method:** | | |
| $\frac{z^n - 1}{z-1} = \frac{e^{\mathrm{i}n\theta}-1}{e^{\mathrm{i}\theta}-1}$ | B1 | |
| $\frac{e^{\mathrm{i}(n-\frac{1}{2})\theta} - e^{-\mathrm{i}\frac{1}{2}\theta}}{e^{\mathrm{i}\frac{1}{2}\theta} - e^{-\mathrm{i}\frac{1}{2}\theta}} = \frac{\cos(n-\frac{1}{2})\theta + \mathrm{i}\sin(n-\frac{1}{2})\theta - \cos\frac{1}{2}\theta + \mathrm{i}\sin(\frac{1}{2}\theta)}{2\mathrm{i}\sin\frac{1}{2}\theta}$ | M1 | |
| $\mathrm{Re}\!\left(\frac{z^n-1}{z-1}\right) = \frac{\sin(n-\frac{1}{2})\theta + \sin\frac{1}{2}\theta}{2\sin\frac{1}{2}\theta}$ | M1 | Takes real part |
| $= \frac{\sin(n-\frac{1}{2})\theta}{2\sin\frac{1}{2}\theta} + \frac{1}{2}$ | A1 | |
| $= \frac{\sin n\theta\cos\frac{1}{2}\theta - \cos n\theta\sin\frac{1}{2}\theta}{2\sin\frac{1}{2}\theta} + \frac{1}{2}$ | M1 | Uses compound angle identity |
| $= \frac{\sin n\theta\cos\frac{1}{2}\theta}{2\sin\frac{1}{2}\theta} - \frac{1}{2}\cos n\theta + \frac{1}{2}$ | M1 | Divides through by denominator |
| $\frac{\sin n\theta\sin\theta}{4\sin^2\frac{1}{2}\theta} - \frac{1}{2}\cos n\theta + \frac{1}{2} = \frac{\sin n\theta\sin\theta}{2(1-\cos\theta)} - \frac{1}{2}\cos n\theta + \frac{1}{2}$ | A1 | AG. $\sin\theta = 2\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta$ and $2\sin^2\frac{1}{2}\theta = 1 - \cos\theta$ |
| | **7** | |
---
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \cos x\, dx < \frac{1}{n}\cos\frac{1}{n} + \frac{1}{n}\cos\frac{2}{n} + \ldots + \frac{1}{n}\cos\frac{n-1}{n}$ | M1 A1 | Forms sum of areas of rectangles given in diagram. A0 if comparison with integral missing or unclear |
| $= \frac{1}{n}\!\left(1 + \cos\frac{1}{n} + \cos\frac{2}{n} + \ldots + \cos\frac{n-1}{n}\right) = \frac{1}{2n}\!\left(1 - \cos\frac{n}{n} + \frac{\sin\frac{n}{n}\sin\frac{1}{n}}{1 - \cos\frac{1}{n}}\right)$ | M1 A1 | Applies result from part (b) with $\theta = \frac{1}{n}$. AG |
| | **4** | |
---
## Question 8(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \cos x\, dx > \frac{1}{n}\cos\frac{1}{n} + \frac{1}{n}\cos\frac{2}{n} + \ldots + \frac{1}{n}\cos\frac{n}{n}$ | M1 A1 | Forms sum of areas of rectangles. A0 if comparison with integral missing or unclear |
| $= \frac{1}{2n}\!\left(1 - \cos 1 + \frac{\sin 1\sin\frac{1}{n}}{1-\cos\frac{1}{n}}\right) + \frac{1}{n}\cos 1 - \frac{1}{n} = \frac{1}{2n}\!\left(\cos 1 - 1 + \frac{\sin 1\sin\frac{1}{n}}{1-\cos\frac{1}{n}}\right)$ | A1 | |
| | **3** | |8
\begin{enumerate}[label=(\alph*)]
\item State the sum of the series $1 + z + z ^ { 2 } + \ldots + z ^ { n - 1 }$, for $z \neq 1$.
\item By letting $z = \cos \theta + i \sin \theta$, where $\cos \theta \neq 1$, show that
$$1 + \cos \theta + \cos 2 \theta + \ldots + \cos ( n - 1 ) \theta = \frac { 1 } { 2 } \left( 1 - \cos n \theta + \frac { \sin n \theta \sin \theta } { 1 - \cos \theta } \right)$$
\includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-15_833_785_214_680}
The diagram shows the curve with equation $\mathrm { y } = \cos \mathrm { x }$ for $0 \leqslant x \leqslant 1$, together with a set of $n$ rectangles of width $\frac { 1 } { n }$.
\item By considering the sum of the areas of these rectangles, show that
$$\int _ { 0 } ^ { 1 } \cos x d x < \frac { 1 } { 2 n } \left( 1 - \cos 1 + \frac { \sin 1 \sin \frac { 1 } { n } } { 1 - \cos \frac { 1 } { n } } \right)$$
\item Use a similar method to find, in terms of $n$, a lower bound for $\int _ { 0 } ^ { 1 } \cos x d x$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q8 [15]}}