| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Sum geometric series with complex terms |
| Difficulty | Challenging +1.2 Part (a) is direct recall of geometric series formula. Part (b) requires recognizing that z^n = 1 for nth roots of unity, making this a standard bookwork result. Part (c) involves applying de Moivre's theorem and equating real parts of an infinite geometric series—a multi-step problem requiring several techniques, but follows a well-established method for this type of question. Overall slightly above average due to the synthesis required in part (c), but remains a fairly standard Further Maths exercise. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks |
|---|---|
| \(\frac{z - z^{n+1}}{1-z}\) or \(\frac{z^{n+1}-z}{z-1}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(z^n = 1\) and \(z \neq 1\) leading to \(z + z^2 + z^3 + \cdots + z^n = 0\) leading to \(1 + z + z^2 + \cdots + z^{n-1} = 0\) | M1 A1 | Must see \(z^n = 1\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{m=1}^{\infty} z^m = \frac{z}{1-z} = \frac{\cos\theta + i\sin\theta}{3 - \cos\theta - i\sin\theta}\) | M1 A1 | Applies sum to infinity and substitutes for \(z\). |
| \(\frac{(\cos\theta + i\sin\theta)(3 - \cos\theta + i\sin\theta)}{(3-\cos\theta)^2 + \sin^2\theta}\) | M1 A1 | Rationalises denominator. |
| \(\frac{3\cos\theta - \cos^2\theta - \sin^2\theta + i\sin\theta\cos\theta + i\sin\theta(3-\cos\theta)}{9 - 6\cos\theta + \cos^2\theta + \sin^2\theta}\) | M1 | Applies \(\sin^2\theta + \cos^2\theta = 1\) or \(e^{i\theta} + e^{-i\theta} = 2\cos\theta\). |
| \(\text{Re}\left(\sum_{m=1}^{\infty} z^m\right) = \frac{3\cos\theta - 1}{10 - 6\cos\theta}\) | M1 A1 | Takes the real part, AG. |
## Question 5(a):
$\frac{z - z^{n+1}}{1-z}$ or $\frac{z^{n+1}-z}{z-1}$ | **B1** |
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## Question 5(b):
$z^n = 1$ and $z \neq 1$ leading to $z + z^2 + z^3 + \cdots + z^n = 0$ leading to $1 + z + z^2 + \cdots + z^{n-1} = 0$ | **M1 A1** | Must see $z^n = 1$.
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## Question 5(c):
$\sum_{m=1}^{\infty} z^m = \frac{z}{1-z} = \frac{\cos\theta + i\sin\theta}{3 - \cos\theta - i\sin\theta}$ | **M1 A1** | Applies sum to infinity and substitutes for $z$.
$\frac{(\cos\theta + i\sin\theta)(3 - \cos\theta + i\sin\theta)}{(3-\cos\theta)^2 + \sin^2\theta}$ | **M1 A1** | Rationalises denominator.
$\frac{3\cos\theta - \cos^2\theta - \sin^2\theta + i\sin\theta\cos\theta + i\sin\theta(3-\cos\theta)}{9 - 6\cos\theta + \cos^2\theta + \sin^2\theta}$ | **M1** | Applies $\sin^2\theta + \cos^2\theta = 1$ or $e^{i\theta} + e^{-i\theta} = 2\cos\theta$.
$\text{Re}\left(\sum_{m=1}^{\infty} z^m\right) = \frac{3\cos\theta - 1}{10 - 6\cos\theta}$ | **M1 A1** | Takes the real part, AG.
---5
\begin{enumerate}[label=(\alph*)]
\item State the sum of the series $z + z ^ { 2 } + z ^ { 3 } + \ldots + z ^ { n }$, for $z \neq 1$.
\item Given that $z$ is an $n$th root of unity and $z \neq 1$, deduce that $1 + z + z ^ { 2 } + \ldots + z ^ { n - 1 } = 0$.
\item Given instead that $z = \frac { 1 } { 3 } ( \cos \theta + \mathrm { i } \sin \theta )$, use de Moivre's theorem to show that
$$\sum _ { m = 1 } ^ { \infty } 3 ^ { - m } \cos m \theta = \frac { 3 \cos \theta - 1 } { 10 - 6 \cos \theta }$$
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q5 [10]}}