| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Standard +0.8 This is a standard Further Maths Core Pure question requiring De Moivre's theorem and binomial expansion to derive a multiple angle formula, then solve a trigonometric equation. Part (a) is routine proof, part (b) requires careful algebraic manipulation of (z + 1/z)^5, and part (c) involves recognizing the connection and solving systematically. While multi-step and requiring several techniques, it follows a well-established pattern for this topic with clear signposting between parts. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(z^n + \frac{1}{z^n} \equiv e^{in\theta} + \frac{1}{e^{in\theta}} \equiv e^{in\theta} + e^{-in\theta}\) | M1 | Substitutes \(z\) into LHS and simplifies powers; allow if going direct to trig expressions |
| \(\equiv \cos n\theta + i\sin n\theta + \cos n\theta - i\sin n\theta \equiv 2\cos n\theta\) | A1* | Converts exponential to trig form correctly; no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\left(z + z^{-1}\right)^5 = 32\cos^5\theta\) | B1 | Do not accept \(2^5\) for 32; may be implied |
| \(\left(z + z^{-1}\right)^5 = z^5 + 5z^3 + 10z + 10z^{-1} + 5z^{-3} + z^{-5}\) | M1, A1 | Correct binomial coefficients required; condone at most one slip in powers |
| \(32\cos^5\theta = \left(z^5 + z^{-5}\right) + 5\left(z^3 + z^{-3}\right) + 10\left(z + z^{-1}\right)\) \(= 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta\) | M1 | Sets expressions equal and applies result from (a); grouping must be shown |
| \(\cos^5\theta = \frac{1}{16}\left(\cos 5\theta + 5\cos 3\theta + 10\cos\theta\right)\) | A1* | Reaches printed answer with no errors and all relevant steps shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\cos 5\theta + 5\cos 3\theta + 10\cos\theta = -2\cos\theta \Rightarrow 16\cos^5\theta = -2\cos\theta\) | B1 | Uses result from (b) to deduce correct equation |
| \(2\cos\theta\left(8\cos^4\theta + 1\right) = 0 \Rightarrow \theta = \ldots\) | M1 | Must use part (b) to obtain \(\alpha\cos^5\theta = \beta\cos\theta\); collects to one side and factorises; dividing by \(\cos\theta\) is M0 |
| \(8\cos^4\theta + 1 = 0\) has no solution so \(\cos\theta = 0\) \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}\) | A1 | Rejects inappropriate solution; must consider \(8\cos^4\theta + 1\) having no solutions |
# Question 4:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $z^n + \frac{1}{z^n} \equiv e^{in\theta} + \frac{1}{e^{in\theta}} \equiv e^{in\theta} + e^{-in\theta}$ | M1 | Substitutes $z$ into LHS and simplifies powers; allow if going direct to trig expressions |
| $\equiv \cos n\theta + i\sin n\theta + \cos n\theta - i\sin n\theta \equiv 2\cos n\theta$ | A1* | Converts exponential to trig form correctly; no errors seen |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\left(z + z^{-1}\right)^5 = 32\cos^5\theta$ | B1 | Do not accept $2^5$ for 32; may be implied |
| $\left(z + z^{-1}\right)^5 = z^5 + 5z^3 + 10z + 10z^{-1} + 5z^{-3} + z^{-5}$ | M1, A1 | Correct binomial coefficients required; condone at most one slip in powers |
| $32\cos^5\theta = \left(z^5 + z^{-5}\right) + 5\left(z^3 + z^{-3}\right) + 10\left(z + z^{-1}\right)$ $= 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta$ | M1 | Sets expressions equal and applies result from (a); grouping must be shown |
| $\cos^5\theta = \frac{1}{16}\left(\cos 5\theta + 5\cos 3\theta + 10\cos\theta\right)$ | A1* | Reaches printed answer with no errors and all relevant steps shown |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $\cos 5\theta + 5\cos 3\theta + 10\cos\theta = -2\cos\theta \Rightarrow 16\cos^5\theta = -2\cos\theta$ | B1 | Uses result from (b) to deduce correct equation |
| $2\cos\theta\left(8\cos^4\theta + 1\right) = 0 \Rightarrow \theta = \ldots$ | M1 | Must use part (b) to obtain $\alpha\cos^5\theta = \beta\cos\theta$; collects to one side and factorises; dividing by $\cos\theta$ is M0 |
| $8\cos^4\theta + 1 = 0$ has no solution so $\cos\theta = 0$ $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$ | A1 | Rejects inappropriate solution; must consider $8\cos^4\theta + 1$ having no solutions |
---\begin{enumerate}
\item The complex number $z = \mathrm { e } ^ { \mathrm { i } \theta }$, where $\theta$ is real.\\
(a) Show that
\end{enumerate}
$$z ^ { n } + \frac { 1 } { z ^ { n } } \equiv 2 \cos n \theta$$
where $n$ is a positive integer.\\
(b) Show that
$$\cos ^ { 5 } \theta = \frac { 1 } { 16 } ( \cos 5 \theta + 5 \cos 3 \theta + 10 \cos \theta )$$
(c) Hence, making your reasoning clear, determine all the solutions of
$$\cos 5 \theta + 5 \cos 3 \theta + 12 \cos \theta = 0$$
in the interval $0 \leqslant \theta < 2 \pi$
\hfill \mbox{\textit{Edexcel CP1 2024 Q4 [10]}}