| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Challenging +1.8 This is a sophisticated multi-part Further Maths question requiring students to connect roots of unity, de Moivre's theorem, and polynomial manipulation in a non-routine way. Part (a) is standard, part (b) is a typical de Moivre application, but part (c) requires genuine insight to recognize how the polynomial relates to the previous results and to extract roots in the form cos(kπ). The synthesis across parts and the non-obvious connection elevates this significantly above average A-level difficulty. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks |
|---|---|
| 4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a) | ei2 5 kπ, k =0,1,2,3,4 | B2 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(b) | ( c+is )4 =c4 +4c3(is)+6c2(is)2 +4c(is)3+(is)4 | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| cos4θ=c4 −6c2s2 +s4 =c4 −6c2(1−c2)+(1−c2)2 | M1 | Applies s2 =1−c2 or 2cosθ=z+z −1. |
| 8cos4θ−8cos2θ+1 | A1 | AG. Can get final A1 without first. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 4(c) | 8x9 −8x7 +x5 −8x4 +8x2 −1=(x5 −1)(8x4 −8x2 +1) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 A1 | Solves cos4θ=0. A1 for 4θ= 1π,3π,5π,7π OE. |
| Answer | Marks | Guidance |
|---|---|---|
| 8 8 8 8 | A1 | Gives exactly 5 distinct, real roots. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | ei2 5 kπ, k =0,1,2,3,4 | B2 | OE. B1 for 1 correct fifth root of unity.
B2 for exactly 5 distinct, correct roots.
2
--- 4(b) ---
4(b) | ( c+is )4 =c4 +4c3(is)+6c2(is)2 +4c(is)3+(is)4 | M1 A1 | Uses binomial expansion. Can also go RHS to LHS
using 2cosθ=z+z −1.
cos4θ=c4 −6c2s2 +s4 =c4 −6c2(1−c2)+(1−c2)2 | M1 | Applies s2 =1−c2 or 2cosθ=z+z −1.
8cos4θ−8cos2θ+1 | A1 | AG. Can get final A1 without first.
SC: Using double angle formulae scores 1/4.
4
Question | Answer | Marks | Guidance
--- 4(c) ---
4(c) | 8x9 −8x7 +x5 −8x4 +8x2 −1=(x5 −1)(8x4 −8x2 +1) | B1
cos4θ=0 leading to4θ= 1(2k+1)π
2 | M1 A1 | Solves cos4θ=0. A1 for 4θ= 1π,3π,5π,7π OE.
2 2 2 2
cos0,cos1π,cos3π,cos5π,cos7π
8 8 8 8 | A1 | Gives exactly 5 distinct, real roots.
Accept 1,±cos1π,±cos3π.
8 8
4
Question | Answer | Marks | Guidance\begin{enumerate}[label=(\alph*)]
\item Write down all the roots of the equation $x^5 - 1 = 0$. [2]
\item Use de Moivre's theorem to show that $\cos 4\theta = 8\cos^4 \theta - 8\cos^2 \theta + 1$. [4]
\item Use the results of parts (a) and (b) to express each real root of the equation
$$8x^9 - 8x^7 + x^5 - 8x^4 + 8x^2 - 1 = 0$$
in the form $\cos k\pi$, where $k$ is a rational number. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q4 [10]}}