| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Standard +0.8 Part (a) is a standard de Moivre's theorem proof requiring recognition that 1/z^n = z^(-n) = cos(nθ) - i·sin(nθ). Part (b) requires expressing cos^6(θ) using the binomial expansion of (z + 1/z)^6, then collecting terms—this is a multi-step algebraic manipulation typical of Further Maths but follows a well-established technique. The 9 marks and algebraic complexity place it moderately above average difficulty. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| a) | \(z^n = \cos n\theta + i \sin n\theta\) | M1 |
| \(\frac{1}{z^n} = z^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \cos(n\theta) - i\sin(n\theta)\) | A1 | Both \(z^n\) and \(z^{-n}\) |
| \(\therefore z^n + \frac{1}{z^n} = \cos n\theta + i\sin n\theta + \cos(n\theta) - i\sin(n\theta)\) | ||
| \(= 2\cos n\theta\) | A1 | Convincing |
| (3) | ||
| b) | \(\left(z + \frac{1}{z}\right)^6\) | M1 |
| \(= z^6 + 6z^4 + 15z^2 + 20 + 15z^{-2} + 6z^{-4} + z^{-6}\) | A2 | A1 at least 4 terms correct; Condone unsimplified \(z\) terms |
| \(= (z^6 + z^{-6}) + (6z^4 + 6z^{-4}) + (15z^2 + 15z^{-2}) + 20\) | m1 | si at least 2 pairs correct |
| \(= 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20\) | A1 | cao |
| \(\therefore (2\cos\theta)^6 = 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20\) | ||
| \((2\cos\theta)^6 = 64\cos^6 \theta\) | ||
| \(32\cos^6 \theta = \cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10\) | A1 | FT cao above |
| (6) | ||
| [9] |
a) | $z^n = \cos n\theta + i \sin n\theta$ | M1 | 1 use of de Moivre's theorem |
| $\frac{1}{z^n} = z^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \cos(n\theta) - i\sin(n\theta)$ | A1 | Both $z^n$ and $z^{-n}$ |
| | | |
| $\therefore z^n + \frac{1}{z^n} = \cos n\theta + i\sin n\theta + \cos(n\theta) - i\sin(n\theta)$ | | |
| $= 2\cos n\theta$ | A1 | Convincing |
| | (3) | |
b) | $\left(z + \frac{1}{z}\right)^6$ | M1 | Attempt to expand |
| $= z^6 + 6z^4 + 15z^2 + 20 + 15z^{-2} + 6z^{-4} + z^{-6}$ | A2 | A1 at least 4 terms correct; Condone unsimplified $z$ terms |
| | | |
| $= (z^6 + z^{-6}) + (6z^4 + 6z^{-4}) + (15z^2 + 15z^{-2}) + 20$ | m1 | si at least 2 pairs correct |
| $= 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20$ | A1 | cao |
| | | |
| $\therefore (2\cos\theta)^6 = 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20$ | | |
| $(2\cos\theta)^6 = 64\cos^6 \theta$ | | |
| | | |
| $32\cos^6 \theta = \cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10$ | A1 | FT cao above |
| | (6) | |
| | [9] | |\begin{enumerate}[label=(\alph*)]
\item Given that $z = \cos\theta + i\sin\theta$, use de Moivre's theorem to show that
$$z^n + \frac{1}{z^n} = 2\cos n\theta .$$ [3]
\item Express $32\cos^6\theta$ in the form $a\cos 6\theta + b\cos 4\theta + c\cos 2\theta + d$, where $a$, $b$, $c$, $d$ are integers whose values are to be determined. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2023 Q3 [9]}}