| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Roots of unity with trigonometric identities |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question on roots of unity requiring several standard techniques: proving closure under powers (routine), summing geometric series (standard), plotting on Argand diagram (straightforward), and extracting real parts to prove a trigonometric identity (moderately challenging but a well-known application). While it requires multiple steps and synthesis of ideas, each component follows established methods taught in Further Maths. The final part (d) is the most demanding, requiring students to connect De Moivre's theorem with the geometric series result, but this is a classic examination technique rather than requiring novel insight. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02n 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 | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((w^n)^7 = w^{7n} = (w^7)^n = 1^n = 1\) \(\therefore w^n\) satisfies the equation \(z^7 = 1\) | R1 | Completes a rigorous argument to show \(w^n\) satisfies \(z^7 = 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The roots of \(z^7 - 1 = 0\) are \(1, w, w^2, w^3, w^4, w^5, w^6\). \(z^6\) term \(= 0\) \(\therefore\) sum of roots \(= 0\) | M1 | Deduces that LHS is the sum of the roots of \(z^7 = 1\) or factorises \(w^7 - 1\) or uses sum of geometric series with values for \(n\) and \(a\) |
| \(1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0\) as required | R1 | Completes a rigorous argument to show \(1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Six points shown with correct arguments, approximately evenly spaced and approximately symmetric in the real axis | M1 | Shows six required points/vectors with correct arguments |
| Points/vectors have modulus 1, PI by "1" marked on an axis | A1 | Clearly shows modulus 1; labelling of points not required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w = \cos\dfrac{2\pi}{7} + i\sin\dfrac{2\pi}{7}\) | B1 | States \(w = \cos\dfrac{2\pi}{7} + \cdots\) |
| \(w^6 = \cos\left(-\dfrac{2\pi}{7}\right) + i\sin\left(-\dfrac{2\pi}{7}\right) = \cos\dfrac{2\pi}{7} - i\sin\dfrac{2\pi}{7}\) which is the conjugate of \(w\) | E1 | Explains that complex conjugate pairs have the same real part |
| \(\therefore w + w^6 = 2\cos\dfrac{2\pi}{7}\). Similarly \(w^2 + w^5 = 2\cos\dfrac{4\pi}{7}\) and \(w^3 + w^4 = 2\cos\dfrac{6\pi}{7}\) | M1 | Deduces that a sum of pairs of powers of \(w\) equals twice the cosine of a correct angle |
| From part (b): \(1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0\) \((w + w^6) + (w^2 + w^5) + (w^3 + w^4) = -1\) \(2\cos\dfrac{2\pi}{7} + 2\cos\dfrac{4\pi}{7} + 2\cos\dfrac{6\pi}{7} = -1\) \(\therefore \cos\dfrac{2\pi}{7} + \cos\dfrac{4\pi}{7} + \cos\dfrac{6\pi}{7} = -\dfrac{1}{2}\) as required | R1 | Completes rigorous argument using \(1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0\) to show \(\cos\dfrac{2\pi}{7} + \cos\dfrac{4\pi}{7} + \cos\dfrac{6\pi}{7} = -\dfrac{1}{2}\) |
# Question 6:
## Part 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(w^n)^7 = w^{7n} = (w^7)^n = 1^n = 1$ $\therefore w^n$ satisfies the equation $z^7 = 1$ | R1 | Completes a rigorous argument to show $w^n$ satisfies $z^7 = 1$ |
## Part 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The roots of $z^7 - 1 = 0$ are $1, w, w^2, w^3, w^4, w^5, w^6$. $z^6$ term $= 0$ $\therefore$ sum of roots $= 0$ | M1 | Deduces that LHS is the sum of the roots of $z^7 = 1$ or factorises $w^7 - 1$ or uses sum of geometric series with values for $n$ and $a$ |
| $1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0$ as required | R1 | Completes a rigorous argument to show $1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0$ |
## Part 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Six points shown with correct arguments, approximately evenly spaced and approximately symmetric in the real axis | M1 | Shows six required points/vectors with correct arguments |
| Points/vectors have modulus 1, PI by "1" marked on an axis | A1 | Clearly shows modulus 1; labelling of points not required |
## Part 6(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \cos\dfrac{2\pi}{7} + i\sin\dfrac{2\pi}{7}$ | B1 | States $w = \cos\dfrac{2\pi}{7} + \cdots$ |
| $w^6 = \cos\left(-\dfrac{2\pi}{7}\right) + i\sin\left(-\dfrac{2\pi}{7}\right) = \cos\dfrac{2\pi}{7} - i\sin\dfrac{2\pi}{7}$ which is the conjugate of $w$ | E1 | Explains that complex conjugate pairs have the same real part |
| $\therefore w + w^6 = 2\cos\dfrac{2\pi}{7}$. Similarly $w^2 + w^5 = 2\cos\dfrac{4\pi}{7}$ and $w^3 + w^4 = 2\cos\dfrac{6\pi}{7}$ | M1 | Deduces that a sum of pairs of powers of $w$ equals twice the cosine of a correct angle |
| From part (b): $1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0$ $(w + w^6) + (w^2 + w^5) + (w^3 + w^4) = -1$ $2\cos\dfrac{2\pi}{7} + 2\cos\dfrac{4\pi}{7} + 2\cos\dfrac{6\pi}{7} = -1$ $\therefore \cos\dfrac{2\pi}{7} + \cos\dfrac{4\pi}{7} + \cos\dfrac{6\pi}{7} = -\dfrac{1}{2}$ as required | R1 | Completes rigorous argument using $1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0$ to show $\cos\dfrac{2\pi}{7} + \cos\dfrac{4\pi}{7} + \cos\dfrac{6\pi}{7} = -\dfrac{1}{2}$ |
---6 Let $w$ be the root of the equation $z ^ { 7 } = 1$ that has the smallest argument $\alpha$ in the interval $0 < \alpha < \pi$
6
\begin{enumerate}[label=(\alph*)]
\item Prove that $w ^ { n }$ is also a root of the equation $z ^ { 7 } = 1$ for any integer $n$.
6
\item Prove that $1 + w + w ^ { 2 } + w ^ { 3 } + w ^ { 4 } + w ^ { 5 } + w ^ { 6 } = 0$\\
6
\item Show the positions of $w , w ^ { 2 } , w ^ { 3 } , w ^ { 4 } , w ^ { 5 }$, and $w ^ { 6 }$ on the Argand diagram below.\\[0pt]
[2 marks]\\
\includegraphics[max width=\textwidth, alt={}, center]{44e22a98-6424-4fb1-8a37-c965773cb7b6-08_835_898_1802_571}
6
\item Prove that
$$\cos \frac { 2 \pi } { 7 } + \cos \frac { 4 \pi } { 7 } + \cos \frac { 6 \pi } { 7 } = - \frac { 1 } { 2 }$$
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2020 Q6 [9]}}