AQA Further Paper 2 2023 June — Question 15 10 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2023
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeSum geometric series with complex terms
DifficultyChallenging +1.2 This is a structured multi-part complex numbers question requiring de Moivre's theorem, geometric series summation, and algebraic manipulation. Part (a) is routine application of de Moivre's theorem (2 marks). Part (b) requires recognizing how to express the sine series using complex exponentials and geometric series, which is a standard Further Maths technique but requires some insight. Part (c) involves summing two geometric series and simplifying to the target form using trigonometric identities. While this requires multiple steps and careful algebra, it's a well-scaffolded question following a standard template seen in Further Maths courses, making it moderately above average difficulty but not exceptionally challenging.
Spec1.04i Geometric sequences: nth term and finite series sum4.02q De Moivre's theorem: multiple angle formulae
  1. Given that \(z = \cos \theta + \text{i} \sin \theta\), use de Moivre's theorem to show that $$z^n - z^{-n} = 2\text{i} \sin n\theta$$ [2 marks]
  2. The series \(S\) is defined as $$S = \sin \theta + \sin 3\theta + \ldots + \sin(2n - 1)\theta$$ Use part (a) to express \(S\) in the form $$S = \frac{1}{2\text{i}}(G_1) - \frac{1}{2\text{i}}(G_2)$$ where each of \(G_1\) and \(G_2\) is a geometric series. [3 marks]
  3. Hence, show that $$S = \frac{\sin^2(n\theta)}{\sin \theta}$$ [5 marks]
Question 15:
AnswerMarks Guidance
15(a)Uses de Moivre’s
theorem.1.1a M1
zn =cosnθ+isinnθ
−n (−nθ) (−nθ)
z =cos +isin
( nθ)−isin ( nθ)
=cos
zn −z−n =2isinnθ
Completes a reasoned
argument to obtain the
required result.
Must see
(−nθ)+isin (−nθ)
cos
AnswerMarks Guidance
and zn −z−n2.1 R1
Subtotal2
QMarking Instructions AO
AnswerMarks
15(b)Uses part (a) to express at
least three terms of S in
AnswerMarks Guidance
terms of z3.1a M1
= z−z −1+z3 −z −3 +...+z2n−1−z −(2n−1)
( −(2n−1))
= z+z3 +...+z2n−1− z −1+z −3 +...+z
S = 1( z+z3+...+z2n−1) − 1( z−1+z−3+...+z −(2n−1))
2i 2i
Expresses S or 2iS as the
AnswerMarks Guidance
difference of two series.1.1a M1
Completes a reasoned
argument to obtain the
AnswerMarks Guidance
required result.2.1 R1
Subtotal3
QMarking Instructions AO
AnswerMarks
15(c)Obtains expressions for the
sums of their geometric
AnswerMarks Guidance
series.3.1a M1
2iS = −
( 1−z2 ) ( 1−z −2 )
z2n −1 1−z −2n z2n +z −2n −2
= − =
z−z −1 z−z −1 z−z −1
( zn −z−n)2 ( 2isinnθ)2 2sin2nθ
= = =−
2isinθ 2isinθ isinθ
2nθ
2sin
−2S =−
sinθ
2nθ
sin
S =
sinθ
Obtains fully correct
expressions for the sums of
G and G
AnswerMarks Guidance
1 21.1b A1
Rearranges to obtain
z−z−1in
the denominator
AnswerMarks Guidance
of any fraction.3.1a B1
Obtains z2n +z−2n −2 in
the numerator of their
AnswerMarks Guidance
single fraction.3.1a B1
Completes a reasoned
argument to obtain the
AnswerMarks Guidance
required result.2.1 R1
Subtotal5
Question total10
QMarking Instructions AO 
Question 15:
--- 15(a) ---
15(a) | Uses de Moivre’s
theorem. | 1.1a | M1 | By de Moivre’s theorem,
zn =cosnθ+isinnθ
−n (−nθ) (−nθ)
z =cos +isin
( nθ)−isin ( nθ)
=cos
zn −z−n =2isinnθ
Completes a reasoned
argument to obtain the
required result.
Must see
(−nθ)+isin (−nθ)
cos
and zn −z−n | 2.1 | R1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical solution
--- 15(b) ---
15(b) | Uses part (a) to express at
least three terms of S in
terms of z | 3.1a | M1 | 2iS =2isinθ+2isin3θ+...+2isin ( 2n−1 )θ
= z−z −1+z3 −z −3 +...+z2n−1−z −(2n−1)
( −(2n−1))
= z+z3 +...+z2n−1− z −1+z −3 +...+z
S = 1( z+z3+...+z2n−1) − 1( z−1+z−3+...+z −(2n−1))
2i 2i
Expresses S or 2iS as the
difference of two series. | 1.1a | M1
Completes a reasoned
argument to obtain the
required result. | 2.1 | R1
Subtotal | 3
Q | Marking Instructions | AO | Marks | Typical solution
--- 15(c) ---
15(c) | Obtains expressions for the
sums of their geometric
series. | 3.1a | M1 | z ( 1−z2n) z −1 ( 1−z −2n)
2iS = −
( 1−z2 ) ( 1−z −2 )
z2n −1 1−z −2n z2n +z −2n −2
= − =
z−z −1 z−z −1 z−z −1
( zn −z−n)2 ( 2isinnθ)2 2sin2nθ
= = =−
2isinθ 2isinθ isinθ
2nθ
2sin
−2S =−
sinθ
2nθ
sin
S =
sinθ
Obtains fully correct
expressions for the sums of
G and G
1 2 | 1.1b | A1
Rearranges to obtain
z−z−1in
the denominator
of any fraction. | 3.1a | B1
Obtains z2n +z−2n −2 in
the numerator of their
single fraction. | 3.1a | B1
Completes a reasoned
argument to obtain the
required result. | 2.1 | R1
Subtotal | 5
Question total | 10
Q | Marking Instructions | AO | Marks | Typical Solution
\begin{enumerate}[label=(\alph*)]
\item Given that $z = \cos \theta + \text{i} \sin \theta$, use de Moivre's theorem to show that
$$z^n - z^{-n} = 2\text{i} \sin n\theta$$
[2 marks]

\item The series $S$ is defined as
$$S = \sin \theta + \sin 3\theta + \ldots + \sin(2n - 1)\theta$$

Use part (a) to express $S$ in the form
$$S = \frac{1}{2\text{i}}(G_1) - \frac{1}{2\text{i}}(G_2)$$

where each of $G_1$ and $G_2$ is a geometric series.
[3 marks]

\item Hence, show that
$$S = \frac{\sin^2(n\theta)}{\sin \theta}$$
[5 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2023 Q15 [10]}}
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