AQA Further Paper 1 2022 June — Question 12 17 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2022
SessionJune
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeRoots of unity with derived equations
DifficultyChallenging +1.2 This is a structured multi-part question on complex numbers and De Moivre's theorem. Parts (a)-(b) are routine (finding 5th roots of unity and recognizing symmetry). Part (c) requires applying De Moivre's theorem and algebraic manipulation but follows a standard 'show that' format. Parts (d)-(e) require geometric insight linking the Argand diagram to the polynomial's structure and then solving to find a pentagon dimension. While it spans multiple techniques and requires some geometric reasoning, the scaffolding through parts and standard Further Maths content places it moderately above average difficulty.
Spec4.02k Argand diagrams: geometric interpretation4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers4.05a Roots and coefficients: symmetric functions
The Argand diagram shows the solutions to the equation \(z^5 = 1\) \includegraphics{figure_3}
  1. Solve the equation $$z^5 = 1$$ giving your answers in the form \(z = \cos\theta + i\sin\theta\), where \(0 \leq \theta < 2\pi\) [2 marks]
  2. Explain why the points on an Argand diagram which represent the solutions found in part (a) are the vertices of a regular pentagon. [2 marks]
  3. Show that if \(c = \cos\theta\), where \(z = \cos\theta + i\sin\theta\) is a solution to the equation \(z^5 = 1\), then \(c\) satisfies the equation $$16c^5 - 20c^3 + 5c - 1 = 0$$ [5 marks]
  4. The Argand diagram on page 22 is repeated below. \includegraphics{figure_4} Explain, with reference to the Argand diagram, why the expression $$16c^5 - 20c^3 + 5c - 1$$ has a repeated quadratic factor. [3 marks]
  5. \(O\) is the centre of a regular pentagon \(ABCDE\) such that \(OA = OB = OC = OD = OE = 1\) unit. The distance from \(O\) to \(AB\) is \(h\) By solving the equation \(16c^5 - 20c^3 + 5c - 1 = 0\), show that $$h = \frac{\sqrt{5} + 1}{4}$$ [5 marks]
Question 12:
AnswerMarks
12(a)Obtains at least one
correct non-zero
AnswerMarks Guidance
argument/solution1.1a M1
5πœƒπœƒ = 2π‘›π‘›πœ‹πœ‹
𝑧𝑧 = cos0+isin0,
2πœ‹πœ‹ 2πœ‹πœ‹
cos +isin ,
5 5
4πœ‹πœ‹ 4πœ‹πœ‹
cos +isin ,
5 5
6πœ‹πœ‹ 6πœ‹πœ‹
cos +isin ,
5 5
8πœ‹πœ‹ 8πœ‹πœ‹
Obtains completely
correct solutions must
be 0≀θ<2Ο€
AnswerMarks Guidance
(condone z=1)1.1b A1
Total2 cos +isin
5 5
AnswerMarks Guidance
QMarking instructions AO
AnswerMarks
12(b)States that all the points
are the same distance
AnswerMarks Guidance
from the origin2.4 E1
arguments increase in steps of
2πœ‹πœ‹
οΏ½
5
States that the angles
between lines from the
origin to adjacent points
AnswerMarks Guidance
are all equal2.1 E1
Total2
QMarking instructions AO
AnswerMarks
12(c)Expands
Condone errors5 in or
o(mcoisssπœƒπœƒio+nsi soinf Iπœƒπœƒm)aginary
AnswerMarks Guidance
part1.1a M1
(Lceot sπœƒπœƒ+isinπœƒπœƒ) = 1
𝑐𝑐 = cosπœƒπœƒ,𝑠𝑠 = sinπœƒπœƒ
5 4 3 2 2 3 4 5
𝑐𝑐 +5𝑐𝑐 iπ‘ π‘ βˆ’10𝑐𝑐 𝑠𝑠 βˆ’10𝑐𝑐 i𝑠𝑠 +5𝑐𝑐𝑠𝑠 +𝑖𝑖𝑠𝑠 = 1
Real parts:
5 3 2 4
𝑐𝑐 βˆ’10𝑐𝑐 𝑠𝑠 +5𝑐𝑐𝑠𝑠 = 1
5 3 2 2 2
𝑐𝑐 βˆ’10𝑐𝑐 (1βˆ’π‘π‘ )+5𝑐𝑐(1βˆ’π‘π‘ ) βˆ’1 = 0
5 3 5 3 5
𝑐𝑐 βˆ’10𝑐𝑐 +1 0𝑐𝑐 +5 π‘π‘βˆ’10𝑐𝑐 +5𝑐𝑐 βˆ’1 = 0
5 3
as required 16𝑐𝑐 βˆ’20𝑐𝑐 +5π‘π‘βˆ’1 = 0
Obtains correct
unsimplified Real part of
AnswerMarks Guidance
expansion1.1b A1
Equates real parts3.1a M1
Uses appropriate trig
identity to express real
AnswerMarks Guidance
part in terms of3.1a M1
𝑐𝑐
Completes a rigorous
argument to obtain the
AnswerMarks Guidance
required result2.1 R1
Total5
QMarking instructions AO
AnswerMarks
12(d)Explains that z is the
3
complex conjugate of z
4
and that z is the
2
complex conjugate of z
AnswerMarks Guidance
52.4 E1
z* = z so
4 3
and
z* = z so cos(arg𝑧𝑧3)= cos(arg𝑧𝑧4)= π‘Žπ‘Ž
5 2
So ancdo s(arg𝑧𝑧 2 a)re= bcooths( darogu𝑧𝑧b 5 le) =roo𝑏𝑏ts of the
equation
and𝑐𝑐, b=y π‘Žπ‘Žth e fa𝑐𝑐ct=o5r 𝑏𝑏theore3m, is a
repeated qua1d6r𝑐𝑐atiβˆ’c f2a0c𝑐𝑐tor+ of5 𝑐𝑐 βˆ’1 = 0
(π‘π‘βˆ’π‘Žπ‘Ž)(π‘π‘βˆ’π‘π‘)
5 3
16𝑐𝑐 βˆ’20𝑐𝑐 +5π‘π‘βˆ’1
Explains that the Real
parts of the points on
the diagram are the
solutions of
AnswerMarks Guidance
5 32.2a M1
16𝑐𝑐 βˆ’20𝑐𝑐 +5π‘π‘βˆ’1=0
Completes a rigorous
argument to obtain the
AnswerMarks Guidance
required result2.1 R1
Total3
QMarking instructions AO
AnswerMarks
12(e)Deduces that h is a
solution of the equation
This may appear
AnswerMarks Guidance
anywhere in the solution2.2a B1
5 3
16𝑐𝑐 βˆ’20𝑐𝑐 +5𝑐𝑐 βˆ’1 = 0
4 3 2
(π‘π‘βˆ’1)(16𝑐𝑐 +16𝑐𝑐 βˆ’4𝑐𝑐 βˆ’4𝑐𝑐+1)= 0
Discard
𝑐𝑐 = 1
2 2
(4𝑐𝑐 +2π‘π‘βˆ’1) = 0
2
4𝑐𝑐 +2π‘π‘βˆ’1 = 0
βˆ’1±√5
𝑐𝑐 =
Select the solution with the4 greater absolute
value; so
βˆ’1βˆ’βˆš5 √5+1
as required β„Ž = οΏ½ οΏ½ =
4 4
Factorises to obtain a
linear factor and a
AnswerMarks Guidance
quartic factor or better3.1a M1
Solves the quartic or
quadratic equation
correctly to get only two
AnswerMarks Guidance
solutions1.1b A1
Selects the correct
AnswerMarks Guidance
solution3.2a E1
Completes a rigorous
argument to explain the
AnswerMarks Guidance
required result2.1 R1
Total5
Question total17
Paper total100 
Question 12:
--- 12(a) ---
12(a) | Obtains at least one
correct non-zero
argument/solution | 1.1a | M1 | cos5πœƒπœƒ = 1
5πœƒπœƒ = 2π‘›π‘›πœ‹πœ‹
𝑧𝑧 = cos0+isin0,
2πœ‹πœ‹ 2πœ‹πœ‹
cos +isin ,
5 5
4πœ‹πœ‹ 4πœ‹πœ‹
cos +isin ,
5 5
6πœ‹πœ‹ 6πœ‹πœ‹
cos +isin ,
5 5
8πœ‹πœ‹ 8πœ‹πœ‹
Obtains completely
correct solutions must
be 0≀θ<2Ο€
(condone z=1) | 1.1b | A1
Total | 2 | cos +isin
5 5
Q | Marking instructions | AO | Marks | Typical solution
--- 12(b) ---
12(b) | States that all the points
are the same distance
from the origin | 2.4 | E1 | Each of the solutions has modulus 1, and their
arguments increase in steps of
2πœ‹πœ‹
οΏ½
5
States that the angles
between lines from the
origin to adjacent points
are all equal | 2.1 | E1
Total | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 12(c) ---
12(c) | Expands
Condone errors5 in or
o(mcoisssπœƒπœƒio+nsi soinf Iπœƒπœƒm)aginary
part | 1.1a | M1 | 5
(Lceot sπœƒπœƒ+isinπœƒπœƒ) = 1
𝑐𝑐 = cosπœƒπœƒ,𝑠𝑠 = sinπœƒπœƒ
5 4 3 2 2 3 4 5
𝑐𝑐 +5𝑐𝑐 iπ‘ π‘ βˆ’10𝑐𝑐 𝑠𝑠 βˆ’10𝑐𝑐 i𝑠𝑠 +5𝑐𝑐𝑠𝑠 +𝑖𝑖𝑠𝑠 = 1
Real parts:
5 3 2 4
𝑐𝑐 βˆ’10𝑐𝑐 𝑠𝑠 +5𝑐𝑐𝑠𝑠 = 1
5 3 2 2 2
𝑐𝑐 βˆ’10𝑐𝑐 (1βˆ’π‘π‘ )+5𝑐𝑐(1βˆ’π‘π‘ ) βˆ’1 = 0
5 3 5 3 5
𝑐𝑐 βˆ’10𝑐𝑐 +1 0𝑐𝑐 +5 π‘π‘βˆ’10𝑐𝑐 +5𝑐𝑐 βˆ’1 = 0
5 3
as required 16𝑐𝑐 βˆ’20𝑐𝑐 +5π‘π‘βˆ’1 = 0
Obtains correct
unsimplified Real part of
expansion | 1.1b | A1
Equates real parts | 3.1a | M1
Uses appropriate trig
identity to express real
part in terms of | 3.1a | M1
𝑐𝑐
Completes a rigorous
argument to obtain the
required result | 2.1 | R1
Total | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 12(d) ---
12(d) | Explains that z is the
3
complex conjugate of z
4
and that z is the
2
complex conjugate of z
5 | 2.4 | E1 | By symmetry
z* = z so
4 3
and
z* = z so cos(arg𝑧𝑧3)= cos(arg𝑧𝑧4)= π‘Žπ‘Ž
5 2
So ancdo s(arg𝑧𝑧 2 a)re= bcooths( darogu𝑧𝑧b 5 le) =roo𝑏𝑏ts of the
equation
and𝑐𝑐, b=y π‘Žπ‘Žth e fa𝑐𝑐ct=o5r 𝑏𝑏theore3m, is a
repeated qua1d6r𝑐𝑐atiβˆ’c f2a0c𝑐𝑐tor+ of5 𝑐𝑐 βˆ’1 = 0
(π‘π‘βˆ’π‘Žπ‘Ž)(π‘π‘βˆ’π‘π‘)
5 3
16𝑐𝑐 βˆ’20𝑐𝑐 +5π‘π‘βˆ’1
Explains that the Real
parts of the points on
the diagram are the
solutions of
5 3 | 2.2a | M1
16𝑐𝑐 βˆ’20𝑐𝑐 +5π‘π‘βˆ’1=0
Completes a rigorous
argument to obtain the
required result | 2.1 | R1
Total | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 12(e) ---
12(e) | Deduces that h is a
solution of the equation
This may appear
anywhere in the solution | 2.2a | B1 | h is a solution to the equation
5 3
16𝑐𝑐 βˆ’20𝑐𝑐 +5𝑐𝑐 βˆ’1 = 0
4 3 2
(π‘π‘βˆ’1)(16𝑐𝑐 +16𝑐𝑐 βˆ’4𝑐𝑐 βˆ’4𝑐𝑐+1)= 0
Discard
𝑐𝑐 = 1
2 2
(4𝑐𝑐 +2π‘π‘βˆ’1) = 0
2
4𝑐𝑐 +2π‘π‘βˆ’1 = 0
βˆ’1±√5
𝑐𝑐 =
Select the solution with the4 greater absolute
value; so
βˆ’1βˆ’βˆš5 √5+1
as required β„Ž = οΏ½ οΏ½ =
4 4
Factorises to obtain a
linear factor and a
quartic factor or better | 3.1a | M1
Solves the quartic or
quadratic equation
correctly to get only two
solutions | 1.1b | A1
Selects the correct
solution | 3.2a | E1
Completes a rigorous
argument to explain the
required result | 2.1 | R1
Total | 5
Question total | 17
Paper total | 100
The Argand diagram shows the solutions to the equation $z^5 = 1$

\includegraphics{figure_3}

\begin{enumerate}[label=(\alph*)]
\item Solve the equation
$$z^5 = 1$$
giving your answers in the form $z = \cos\theta + i\sin\theta$, where $0 \leq \theta < 2\pi$ [2 marks]

\item Explain why the points on an Argand diagram which represent the solutions found in part (a) are the vertices of a regular pentagon. [2 marks]

\item Show that if $c = \cos\theta$, where $z = \cos\theta + i\sin\theta$ is a solution to the equation $z^5 = 1$, then $c$ satisfies the equation
$$16c^5 - 20c^3 + 5c - 1 = 0$$ [5 marks]

\item The Argand diagram on page 22 is repeated below.

\includegraphics{figure_4}

Explain, with reference to the Argand diagram, why the expression
$$16c^5 - 20c^3 + 5c - 1$$
has a repeated quadratic factor. [3 marks]

\item $O$ is the centre of a regular pentagon $ABCDE$ such that $OA = OB = OC = OD = OE = 1$ unit.
The distance from $O$ to $AB$ is $h$

By solving the equation $16c^5 - 20c^3 + 5c - 1 = 0$, show that
$$h = \frac{\sqrt{5} + 1}{4}$$ [5 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2022 Q12 [17]}}
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