AQA Further Paper 2 2023 June — Question 9 7 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2023
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeComplex number arithmetic and simplification
DifficultyChallenging +1.2 Part (a) is a routine complex number division requiring multiplication by conjugate—standard Further Maths technique. Part (b) requires recognizing that z can be expressed in modulus-argument form, identifying the argument as 7π/12, then using cos(7π/12) = Re(z)/|z|. While this involves multiple steps and connecting complex numbers to trigonometry, it's a well-signposted proof with clear direction. The calculation is moderately involved but follows established methods for Further Maths students.
Spec1.05g Exact trigonometric values: for standard angles4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide
The complex number \(z\) is such that $$z = \frac{1 + \text{i}}{1 - k\text{i}}$$ where \(k\) is a real number.
  1. Find the real part of \(z\) and the imaginary part of \(z\), giving your answers in terms of \(k\) [2 marks]
  2. In the case where \(k = \sqrt{3}\), use part (a) to show that $$\cos \frac{7\pi}{12} = \frac{\sqrt{2} - \sqrt{6}}{4}$$ [5 marks]
Question 9:
AnswerMarks
9(a)Multiplies numerator and
denominator by conjugate
AnswerMarks Guidance
of denominator.1.1a M1
z= × = +i 
1−ki 1+ki 1+k2 1+k2 
1−k
Real part =
1+k2
1+k
Imaginary part =
1+k2
Obtains correct real part
and correct imaginary part.
1+k
Condone i
AnswerMarks Guidance
1+k21.1b A1
Subtotal2
QMarking Instructions AO
AnswerMarks
9(b)Substitutes 3 into z and
finds z
AnswerMarks Guidance
𝑘𝑘 = √1.1a M1
When 3, Re(z)=
4
𝑘𝑘 = √
1+i 2
z = =
1− 3i 2
( ) π  π 7π
argz=arg ( 1+i )−arg 1− 3i = −  −  =
4  3 12
2 7π 1− 3
cos =
2  12  4
( )
7π 2 1− 3 2− 6
cos = =
12 4 4
Obtains the correct value for
AnswerMarks Guidance
z1.1b A1
7π
Obtains argz = by a fully
12
AnswerMarks Guidance
correct method.3.1a B1
Forms an equation of the
form
AnswerMarks Guidance
z cos(arg(z))=Re(z)3.1a M1
Completes a reasoned
argument to show the
AnswerMarks Guidance
required result.2.1 R1
Subtotal5
Question total7
QMarking Instructions AO 
Question 9:
--- 9(a) ---
9(a) | Multiplies numerator and
denominator by conjugate
of denominator. | 1.1a | M1 | 1+i 1+ki 1−k  1+k 
z= × = +i 
1−ki 1+ki 1+k2 1+k2 
1−k
Real part =
1+k2
1+k
Imaginary part =
1+k2
Obtains correct real part
and correct imaginary part.
1+k
Condone i
1+k2 | 1.1b | A1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | Substitutes 3 into z and
finds z
𝑘𝑘 = √ | 1.1a | M1 | 1− 3
When 3, Re(z)=
4
𝑘𝑘 = √
1+i 2
z = =
1− 3i 2
( ) π  π 7π
argz=arg ( 1+i )−arg 1− 3i = −  −  =
4  3 12
2 7π 1− 3
cos =
2  12  4
( )
7π 2 1− 3 2− 6
cos = =
12 4 4
Obtains the correct value for
z | 1.1b | A1
7π
Obtains argz = by a fully
12
correct method. | 3.1a | B1
Forms an equation of the
form
z cos(arg(z))=Re(z) | 3.1a | M1
Completes a reasoned
argument to show the
required result. | 2.1 | R1
Subtotal | 5
Question total | 7
Q | Marking Instructions | AO | Marks | Typical solution
The complex number $z$ is such that
$$z = \frac{1 + \text{i}}{1 - k\text{i}}$$

where $k$ is a real number.

\begin{enumerate}[label=(\alph*)]
\item Find the real part of $z$ and the imaginary part of $z$, giving your answers in terms of $k$
[2 marks]

\item In the case where $k = \sqrt{3}$, use part (a) to show that
$$\cos \frac{7\pi}{12} = \frac{\sqrt{2} - \sqrt{6}}{4}$$
[5 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2023 Q9 [7]}}
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