AQA Further Paper 2 2024 June — Question 17 9 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2024
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeOptimization of argument on loci
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring geometric understanding of loci in the complex plane, finding the point with minimum argument on a circle (requiring tangent line reasoning), and exact calculation involving trigonometry. Part (a) is straightforward reading from a diagram, but parts (b)(i)-(ii) require non-trivial geometric insight about where the minimum argument occurs and careful algebraic manipulation to reach the exact forms requested. The 9 total marks and proof element place this above average difficulty but not at the extreme end for Further Maths.
Spec4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines
The Argand diagram below shows a circle \(C\) \includegraphics{figure_17}
  1. Write down the equation of the locus of \(C\) in the form $$|z - w| = a$$ where \(w\) is a complex number whose real and imaginary parts are integers, and \(a\) is an integer. [2 marks]
  2. It is given that \(z_1\) is a complex number representing a point on \(C\). Of all the complex numbers which represent points on \(C\), \(z_1\) has the least argument.
    1. Find \(|z_1|\) Give your answer in an exact form. [3 marks]
    2. Show that \(\arg z_1 = \arcsin\left(\frac{6\sqrt{3} - 2}{13}\right)\) [4 marks]
Question 17:
AnswerMarks Guidance
17(a)Obtains 4+6i
Allow z−4−6i1.2 B1
Obtains a=21.2 B1
Subtotal2
QMarking instructions AO
AnswerMarks
17(b)(i)Correctly identifies the point
representingz
1
PI by correct method.
or
6−2 3
Obtains m= as the
3
gradient of the tangent at z
AnswerMarks Guidance
12.2a B1
1
OP2 = 52
PQ2 = 4
OQ2 = 48
OQ = 4 3
z = 4 3
1
Uses Pythagoras or other
correct method to obtain z
AnswerMarks Guidance
13.1a M1
Obtains 4 3
Accept any exact correct value
AnswerMarks Guidance
eg 481.1b A1
Subtotal3
QMarking instructions AO
17(b )(ii)Deduces that
 
argz = POR−POQ
1
or
6−2 3
Uses tan(arg(z ))=
1 3
or
48+12 3
Obtains x= or
13
72−8 3
y= where
13
z =x+iy
AnswerMarks Guidance
12.2a B1
sinPOQ= =
52 13
 4 3 2 3
cosPOQ= =
52 13
 6 3
sinPOR= =
52 13
 4 2
cosPOR= =
52 13
 
argz = POR−POQ
1
3 2 3 2 1 6 3–2
sin(arg (z ) )= × – × =
1
13 13 13 13 13
( )
arg z is acute, so
1
 6 3 –2 
( )
arg z = arcsin 
1  
 13 
Uses a suitable
trigonometric identity
or
Uses a correct method to
( )
obtain sin(arg z ) from
1
tan(arg(z ))
1
or
48+12 3
Obtains x= and
13
72−8 3
y= where z =x+iy
AnswerMarks Guidance
13 11.1a M1
Obtains sines and cosines
of P  OR and P  OQ(at least
three correct)
or
sin2(arg(z ))
Obtains or
1
cos2(arg(z ))
1
or
Obtains the values of the
sides of a right-angled
triangle with an angle equal
( )
to arg z
AnswerMarks Guidance
13.1a M1
Uses correct reasoning to
obtain the required result.
Condone omission of
( )
“arg z is acute”.
1
AnswerMarks Guidance
AG2.1 R1
Subtotal4
Question total9
QMarking
instructionsAO Marks 
Question 17:
--- 17(a) ---
17(a) | Obtains 4+6i
Allow z−4−6i | 1.2 | B1 | z–(4+6i) =2
Obtains a=2 | 1.2 | B1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 17(b)(i) ---
17(b)(i) | Correctly identifies the point
representingz
1
PI by correct method.
or
6−2 3
Obtains m= as the
3
gradient of the tangent at z
1 | 2.2a | B1 | Q is the point representing z
1
OP2 = 52
PQ2 = 4
OQ2 = 48
OQ = 4 3
z = 4 3
1
Uses Pythagoras or other
correct method to obtain z
1 | 3.1a | M1
Obtains 4 3
Accept any exact correct value
eg 48 | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
17(b )(ii) | Deduces that
 
argz = POR−POQ
1
or
6−2 3
Uses tan(arg(z ))=
1 3
or
48+12 3
Obtains x= or
13
72−8 3
y= where
13
z =x+iy
1 | 2.2a | B1 |  2 1
sinPOQ= =
52 13
 4 3 2 3
cosPOQ= =
52 13
 6 3
sinPOR= =
52 13
 4 2
cosPOR= =
52 13
 
argz = POR−POQ
1
3 2 3 2 1 6 3–2
sin(arg (z ) )= × – × =
1
13 13 13 13 13
( )
arg z is acute, so
1
 6 3 –2 
( )
arg z = arcsin 
1  
 13 
Uses a suitable
trigonometric identity
or
Uses a correct method to
( )
obtain sin(arg z ) from
1
tan(arg(z ))
1
or
48+12 3
Obtains x= and
13
72−8 3
y= where z =x+iy
13 1 | 1.1a | M1
Obtains sines and cosines
of P  OR and P  OQ(at least
three correct)
or
sin2(arg(z ))
Obtains or
1
cos2(arg(z ))
1
or
Obtains the values of the
sides of a right-angled
triangle with an angle equal
( )
to arg z
1 | 3.1a | M1
Uses correct reasoning to
obtain the required result.
Condone omission of
( )
“arg z is acute”.
1
AG | 2.1 | R1
Subtotal | 4
Question total | 9
Q | Marking
instructions | AO | Marks | Typical solution
The Argand diagram below shows a circle $C$

\includegraphics{figure_17}

\begin{enumerate}[label=(\alph*)]
\item Write down the equation of the locus of $C$ in the form
$$|z - w| = a$$

where $w$ is a complex number whose real and imaginary parts are integers, and $a$ is an integer.
[2 marks]

\item It is given that $z_1$ is a complex number representing a point on $C$. Of all the complex numbers which represent points on $C$, $z_1$ has the least argument.

\begin{enumerate}[label=(\roman*)]
\item Find $|z_1|$

Give your answer in an exact form.
[3 marks]

\item Show that $\arg z_1 = \arcsin\left(\frac{6\sqrt{3} - 2}{13}\right)$
[4 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2024 Q17 [9]}}
This paper (20 questions)
View full paper
Q1 1 Q2 1 Q3 1 Q4 1 Q5 3 Q6 3 Q7 4 Q8 4 Q9 4 Q10 4 Q11 3 Q12 5 Q13 8 Q14 10 Q15 7 Q16 9 Q17 9 Q18 4 Q19 10 Q20 9