AQA Further Paper 1 2024 June — Question 10 6 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeComplex number arithmetic and simplification
DifficultyStandard +0.8 This is a Further Maths question requiring students to multiply complex numbers in polar form, apply De Moivre's theorem to find the argument of the product (5π/12), then extract tan(5π/12) by converting to Cartesian form and manipulating surds. While the individual steps are standard, the multi-stage process (multiply → simplify trigonometric expressions → rationalize/simplify surds → extract tangent) and the need to work fluently with exact values makes this moderately challenging, though still within expected Further Maths territory.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=14.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae

The complex numbers \(z\) and \(w\) are defined by $$z = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4}$$ and $$w = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6}$$ By evaluating the product \(zw\), show that $$\tan\frac{5\pi}{12} = 2 + \sqrt{3}$$ [6 marks]

Question 10:
AnswerMarks Guidance
10Forms the product zw 1.1a
2 2 3 1
z w = + i + i
2 2 2 2
 
6 2 6 2
= – + i +
4 4 4 4
Also
 π π   π π 
z w = c o s  + + i s in  +
4 6 4 6
5 π 5 π
= c o s + i s in
1 2 1 2
π 5 6 2
s in +
5 π
1 2π 4 4
t a n = =
1 2 5 6 2
c o s –
1 2 4 4
= 2 + 3
Obtains
 
6 2 6 2
– + i +
AnswerMarks Guidance
4 4 4 41.1b A1
States or uses
AnswerMarks Guidance
a r g ( z w ) = a r g ( z ) + a r g ( w )3.1a M1
π π 5 π
States + =
AnswerMarks Guidance
4 6 1 21.1b B1
Uses their
 
6 2 6 2
z w = – + i +
4 4 4 4
to deduce an expression for
tan
AnswerMarks Guidance
122.2a M1
Completes a reasoned
argument to obtain
5 π
t a n = 2 + 3
1 2
AnswerMarks Guidance
AG2.1 R1
Question total6
QMarking instructions AO
Question 10:
10 | Forms the product zw | 1.1a | M1 |    
2 2 3 1
z w = + i + i
2 2 2 2
 
6 2 6 2
= – + i +
4 4 4 4
Also
 π π   π π 
z w = c o s  + + i s in  +
4 6 4 6
5 π 5 π
= c o s + i s in
1 2 1 2
π 5 6 2
s in +
5 π
1 2π 4 4
t a n = =
1 2 5 6 2
c o s –
1 2 4 4
= 2 + 3
Obtains
 
6 2 6 2
– + i +
4 4 4 4 | 1.1b | A1
States or uses
a r g ( z w ) = a r g ( z ) + a r g ( w ) | 3.1a | M1
π π 5 π
States + =
4 6 1 2 | 1.1b | B1
Uses their
 
6 2 6 2
z w = – + i +
4 4 4 4
to deduce an expression for
5π
tan
12 | 2.2a | M1
Completes a reasoned
argument to obtain
5 π
t a n = 2 + 3
1 2
AG | 2.1 | R1
Question total | 6
Q | Marking instructions | AO | Marks | Typical solution
The complex numbers $z$ and $w$ are defined by
$$z = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4}$$
and
$$w = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6}$$

By evaluating the product $zw$, show that
$$\tan\frac{5\pi}{12} = 2 + \sqrt{3}$$
[6 marks]

\hfill \mbox{\textit{AQA Further Paper 1 2024 Q10 [6]}}