Edexcel Further Pure 2
Chapter 3 – Further Complex Numbers:
• Write
down a complex number, z, in modulus-argument
form as either z=r(cos ϴ + isin ϴ) or z = reiϴ .
apply de Moivre’s theorem
• to find trigonometric identities
•To find the nth roots of a complex number.
Chapter 3 – Further Complex Numbers: FP1 Recap
In FP1, we have calculated the modulus and argument, now we are
going to use this in the form z=r(cos ϴ + i sin ϴ).
Chapter 3 – Further Complex Numbers
Consider the function eiθ
Remember
Maclaurin/Taylor’s
Comparing with the other series expansions – we see that we have cos and sin
This is called Euler’s relation
Chapter 3 – Further Complex Numbers
Complex numbers written in the modulus argument form
R(cos  + i sin ) can now be rewritten using Euler’s relation
m
as Rei
5i
Therefore
z=
Z
4i
3i
π/3
i
5e
represents the
complex number with
modulus 5 and
argument = π/3
2i
i
-5 -4 -3
-2
-1
0
-i
-2i
-3i
-4i
-5i
1
2
3
4
5
e
Chapter 3 – Further Complex Numbers
Chapter 3 – Further Complex Numbers: Proof
when n = 0  (cos θ + i sin θ)0 = cos 0 + i sin 0

when n < 0  (cos θ + i sin θ)n = (cos θ + i sin θ)-m
1
where n = – m, m > 0
=
m
(cos θ + i sin θ)
1
using De Moivre’s theorem
=
cos mθ + i sin mθ
Rationalising the
cos mθ – i sin mθ
=
(cos mθ + i sin mθ)(cos mθ – i sin mθ) denominator
= cos mθ – i sin mθ
= cos (– mθ) + i sin (– mθ)
= cos (nθ) + i sin (nθ)
(cos θ + i sin θ)n = cos nθ + i sin nθ true for
Exercise 3C, Page 31

Use de Moivre’s theorem to simplify question
1.
Chapter 3 – Further Complex Numbers: Binomial
You need to be able to apply the following binomial
expansion found in C2:
Chapter 3 – Further Complex Numbers: Binomial
Use De Moivre’s theorem to show that
cos 4θ = cos4θ – 6cos2θ sin²θ + sin4θ
and
sin 4θ = 4cos3θ sin θ – 4cos θ sin3θ
cos 4θ + i sin 4θ = (cos θ + i sin θ)4
= cos4θ + 4 cos3θ (i sin θ) + 6 cos2θ (i sin θ)² +
4 cos θ (i sin θ)3 + (i sin θ)4
= cos4θ + i 4 cos3θ sin θ – 6 cos2θ sin²θ –
i 4 cos θ sin3θ + sin4θ
= cos4θ – 6 cos2θ sin²θ + sin4θ +
i (4 cos3θ sin θ – 4 cos θ sin3θ)
Comparing real parts
cos 4θ = cos4θ – 6 cos2θ sin²θ + sin4θ
Comparing imaginary parts
sin 4θ = 4 cos3θ sin θ – 4 cos θ sin3θ
Chapter 3 – Further Complex Numbers
Now
and so
z = cos θ + i sin θ = eiθ
z-1 = cos θ – i sin θ = e-iθ
By alternately adding and subtracting these equation we
obtain
z + 1/z = 2 cos θ
and also (z + 1/z)n = 2 cos nθ
z - 1/z = 2i sin θ
and also (z - 1/z)n = 2i sin nθ
cos θ = 1/2(eiθ + e-iθ)
sin θ = 1/2i (eiθ – e-iθ)
Chapter 3 – Further Complex Numbers
Exercise 3D, Page 36

Answer the following questions:
Questions 1, 2, 3, 4 and 5.

Extension Task:
Question 7.
Chapter 3 – Further Complex Numbers: Nth roots
Solve the equation z3 = 1
Chapter 3 – Further Complex Numbers: Nth roots
Chapter 3 – Further Complex Numbers: Nth roots
Exam Questions
1.
(a) Express 32 cos 6θ in the form p cos 6θ + q cos 4θ + r cos 2θ + s, where p, q,
r and s are integers.
(5)
(b) Show that sin 5q = (sin 5q – 5 sin 3q + 10 sin q ).
(5)
(Total 10 marks)
Exam Answers
1.
(a)
1

z 
z

6
 z
6
 6z
4
 15 z
2
 20  15 z
2
 6z
4
z
6
M1
= z6 + z–6 + 6(z4 + z–4) + 15(z2 + z–2) + 20
M1
64cos6 ϴ = 2 cos 6ϴ + 12 cos 4ϴ + 30 cos 2ϴ + 20
M1
32 cos6ϴ = cos 6ϴ + 6 cos 4ϴ + 15 cos 2ϴ + 10
(p = 1, q = 6, r = 15, s = 10)
A1 any two correct
A1, A1
(5)
Exam Answers
(b)
1

z  
z

=
5
= z5 – 5z3 + 10z –
10
z

5
z
3

1
z
M1, A1
5
1 
1 
1
 5
 3

 z  5   5  z  3   10  z  
z
z 
z 



(2i sin ϴ)5 = 32i sin5ϴ = 2i sin 5ϴ – 10i sin 3ϴ + 20i sin ϴ
sin5 ϴ = (sin 5ϴ – 5sin 3ϴ + 10 sin ϴ)
M1, A1
A1
(Total 10 marks)