FP2 Chapter 3 – Further Complex
Numbers
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 11th December 2015
OVERVIEW
This chapter is divided up into two main sections.
You are pretty much guaranteed one exam question on each of the two halves:
1
Different forms of a complex number:
• Cartesian 𝑧 = 𝑥 + 𝑖𝑦
• Modulus-Argument Form 𝑧 = 𝑟 cos 𝜃 + 𝑖 sin 𝜃 and
• NEW! Exponential form 𝑧 = 𝑟𝑒 𝜃𝑖
Multiplying, dividing and raising to a power of complex
numbers (the last using De Moivre’s Theorem)
Uses of De Moivre’s Theorem (trig identities, roots)
Loci with Complex Numbers
e.g. “Draw 𝑧 − 2𝑖 = 9 on an Argand diagram.”
2
2
-1
RECAP :: Modulus-Argument Form
Im[z]
(𝑥, 𝑦)
If 𝒛 = 𝒙 + 𝒊𝒚 (and suppose
in this case 𝑧 is in the first
quadrant), what was:
𝜃
Re[z]
modulus
?
argument
?
Then In terms of 𝑟 and 𝜃:
𝒙 = 𝒓 𝐜𝐨𝐬?𝜽
𝒚 = 𝒓 𝐬𝐢𝐧?𝜽
𝒛 = 𝒓 𝐜𝐨𝐬 𝜽 +? 𝒊 𝐬𝐢𝐧 𝜽
𝑟= 𝑧 =
𝒙𝟐 +? 𝒚𝟐
𝜃 = arg 𝑧 =
𝒚
?
𝒙
𝐭𝐚𝐧−𝟏
When −𝜋 < 𝜃 < 𝜋 it is known as
the principal?
argument.
This is known as the
modulus-argument
form of 𝑧.
?
RECAP :: Modulus-Argument Form
𝒙 + 𝒊𝒚
−1
𝑖
𝒓
1
?
1
1+𝑖
?2
− 3+𝑖
2
1−𝑖 3
2
?
−1 − 𝑖
?2
?
?
𝜽 Mod-arg form
𝜋
𝑧 = cos 𝜋? + 𝑖 sin 𝜋
?
𝜋
𝜋
𝜋
𝑧 = cos ? + 𝑖 sin
?
2
2
2
𝜋
𝜋
𝜋
𝑧 = 2 cos? + 𝑖 sin
?
4
4
4
5𝜋
5𝜋
5𝜋
?
𝑧 = 2 cos ? + 𝑖 sin
6
6
6
𝜋
𝜋
𝜋
−?
𝑧 = 2 cos − ? + 𝑖 sin −
3
3
3
3𝜋
3𝜋
3𝜋
𝑧 = 2 cos − ? + 𝑖 sin −
−?
4
4
4
Exponential Form
We’ve seen the Cartesian form a complex number 𝑧 = 𝑥 + 𝑦𝑖 and the modulus-argument
form 𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃). But wait, there’s a third form!
In the later chapter on Taylor expansions, you’ll see that you that you can write functions
as an infinitely long polynomial:
cos 𝑥 = 1
𝑥2
−
2!
𝑥4
+
4!
𝑥6
−
6!
+⋯
𝑥3
𝑥5
sin 𝑥 =
𝑥
−
+
−⋯
3!
5!
𝑥2 𝑥3 𝑥4 𝑥5 𝑥6
𝑥
𝑒 = 1+𝑥+
+
+
+
+
+
2! 3! 4! 5! 6!
It looks like the cos 𝑥 and sin 𝑥 somehow add to give 𝑒 𝑥 . The one problem is that the
signs don’t quite match up. But 𝑖 changes sign as we raise it to higher powers.
𝑖𝜃 2
𝑖𝜃 3
𝑖𝜃 4
𝑒 = 1 + 𝑖𝜃 +
+
+
+⋯
2!
3!
4!
𝜃 2 𝑖𝜃 3 𝜃 4
= 1 + 𝑖𝜃 −
−
+
+⋯
2!
3!
4!
𝜃2 𝜃4 𝜃6
𝜃3 𝜃5
= 1−
+
−
+⋯ +𝑖 𝜃−
+
−⋯
2! 4! 6!
3! 5!
= cos 𝜃 + 𝑖 sin 𝜃
?
i𝜃
?
?
Holy smokes Batman!
?
Exponential Form
! Exponential form
𝑧 = 𝑟𝑒 𝑖𝜃
You need to be able to convert to and from exponential form.
𝒙 + 𝒊𝒚
−1
Mod-arg form
𝑧 = cos 𝜋?+ 𝑖 sin 𝜋
?
𝑧 =?
𝑒 𝑖𝜋
? −0.98𝑖
𝑧 = 13𝑒
2 − 3𝑖
−1 + 𝑖
Exp Form
𝜋
𝜋
2 cos + 𝑖 sin
10
10
3𝜋
3𝜋
2 cos ?+ 𝑖 sin
4
4
3𝜋
3𝜋
2 cos ?
+ 𝑖 sin
5
5
𝑧=
𝜋𝑖
2𝑒 10
?
𝑧=
3𝜋𝑖
2𝑒 4
𝑧=
23𝜋𝑖
2𝑒 5
𝒆𝒊𝝅 + 𝟏 = 𝟎
This is Euler’s identity.
It relates the five
most fundamental
constants in maths!
To get Cartesian form,
put in modulusargument form first.
Notice this
is not a
principal
argument.
A Final Example
1
2
Use 𝑒 𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃 to show that cos 𝜃 = (𝑒 𝑖𝜃 + 𝑒 −𝑖𝜃 )
𝑒 𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃
𝑒 −𝑖𝜃 = cos −𝜃 + 𝑖 sin −𝜃 = cos 𝜃 − 𝑖 sin 𝜃
?
∴ 𝑒 𝑖𝜃 + 𝑒 −𝑖𝜃 = 2 cos 𝜃
1
∴ cos 𝜃 = (𝑒 𝑖𝜃 + 𝑒 −𝑖𝜃 )
2
Exercise 3A
Multiplying and Dividing Complex Numbers
FP1 recap:
If 𝑧1 = 𝑟1 cos 𝜃1 + 𝑖 sin 𝜃1 and 𝑧2 = 𝑟2 cos 𝜃2 + 𝑖 sin 𝜃2
Then:
𝑧1 𝑧2 = 𝒓𝟏 𝒓𝟐 (𝐜𝐨𝐬 𝜽𝟏 + 𝜽𝟐 + 𝒊?𝐬𝐢𝐧(𝜽𝟏 + 𝜽𝟐 ))
𝑧1 𝒓𝟏
=
(𝐜𝐨𝐬 𝜽𝟏 − 𝜽𝟐 + 𝒊 𝐬𝐢𝐧(𝜽𝟏 − 𝜽𝟐 ))
𝑧2 𝒓𝟐
?
i.e. If you multiply two complex numbers, you multiply the moduli
and add the arguments, and if you divide them, you divide the
moduli and subtract the arguments.
Similarly if 𝑧1 = 𝑟1 𝑒 𝑖𝜃1 and 𝑧2 = 𝑟2 𝑒 𝑖𝜃2
Then:
?𝟐
𝑧1 𝑧2 = 𝒓𝟏 𝒓𝟐 𝒆𝒊 𝜽𝟏 +𝜽
𝑧1 𝒓𝟏 𝒊 𝜽 −𝜽
= 𝒆 𝟏 ?𝟐
𝑧2 𝒓𝟐
Examples
5𝜋
5𝜋
𝜋
𝜋
3 cos
+ 𝑖 sin
× 4 cos + 𝑖 sin
12
12
12
12
𝝅
𝝅
= 𝟏𝟐 𝐜𝐨𝐬 + 𝒊 𝒔𝒊𝒏
𝟐?
𝟐
= 𝟏𝟐 𝟎 + 𝒊 𝟏 = 𝟏𝟐𝒊
2 cos
𝜋
𝜋
2𝜋
2𝜋
+ 𝑖 sin
× 3 cos
− 𝑖 sin
15
15
5
5
𝜋
𝜋
2𝜋
2𝜋
= 2 cos + 𝑖 sin
× 3 cos −
+ 𝑖 sin −
15
15
5
5
𝝅
𝝅
= 𝟔 𝐜𝐨𝐬 −
+ 𝒊 𝒔𝒊𝒏 −
𝟑 ?
𝟑
=𝟑−𝟑 𝟑𝒊
𝜋
𝜋
2 cos 12 + 𝑖 sin 12
𝟐
𝟑𝝅
𝟑𝝅
=
𝐜𝐨𝐬 −
+ 𝒊 𝐬𝐢𝐧 −
5𝜋
5𝜋
𝟐
𝟒
𝟒
2 cos
+ 𝑖 sin
?
6
6
𝟏 𝟏
=− − 𝒊
𝟐 𝟐
Test Your Understanding
FP2 June 2013 Q2
𝑧 = 5 3 − 5𝑖
Find
(a) |𝑧|
(b) arg(𝑧) in terms of 𝜋
𝜋
𝜋
𝑤 = 2 cos + 𝑖 sin
4
4
Find
(c)
𝑤
𝑧
(d) arg
(1)
(2)
?
?
(1)
𝑤
𝑧
?
(2)
?
Exercise 3B
De Moivre’s Theorem
We saw that:
𝑧1 𝑧2 = 𝒓𝟏 𝒓𝟐 (𝐜𝐨𝐬 𝜽𝟏 + 𝜽𝟐 + 𝒊 𝐬𝐢𝐧(𝜽𝟏 + 𝜽𝟐 ))
Can you think therefore what 𝑧 𝑛 is going to be?
𝒛𝒏 = 𝒓𝒏 𝐜𝐨𝐬?𝒏𝜽 + 𝒊 𝐬𝐢𝐧 𝒏𝜽
This is known as De Moivre’s Theorem.
FP2 June 2013 Q4
Prove by induction that 𝑧 𝑛 = 𝑟 𝑛 cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃
?
They so went there...
Secret Alternative Way
?
De Moivre’s Theorem for Exponential Form
If 𝑧 =
𝑟𝑒 𝑖𝜃
then
𝒛𝒏
=
𝒏
𝒊𝜽
𝒓𝒆
= 𝒓𝒏?𝒆𝒊𝒏𝜽
Examples
Simplify
9𝜋
9𝜋 5
cos +𝑖 sin
17
17
2𝜋
2𝜋 3
cos 17 − 𝑖 sin 17
𝟗𝝅
𝟗𝝅 𝟓
𝒄𝒐𝒔
+ 𝒊 𝒔𝒊𝒏
𝟏𝟕
𝟏𝟕
=
𝟐𝝅
𝟐𝝅
𝒄𝒐𝒔 −
+ 𝒊 𝒔𝒊𝒏 −
𝟏𝟕
𝟏𝟕
= 𝐜𝐨𝐬 𝟑𝝅 + 𝒊 𝐬𝐢𝐧 𝟑𝝅
= 𝐜𝐨𝐬 𝝅 + 𝒊 𝐬𝐢𝐧 𝝅 = −𝟏
Express 1 + 3 𝑖
7
𝟒𝟓𝝅
𝟒𝟓𝝅
+ 𝒊 𝐬𝐢𝐧
𝟏𝟕
𝟏𝟕
=
𝟑
𝟔𝝅
𝟔𝝅
? 𝒄𝒐𝒔 − 𝟏𝟕 + 𝒊 𝒔𝒊𝒏 − 𝟏𝟕
𝐜𝐨𝐬
in the form 𝑥 + 𝑖𝑦 where 𝑥, 𝑦 ∈ ℝ.
𝝅
𝝅
𝟏 + 𝟑 𝒊 = 𝟐(𝐜𝐨𝐬 + 𝒊 𝐬𝐢𝐧 )
𝟑
𝟑
𝟕
𝟕𝝅
𝟕𝝅
Therefore 𝟏 + 𝟑 𝒊 = 𝟐𝟕 𝐜𝐨𝐬? + 𝒊 𝐬𝐢𝐧
𝟑
𝟑
𝝅
𝝅
= 𝟏𝟐𝟖 𝐜𝐨𝐬 + 𝒊 𝐬𝐢𝐧
= 𝟔𝟒 + 𝟔𝟒 𝟑 𝒊
𝟑
𝟑
Exercise 3C
Applications of de Moivre #1: Trig identities
Express cos 3𝜃 in terms of powers of cos 𝜃
Is there someway we could use de Moivre’s theorem to get a cos 3𝜃 term?
𝐜𝐨𝐬 𝜽 + 𝒊 𝐬𝐢𝐧 𝜽
𝟑
?= 𝐜𝐨𝐬 𝟑𝜽 + 𝒊 𝐬𝐢𝐧 𝟑𝜽
Could we use the LHS to get another expression?
Use a Binomial expansion!
cos 𝜃 + 𝑖 sin 𝜃 3 = 𝟏 𝐜𝐨𝐬 𝟑 𝜽 + 𝟑 𝐜𝐨𝐬 𝟐 𝜽 𝒊 𝐬𝐢𝐧 𝜽 + 𝟑 𝐜𝐨𝐬 𝜽 𝒊 𝐬𝐢𝐧 𝜽 𝟐 + 𝒊 𝐬𝐢𝐧 𝜽
?
= 𝐜𝐨𝐬 𝟑 𝜽 + 𝟑𝒊 𝐜𝐨𝐬 𝟐 𝜽 𝐬𝐢𝐧 𝜽 − 𝟑 𝐜𝐨𝐬 𝜽 𝐬𝐢𝐧𝟐 𝜽 − 𝒊 𝐬𝐢𝐧𝟑 𝜽
How could we then compare the two expressions?
Equating real parts:
𝐜𝐨𝐬 𝟑𝜽 = 𝐜𝐨𝐬 𝟑 𝜽 − 𝟑 𝐜𝐨𝐬 𝜽 𝐬𝐢𝐧𝟐 𝜽
?
=⋯
= 𝟒 𝐜𝐨𝐬 𝟑 𝜽 − 𝟑 𝐜𝐨𝐬 𝜽
𝟑
Applications of de Moivre #1: Trig identities
Express
(a) cos 6𝜃 in terms of cos 𝜃.
(b)
sin 6𝜃
,𝜃
sin 𝜃
≠ 𝑛𝜋, in terms of cos 𝜃.
cos 𝜃 + 𝑖 sin 𝜃 6 = cos 6𝜃 + 𝑖 sin 6𝜃
= cos 6 𝜃 + 6 cos 5 𝜃 𝑖 sin 𝜃 + ⋯
= cos6 𝜃 + 6𝑖 cos 5 𝜃 sin 𝜃 − 15 cos 4 𝜃 sin2 𝜃 − 20𝑖 cos 3 𝜃 sin3 𝜃 + 15 cos 2 𝜃 sin4 𝜃
+ 6𝑖 cos 𝜃 sin5 𝜃 − sin6 𝜃
Equating real parts:
cos 6𝜃 = cos 6 𝜃 − 15 cos4 𝜃 sin2 𝜃 + 15 cos 2 𝜃 sin4 𝜃 − sin6 𝜃
= 32 cos6 𝜃 − 48 cos 4 𝜃 + 18 cos2 𝜃 − 1
Equating imaginary parts: (and simplifying)
sin 6𝜃 = 32 cos 5 𝜃 − 32 cos 3 𝜃 + 6 cos 𝜃
Test Your Understanding
FP2 June 2011 Q7
(a) Use de Moivre’s theorem to show that
sin 5𝜃 = 16 sin5 𝜃 − 20 sin3 𝜃 + 5 sin 𝜃
(5)
Hence, given also that sin 3𝜃 = 3 sin 𝜃 − 4 sin3 𝜃
(b) Find all the solutions of
sin 5𝜃 = 5 sin 3𝜃
in the interval 0 ≤ 𝜃 < 2𝜋. Give your answers to 3 decimal places. (6)
?
?
Finding identities for sin𝑛 𝜃 and cos 𝑛 𝜃
The technique we’ve seen allows us to write say cos 3𝜃 in terms of powers of
cos 𝜃 (e.g. cos 3 𝜃).
Is it possible to do the opposite, to say express cos 3 𝜃 in terms of a linear
combination of cos 3𝜃 and cos 𝜃 (with no powers)?
1
1
If 𝑧 = cos 𝜃 + 𝑖 sin 𝜃, what is 𝑧 + 𝑧 and 𝑧 − 𝑧 ?
𝟏
= 𝒛−𝟏 = 𝐜𝐨𝐬 𝜽 + 𝒊 𝐬𝐢𝐧 𝜽 −𝟏 = 𝐜𝐨𝐬 −𝜽 + 𝒊 𝐬𝐢𝐧 −𝜽 = 𝐜𝐨𝐬 𝜽 − 𝒊 𝐬𝐢𝐧 𝜽
𝒛
𝟏
𝟏
∴ 𝒛 + 𝒛 = ⋯ = 𝟐 𝐜𝐨𝐬 𝜽
𝒛−𝒛 =?
𝟐𝒊 𝐬𝐢𝐧 𝜽
1
And what is 𝑧 𝑛 + 𝑛 ?
𝑧
Just using de Moivre’s theorem again:
𝟏?
𝒏
𝒛 + 𝒏 = 𝟐 𝐜𝐨𝐬 𝒏𝜽
𝒛
! Results you need to use (but you don’t need to memorise)
1
1
𝑧 + = 2 cos 𝜃
𝑧 𝑛 + 𝑛 = 2 cos 𝑛𝜃
𝑧
𝑧
1
1
𝑛
𝑧 − = 2𝑖 sin 𝜃
𝑧 − 𝑛 = 2𝑖 sin 𝑛𝜃
𝑧
𝑧
Finding identities for sin𝑛 𝜃 and cos 𝑛 𝜃
Express cos5 𝜃 in the form 𝑎 cos 5𝜃 + 𝑏 cos 3𝜃 + 𝑐 cos 𝜃
5
2 cos 𝜃
5
1
= 𝑧 + Starting
𝑧
point?
10 5
1
+ 3+ 5
𝑧
𝑧
𝑧
1
1
1
32 cos5 𝜃 = 𝑧 5 + 5 + 5 𝑧?3 + 3 + 10 𝑧 +
𝑧
𝑧
𝑧
= 2 cos 5𝜃 + 5(2 cos
3𝜃) + 10(2 cos 𝜃)
?
1
5
5
Therefore cos 5 𝜃 = 16 cos 5𝜃 + 16 cos 3𝜃 + 8 cos 𝜃 ?
32 cos5 𝜃 = 𝑧 5 + 5𝑧 3 + 10𝑧 +
?
1
𝑧 + = 2 cos 𝜃
𝑧
1
𝑧 − = 2𝑖 sin 𝜃
𝑧
1
+ 𝑛 = 2 cos 𝑛𝜃
𝑧
1
𝑛
𝑧 − 𝑛 = 2𝑖 sin 𝑛𝜃
𝑧
𝑧𝑛
Notice how the powers
descend by 2 each time.
Using identities below.
Finding identities for sin𝑛 𝜃 and cos 𝑛 𝜃
1
4
3
4
Prove that sin3 𝜃 = − sin 3𝜃 + sin 𝜃
3
1
Starting
2𝑖 sin 𝜃 3 = 𝑧 −
𝑧
point?
3 1
=
− 3𝑧?+ − 3
𝑧 𝑧
1
1
3
= 𝑧 − 3? − 3 𝑧 −
𝑧
𝑧
= 2𝑖 sin 3𝜃 − 3(2𝑖 sin 𝜃)
−8𝑖 sin3 𝜃
𝑧3
?
Bro Speed Tip: Remember that
terms in such an expansion oscillate
between positive and negative.
Again using identities.
Therefore dividing by −8𝑖:
1
3
sin3 𝜃 = − sin 3𝜃
+
? 4 sin 𝜃
4
1
𝑧 + = 2 cos 𝜃
𝑧
1
𝑧 − = 2𝑖 sin 𝜃
𝑧
1
+ 𝑛 = 2 cos 𝑛𝜃
𝑧
1
𝑛
𝑧 − 𝑛 = 2𝑖 sin 𝑛𝜃
𝑧
𝑧𝑛
Test Your Understanding
(a) Express sin4 𝜃 in the form 𝑎 cos 4𝜃 + 𝑏 cos 2𝜃 + 𝑐
(b) Hence find the exact value of
𝜋
2
0
sin4 𝜃 𝑑𝜃
4
2𝑖 sin 𝜃
4
1
= Starting
𝑧−
𝑧
point?
4
1
16 sin 𝜃 = 𝑧 − 4𝑧 + ?
6− 2+ 4
𝑧
𝑧
1
1
= 𝑧 4 + 4 −?4 𝑧 2 + 2 + 6
𝑧
𝑧
= 2 cos 4𝜃 − 4 ?
2 cos 2𝜃 + 6
1
1
3
∴ sin4 𝜃 = cos 4𝜃 − cos 2𝜃?+
8
2
8
4
𝜋
2
0
sin4 𝜃 𝑑𝜃 =
4
2
1
1
3
sin 4𝜃 − sin 2𝜃?+ 𝜃
32
4
8
1
𝑧 + = 2 cos 𝜃
𝑧
1
𝑧 − = 2𝑖 sin 𝜃
𝑧
𝜋
2
0
=
3𝜋
16
1
+ 𝑛 = 2 cos 𝑛𝜃
𝑧
1
𝑛
𝑧 − 𝑛 = 2𝑖 sin 𝑛𝜃
𝑧
𝑧𝑛
Exercise 3D
Applications of de Moivre #2: Roots
𝑧 𝑛 = 𝑟 𝑛 cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃
We have so far used de Moivre’s theorem when 𝑛 was an integer.
It also works however when 𝑛 is a rational number! (proof not required)
Solve 𝑧 3 = 1
Plot these roots on an Argand diagram.
The modulus-argument form of 1 is
1(𝑐𝑜𝑠 0 + 𝑖 𝑠𝑖𝑛 0) thus:
𝑧 3 = 1 cos 0 + 𝑖 sin 0
= 1(cos 0 + 2𝑘𝜋 + 𝑖 sin 0 + 2𝑘𝜋
1
3
𝑧
2𝜋
3
4𝜋
cos 3
2𝜋
3
4𝜋
𝑖 sin 3
+ 𝑖 sin
2𝜋
3
2𝜋
3
= cos 2𝑘𝜋 + 𝑖 sin 2𝑘𝜋
2𝑘𝜋
2𝑘𝜋
= cos
+?
𝑖 sin
3
3
𝑘 = 0, 𝑧1 = cos 0 + 𝑖 sin 0 = 1
𝑘 = 1, 𝑧2 = cos
𝑧2
?
1
= −2 + 𝑖
1
𝑘 = 2, 𝑧3 =
+
= −2 − 𝑖
These are known as the ‘cube roots of
unity’.
3
2
3
2
𝑧1
2𝜋
3
𝑧3
1 is always a root – the rest are spread
out equally distributed.
Can write three roots as 1, 𝜔, 𝜔2 .
Note that 1 + 𝜔 + 𝜔2 = 0
Applications of de Moivre #2: Roots
Solve 𝑧 4 = 2 + 2 3 𝑖
𝜋
𝜋
+ 𝑖 sin
3
3
𝜋
𝜋
= 4 cos + 2𝑘𝜋 + 𝑖 sin( + 2𝑘𝜋)
3
3
𝜋 2𝑘𝜋
𝜋 2𝑘𝜋
𝑧 = 2 cos
+
+ 𝑖 sin
+
12
4
12
4
𝑧 4 = 4 cos
𝜋
𝜋
+ 𝑖 sin
12
12
7𝜋
7𝜋
cos + 𝑖 sin
12
12
13𝜋
13𝜋
cos
+ 𝑖 sin
12
12
𝑘 = 0, 𝑧 = 2 cos
𝑘 = 1, 𝑧 = 2
𝑘 = 2, 𝑧 = 2
= 2 cos −
?11𝜋
12
+ 𝑖 sin −
11𝜋
12
𝑘 = 3, 𝑧 = ⋯
We need principal roots. Alternative is
to use 𝑘 = 0, 1, −1, −2 instead of 0, 1,
2, 3, which will give your principal roots
directly.
Test Your Understanding
FP2 June 2012 Q3
a) Express the complex number −2 + 2 3 𝑖 in the form 𝑟 cos 𝜃 + 𝑖 sin 𝜃 ,
− 𝜋 < 𝜃 ≤ 𝜋.
(3)
b) Solve the equation
𝑧 4 = −2 + 2 3 i
giving the roots in the form 𝑟 cos 𝜃 + 𝑖 sin 𝜃 , −𝜋 < 𝜃 ≤ 𝜋.
(5)
?
?
You can avoid the
previous large amount
of working by just
dividing your angle by
the power (4), and
adding/subtracting
increments of 2𝜋 again
divided by the power.
Exercise 3E
Chapter 3 Part II – Locus of
Points
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Loci
You have already encountered loci in FP1 as a set of points which satisfy some
restriction.
For example, the definition of a parabola:
A set of points equidistant from a line (the directrix) and a point (the focus).
We often want to find an equation which defines the set of points or the
opposite.
𝑦
3
-3
3
-3
𝑥
𝑧 = 3 means that the modulus
of the complex number has to
be 3. What points does this give
us on the Argand diagram?
Click to
Brosketch >
On quick reminder…
𝑧 = 𝒙 + 𝒊𝒚
= 𝒙𝟐 +? 𝒚𝟐
𝑧 − 3 = 𝒙 + 𝒊𝒚 − 𝟑
= 𝒙 − 𝟑 + 𝒊𝒚 =?
𝒙−𝟑
𝟐
+ 𝒚𝟐
Loci of form 𝑧 − 𝑧1 = 𝑟
𝑦
What does 𝑧 − 𝑧1 = 𝑟 mean?
The difference between some point 𝑧
from a fixed point 𝑧?1 is of length 𝑟.
i.e. The point 𝑧 is a distance 𝑟 from 𝑧1 .
𝑟
𝑧1
𝑥
! 𝑧 − 𝑧1 = 𝑟 is represented by a
circle centre (𝑥1 , 𝑦1 ) with radius 𝑟,
where 𝑧1 = 𝑥1 + 𝑖𝑦1
𝑦
Sketch the locus of points represented
by 𝑧 − 5 − 3𝑖 = 3
5,3
𝑥
𝑧 − 5 − 3𝑖 = |𝑧?− 5 + 3𝑖 |
Find the Cartesian equation of this locus.
𝑧 − 5 − 3𝑖 = 3
𝑥 + 𝑖𝑦 − 5 − 3𝑖 = 3
𝑥−5 +𝑖 𝑦−3 =3
Group by
real/imaginary.
?
Find modulus and square both sides.
𝑥−5
2
+ 𝑦−3
2
=9
Loci of form 𝑧 − 𝑧1 = 𝑧 − 𝑧2
What does 𝑧 − 𝑧1 = 𝑧 − 𝑧2 mean?
𝑦
The distance of 𝑧 from 𝑧1 is the same
as the distance from?𝑧2 .
𝑧2
𝑧1
𝑥
! 𝑧 − 𝑧1 = |𝑧 − 𝑧2 | is represented
is represented by a perpendicular
bisector of the line segment joining
the points 𝑧1 to 𝑧2 .
Sketch the locus of points represented
by 𝑧 = |𝑧 − 6𝑖|. Write its equation.
𝑦
(0,6)
Click to Brosketch >
(The equation is obviously 𝑦 = 3, but
let’s do it the same algebraic way)
𝑥 + 𝑖𝑦 = Equation
𝑥+𝑖 𝑦−6 ?
𝑥 2 + 𝑦 2 = 𝑥 2 + 𝑦 − 6 2 Note that both sides
were squared.
𝑦=3
(0,0)
𝑥
Test Your Understanding
Find the Cartesian equation of the locus of 𝑧 if 𝑧 − 3 = |𝑧 + 𝑖|, and
sketch the locus of 𝑧 on an Argand diagram.
𝑦
𝑥 + 𝑖𝑦 − 3 = 𝑥 + 𝑖𝑦 + 𝑖
𝑥 − 3 + 𝑖𝑦 = 𝑥 + 𝑖 𝑦 + 1
𝑥 − 3 2 + 𝑦2 = 𝑥2 + 𝑦 + 1 2
𝑦 = −3𝑥
+4
Equation
?
4
Argand Diagram ?
(0, −1)
4
3
(3,0)
Loci of form 𝑧 − 𝑧1 = 𝜆 𝑧 − 𝑧2
Find the Cartesian equation of the locus of 𝑧 if 𝑧 − 6 = 2|𝑧 + 6 − 9𝑖|,
and sketch the locus of 𝑧 on an Argand diagram.
𝑥 + 𝑖𝑦 − 6 = 2 𝑥 + 𝑖𝑦 + 6 − 9𝑖
𝑥 − 6 + 𝑖𝑦 = 2 𝑥 + 6 + 𝑖 𝑦 − 9
𝑥 − 6 2 + 𝑦 2 = 2Equation
𝑥+6 2+ ?
𝑦−9
𝑥 − 6 2 + 𝑦2 = 4 𝑥 + 6 2 + 𝑦 − 9 2
…
𝑥 + 10 2 + 𝑦 − 12 2 = 100
𝑦
−10,12
12
𝐵
Argand
2
You can see that at any point on the
circle, you’ll be twice the distance
from 𝐴 as you will be for 𝐵 (this is
Diagram
?
not required
on your sketch).
It’s a little surprising we got a circle,
so an algebraic method is definitely
needed here.
𝑥
𝐴
Test Your Understanding
FP2 June 2012 Q8a
The point 𝑃 represents a complex number 𝑧 on an Argand diagram such
that
𝑧 − 6𝑖 = 2|𝑧 − 3|
(a) Show that, as 𝑧 varies, the locus of 𝑃 is a circle, stating the radius and
the coordinates of the centre of this circle.
?
arg 𝑧 − 𝑧1 = 𝜃
𝑦
Click to Brosketch >
𝜋
6
arg 𝑧 = ?
At any point, the angle made
𝜋
relative to the origin is
6
𝜋
6
𝑥
𝜋
arg 𝑧 =
6
𝜋
arg 𝑥 + 𝑖𝑦 =
6
𝑦
𝜋
1
= tan
=
𝑥
6
3
1
𝑦=
𝑥,
𝑥 > 0, 𝑦 > 0
3
This bit is
important. The
locus is referred
to as a ‘half line’.
Equation ?
𝑧 + 3 + 2𝑖 can
be rewritten as
𝑧 − −3 − 2𝑖
3𝜋
4
(−3, −2)
𝑦
Click to Brosketch >
𝑥
One way of thinking about it: If
we were to add 3 + 2𝑖 to 𝑧 −
(−3 − 2𝑖), we’d then have had
3𝜋
arg 𝑧 = 4 , which would be
centred at the origin.
arg 𝑧 + 3 + 2𝑖 =
arg 𝑥 + 𝑖𝑦 + 3 + 2𝑖 =
3𝜋
4
arg 𝑥 + 3 + 𝑖 𝑦 + 2
=
3𝜋
4
3𝜋
4
Equation ?
𝑦+2
3𝜋
= tan
= −1
𝑥+3
4
𝑦 = −𝑥 − 5,
𝑥 < −3, 𝑦 ≻ 2
?
Harder 𝑎𝑟𝑔 ones
arg
𝑧−6
𝑧−2
=
𝜋
4
?
𝜋
4
Suppose 𝐿1 is half-line such that arg 𝑧 − 6 = 𝜃,
and L2 the half-line such that arg 𝑧 − 2 = 𝜙.
𝜋
It follows 𝜃 − 𝜙 = 4
arg 𝑧 − 6 − arg 𝑧 − 2 =
𝑦
𝐿1
𝐿2
𝜋
4
𝑃(𝑥, 𝑦)
𝜋
2
𝜋
Where is the 4 in this diagram?
Exterior angle of triangle is sum of other two
? on right.
interior, so where indicated
As the two angles 𝜃 and 𝜙 vary (as 𝑧 varies), but
𝜋
the 4 is constant, 𝑃 forms an arc, given that
angles in same segment are equal.
𝜋
4
𝜃
𝜙
𝑥
6
2
How could we find the centre of the circle?
Angle at centre will be double. Resulting
triangle will need to be isosceles as radius
of circle is constant.
?
Equation: 𝒙 − 𝟒
Recall that arg 𝑧1 𝑧2 = arg 𝑧1 + arg 𝑧2
𝑧1
arg
= arg 𝑧1 − arg 𝑧2
𝑧2
𝟐
?
+ 𝒚−𝟐
𝟐
= 𝟖, 𝒚 > 𝟎
Another Example
𝑧
𝜋
If arg
= , sketch the locus of 𝑃(𝑥, 𝑦) which is
𝑧−4𝑖
2
represented by 𝑧 on an Argand diagram.
𝑧
= arg 𝑧 − arg 𝑧 − 4𝑖 = 𝜃 − 𝜙
𝑧 − 4𝑖
𝜋
where 𝜃 − 𝜙 =
arg
2
𝜋
Since the angle at 𝑃 is always 2 , it follows
that?the line segment (0,0) to (4,0) must be
the diameter, so the equation of the circle is
𝑥2 + 𝑦 − 2 2 = 4
𝑦
4
𝜋
2
𝑥
Bro Exam Note: No questions have appeared of the
𝑧−𝑎
form arg 𝑧−𝑏 = 𝑘 since the new syllabus came into
play, but it IS explicitly referenced in the specification.
Another Example
Given that the complex number 𝑧 satisfies the
equation 𝑧 − 12 − 5𝑖 = 3, find the minimum value
of |𝑧| and the maximum.
Since |𝑧| is the
distance to the
origin, this is
minimised and
maximised on the
line which connects
the origin to the
centre of the circle.
𝑦
(12,5)
13
3
?
𝑥
Minimum = 10
Maximum = 16
Exercise 3F
Regions
How would you describe each of the following in words? Therefore draw each of the
regions on an Argand diagram.
𝑧 − 4 − 2𝑖 ≤ 2
𝑧 − 4 < |𝑧 − 6|
“The distance from 4 + 2𝑖 is less than 2”.
𝑦
?
(4,2)
4
𝜋
0 ≤ arg 𝑧 − 2 − 2𝑖 ≤
4
𝜋
? 4
(2,2)
𝑥
“The distance from 4 is less than from 6.”
?
𝑥=5
4
6
𝑧 − 4 − 2𝑖 ≤ 2, 𝑧 − 4 < |𝑧 − 6|
𝜋
and 0 ≤ arg 𝑧 − 2 − 2𝑖 ≤
4
An intersection of the three other regions.
(I couldn’t be ?
bothered to
draw this. Sorry)
Test Your Understanding
P6 June 2003 Q4(i)(b)
Shade the region for which
𝜋
𝜋
𝑧 − 1 ≤ 1 and ≤ arg 𝑧 + 1 ≤
12
2
?
Transformations
If 𝑧 is used to represent the complex number which forms a locus of points,
what do you think will happen for each of the following 𝑤:
𝑤 = 𝑧 + 𝑎 + 𝑖𝑏
𝑤 = 𝑘𝑧
𝑤 = 𝑘𝑧 + 𝑎 + 𝑖𝑏
𝑎
Translation by ?
𝑏
Enlargement by scale factor 𝑘
centre 0,0 ?
Enlargement followed by
translation. ?
Example: If 𝑃 represents the complex number 𝑧 where 𝑧 = 2, describe
1
the locus of the image of 𝑃 under the transformation 𝑤 = 2 𝑧 + 𝑖
1
𝑧
2
1
𝑤−𝑖 = 𝑧
2
1
𝑤−𝑖 =
𝑧
2
1
𝑤−𝑖 = 2 =1
2
𝑤−𝑖 =
?
STEP 1: Make 𝑧 (or multiple of 𝑧) the subject)
STEP 2: Apply |..| to both sides (because we
know what |𝑧| is)
STEP 3: Use 𝑧1 𝑧2 = 𝑧1 |𝑧2 | if necessary.
STEP 4: Substitute using original equation.
𝑎 + 𝑖𝑏
𝑧
𝑤
More Difficult Example
Show that effect on the circle 𝑧 = 1 of the transformation 𝑤 =
𝑖−𝑤
?
𝑤 − 5𝑖
𝑖−𝑤
|𝑧| =
𝑤 −?5𝑖
5𝑖𝑧+𝑖
𝑧+1
STEP 1: Make 𝑧 (or multiple of 𝑧) the subject)
𝑧=
STEP 2: Apply |..| to both sides (because we
know what |𝑧| is)
𝑖−𝑤
=
?
𝑤 − 5𝑖
STEP 3: Use 𝑧1 𝑧2 = 𝑧1 |𝑧2 | or
𝑧1
𝑧2
=
𝑧1
𝑧2
necessary.
𝑖−𝑤
1=
𝑤?
− 5𝑖
STEP 4: Substitute using original equation.
𝑤 − 5𝑖 = 𝑖 − 𝑤
𝑤 − 5𝑖 =?|𝑤 − 𝑖|
STEP 5: Subsequent manipulation to put in
form we recognise the loci of.
𝑦
(0,5)
𝑦=3
Sketch ?
(0,1)
𝑥
if
Test Your Understanding
FP2 June 2009 Q6
A transformation 𝑇 from the 𝑧-plane to the 𝑤-plane is given by
𝑧
𝑤=
,
𝑧 ≠ −𝑖
𝑧+𝑖
The circle with equation 𝑧 = 3 is mapped by 𝑇 onto the curve 𝐶.
(a) Show that 𝐶 is a circle and find its centre and radius. (8)
(b) The region 𝑧 < 3 in the 𝑧-plane is mapped by 𝑇 onto the region 𝑅
in the 𝑤-plane. Shaded the region 𝑅 on an Argand diagram. (2)
(Hint: be careful and don’t guess! Try a value of 𝑧 where you know that 𝑧 < 3, and
then transform it to see where it ends up.)
?
?
𝑤 = 𝑧2
For the transformation 𝑤 = 𝑧 2 , where 𝑧 = 𝑥 + 𝑖𝑦 and 𝑤 = 𝑢 + 𝑖𝑣, find
the locus of 𝑤 when 𝑧 lies on a circle with equation 𝑥 2 + 𝑦 2 = 16
𝑧 =4
?
STEP 0: Represent Cartesian equation as
equation involving 𝑧.
(note: 𝑧 2 is also fine, so don’t do this step)
STEP 1: Make 𝑧 (or multiple of 𝑧) the subject
𝑤 = 𝑧2
𝑤 = |𝑧 2 |
?
STEP 2: Apply |..| to both sides (because we
know what |𝑧| is)
?
STEP 3: Use 𝑧1 𝑧2 = 𝑧1 |𝑧2 | or
𝑤 = 𝑧 𝑧
𝑤 = 16
? 0,0 radius 4.
i.e. A circle with centre
𝑧1
𝑧2
=
𝑧1
𝑧2
necessary.
STEP 4: Substitute using original equation.
if
A slightly different one
Bro Tip: Sometimes we need to represent
𝑧 as 𝑥 + 𝑦𝑖 and 𝑤 as 𝑢 + 𝑖𝑣 so we can
reason about real and imaginary parts.
𝑖𝑧−2
A transformation 𝑇 on the 𝑧-plane to the 𝑤-plane is given by 𝑤 = 1−𝑧 , 𝑧 ≠ 1.
Show that as 𝑧 lies on the real axis in the 𝑧-plane, then 𝑤 lies on the line 𝑙 in the 𝑤plane. Sketch 𝑙 on an Argand diagram.
𝑤+2
?
𝑤+𝑖
𝑤+2
𝑥= ?
𝑤+𝑖
𝑧=
STEP 1: Make 𝑧 (or multiple of 𝑧) the subject.
We’re told 𝑧 lies on the real-axis, so what can we
replace it with? (given that 𝑧 = 𝑥 + 𝑖𝑦)
𝑢 + 𝑖𝑣 + 2
𝑢 + 2 + 𝑖𝑣
=
𝑢 + 𝑖𝑣 + 𝑖 𝑢 + 𝑖 𝑣 + 1
𝑢 + 2 + 𝑖𝑣 𝑢 − 𝑖 𝑣 + 1
=
×
𝑢+𝑖 𝑣+1
𝑢−𝑖 𝑣+1
𝑢 𝑢+2 +𝑣 𝑣+1
𝑢𝑣 − 𝑢 + 2 𝑣 + 1
=
+
𝑖
𝑢2 + 𝑣 + 1 2
𝑢2 + 𝑣 + 1 2
It would be nice to compare real and
imaginary components on each side. First
replace 𝑤 = 𝑢 + 𝑖𝑣 and do the complex
number division on the RHS (as per FP1).
𝑥=
?
𝑢𝑣 − 𝑢 + 2 𝑣 + 1
0=
𝑢2 + 𝑣 + 1 2
0 = 𝑢𝑣 − 𝑢?+ 2 𝑣 + 1
1
𝑣 =− 𝑢−1
2
Now compare imaginary components
(since we get rid of 𝑥)
𝑣
Sketch ?
−2
−1
𝑢
Exercise 3H
Summary
Euler’s relation and de Moivre’s theorem/applications
You’ve been asked to
prove these before in
exams.
Note: both cos 𝑛𝜃 and
sin 𝑛𝜃 to powers and
vice versa.
(this is a screenshot from the official specification)
Loci/Transformations
|𝑧 − 𝑎| = 𝑏 was a circle centred at 𝑎 with
radius 𝑏.
𝑧 − 𝑎 = 𝑘|𝑧 − 𝑏| was a circle but required
an algebraic approach.
arg 𝑧 − 𝑎 = 𝛽 was a half-line starting at 𝑎.
𝑧−𝑎
arg
= 𝛽 required using circle
𝑧−𝑏
theorems. It will probably be the case that
𝜋
𝜋
𝛽 = or .
4
2