Transcript Further complex numbers

Introduction
• This chapter extends on what you have learnt in FP1
• You will learn how to find the complex roots of numbers
• You will learn how to use De Moivre’s theorem in solving
equations
• You will see how to plot the loci of points following a rule on an
Argand diagram
• You will see how to solve problems involving transforming a set
of values in one plane into another plane
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
Further complex numbers
y
You can express a complex number
in the form z = r(cosθ + isinθ)
z (x,y)
r
You should hopefully remember the
modulus-argument form of a complex
number z = x + iy from FP1
The value r is the modulus of the
complex number, its distance from
the origin (0,0)
The argument is the angle the
complex number makes with the
positive x-axis, where:
-π < θ ≤ π
To show this visually…
y
θ
x
x
r is the modulus of z, its
absolute value
 This can be calculated using
Pythagoras’ Theorem
𝑟= 𝑧 =
By GCSE trigonometry, length
x = rcosθ and length y = rsinθ
𝑥 = 𝑟𝑐𝑜𝑠𝜃
𝑦 = 𝑟𝑠𝑖𝑛𝜃
𝑥2 + 𝑦2
𝑧 = 𝑥 + 𝑖𝑦
𝑧 = 𝑟𝑐𝑜𝑠𝜃 + 𝑖𝑟𝑠𝑖𝑛𝜃
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
Replace x and y using
the values above
Factorise by
taking out r
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
Further complex numbers
y
You can express a complex number
in the form z = r(cosθ + isinθ)
Express the following complex
number in the modulus-argument
form:
To do this you need to find both the
argument and the modulus of the
complex number
 Start by sketching it on an Argand
diagram
𝑧 = 2 𝑐𝑜𝑠
5𝜋
5𝜋
+ 𝑖𝑠𝑖𝑛
6
6
This is the
argument
1
r
θ
x
√3
 The ‘y’ part is positive so will
go upwards
z = -√3 + i
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
Pay attention to the directions
 The ‘x’ part is negative so
will go in the negative
direction horizontally
Replace r
and θ
 Once sketched you can then find the modulus and
argument using GCSE Pythagoras and Trigonometry
𝑟=
3
2
+ (1)2
𝑇𝑎𝑛𝜃 =
Calculate
𝑟=2
𝜃=
Remember that the
argument is
measured from the
positive x-axis!
arg 𝑧 =
1
3
𝜋
6
5𝜋
6
Inverse
Tan
Subtract
from π
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
Further complex numbers
You can express a complex number
in the form z = r(cosθ + isinθ)
Express the following complex
number in the modulus-argument
form:
z = -√3 + i
𝑧 = 2 𝑐𝑜𝑠
5𝜋
5𝜋
+ 𝑖𝑠𝑖𝑛
6
6
Remember that the argument is not unique
 We could add 2π to them and the result would be
the same, because 2π radians is a complete turn
To do this you need to find both the
argument and the modulus of the
complex number
 Start by sketching it on an Argand
diagram
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
Further complex numbers
y
You can express a complex number
in the form z = r(cosθ + isinθ)
Express the following complex
number in the modulus-argument
form:
To do this you need to find both the
argument and the modulus of the
complex number
 Start by sketching it on an Argand
diagram
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛 −
4
4
1
θ
r
 The ‘y’ part is negative so will
go downwards
z=1-i
𝑧 = 2 𝑐𝑜𝑠 −
Pay attention to the directions
 The ‘x’ part is positive so will
go in the positive direction
horizontally
Replace r
and θ
x
1
 Once sketched you can then find the modulus and
argument using GCSE Pythagoras and Trigonometry
𝑟=
1
𝑟= 2
2
+ (1)2
𝑇𝑎𝑛𝜃 =
Calculate
1
1
𝜃=
Inverse
Tan
𝜋
4
arg 𝑧 = −
𝜋
4
Negative as
below the
x-axis
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 = 𝑟𝑒 𝑖𝜃
Further complex numbers
You can express a complex number
in the form z = reiθ
In chapter 6 you will meet series
expansions of cosθ and sinθ
This can be used to prove the
following result (which we will do
when we come to chapter 6)
If z = x + iy then the complex number
can also be written in this way:
z = reiθ
As before, r is the modulus of the
complex number and θ is the
argument
 This form is known as the
‘exponential form’
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 = 𝑟𝑒 𝑖𝜃
Further complex numbers
y
You can express a complex number
in the form z = reiθ
Express the following complex
number in the form reiθ, where
-π < θ ≤ π
z = 2 – 3i
As with the modulus-argument form,
you should start by sketching an
Argand diagram and use it to find r
and θ
Pay attention to the directions
 The ‘x’ part is positive so will
go in the positive direction
horizontally
 The ‘y’ part is negative so will
go downwards
𝑟=
2
𝑟 = 13
𝑧 = 13𝑒
−0.98𝑖
Replace r
and θ
r
x
3
 Once sketched you can then find the modulus and
argument using GCSE Pythagoras and Trigonometry
2
+ (3)2
𝑇𝑎𝑛𝜃 =
Calculate
𝑧 = 𝑟𝑒 𝑖𝜃
2
θ
3
2
𝜃 = 0.98
arg 𝑧 = −0.98
Inverse
Tan
Negative as
below the
x-axis
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 = 𝑟𝑒 𝑖𝜃
𝑐𝑜𝑠 −𝜃 = 𝑐𝑜𝑠𝜃
𝑠𝑖𝑛 −𝜃 = −𝑠𝑖𝑛𝜃
Further complex numbers
y
You can express a complex number
in the form z = reiθ
In Core 2, you will have seen the
following:
cos(-θ) = cosθ
y = cosθ
1
-360º
-270º
-θ
-180º
0
-90º -θ
θ
θ
90º
180º
270º
θ
-1
You can see that cos(-θ) = cosθ anywhere on the graph
y
sin(-θ) = -sinθ
y = sinθ
1
-θ
-360º
-270º
-180º
-90º
0
θ 90º
180º
270º
θ
-1
You can see that sin(-θ) = -sinθ anywhere on the graph
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 = 𝑟𝑒 𝑖𝜃
𝑐𝑜𝑠 −𝜃 = 𝑐𝑜𝑠𝜃
𝑠𝑖𝑛 −𝜃 = −𝑠𝑖𝑛𝜃
Further complex numbers
You can express a complex number
in the form z = reiθ
Express the following in the form
z = reiθ where –π < θ ≤ π
𝑧 = 2 𝑐𝑜𝑠
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
10
10
𝑧 = 2 𝑐𝑜𝑠
You can see
from the form
that r = √2
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
10
10
You can see
from the form
that θ = π/10
𝜃=
𝑟= 2
𝜋
10
𝑧 = 𝑟𝑒 𝑖𝜃
𝑧=
𝜋
2𝑒 10𝑖
Replace r and θ
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 = 𝑟𝑒 𝑖𝜃
𝑐𝑜𝑠 −𝜃 = 𝑐𝑜𝑠𝜃
𝑠𝑖𝑛 −𝜃 = −𝑠𝑖𝑛𝜃
Further complex numbers
You can express a complex number
in the form z = reiθ
𝑧 = 5 𝑐𝑜𝑠
𝜋
𝜋
− 𝑖𝑠𝑖𝑛
8
8
Express the following in the form
z = reiθ where –π < θ ≤ π
𝑧 = 5 𝑐𝑜𝑠
𝜋
𝜋
− 𝑖𝑠𝑖𝑛
8
8
We need to adjust this first
 The sign in the centre is negative, we need it to
be positive for the ‘rules’ to work
 We also need both angles to be identical. In this
case we can apply the rules we saw a moment ago…
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 = 𝑟𝑒 𝑖𝜃
𝑐𝑜𝑠 −𝜃 = 𝑐𝑜𝑠𝜃
𝑠𝑖𝑛 −𝜃 = −𝑠𝑖𝑛𝜃
Further complex numbers
You can express a complex number
in the form z = reiθ
Express the following in the form
z = reiθ where –π < θ ≤ π
𝑧 = 5 𝑐𝑜𝑠
𝑧 = 5 𝑐𝑜𝑠
𝜋
𝜋
− 𝑖𝑠𝑖𝑛
8
8
𝑧 = 5 𝑐𝑜𝑠 −
Apply cosθ = cos(-θ)
Apply sin(-θ) = -sin(θ)
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛 −
8
8
𝜋
𝜋
− 𝑖𝑠𝑖𝑛
8
8
You can see
You can see
from the form from the form
that r = 5
that θ = -π/8
𝑟=5
𝜃=−
𝜋
8
𝑧 = 𝑟𝑒 𝑖𝜃
𝑧=
𝜋
− 𝑖
5𝑒 8
Replace r and θ
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 = 𝑟𝑒 𝑖𝜃
𝑐𝑜𝑠 −𝜃 = 𝑐𝑜𝑠𝜃
𝑠𝑖𝑛 −𝜃 = −𝑠𝑖𝑛𝜃
Further complex numbers
You can express a complex number
in the form z = reiθ
Express the following in the form
z = x + iy where 𝑥 ∈ ℝ and 𝑦 ∈ ℝ
𝑧=
3𝜋
2𝑒 4 𝑖
𝑧=
3𝜋
2𝑒 4 𝑖
You can see
from the form
that r = √2
You can see
from the form
that θ = 3π/4
𝑟= 2
𝜃=
3𝜋
4
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
This means that x and
y have to be real
numbers (ie not
complex)
Replace r and θ
𝑧 = 2 𝑐𝑜𝑠
3𝜋
3𝜋
+ 𝑖𝑠𝑖𝑛
4
4
You can calculate all of
this! Leave the second part
in terms of i
𝑧 = −1 + 𝑖
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 = 𝑟𝑒 𝑖𝜃
𝑐𝑜𝑠 −𝜃 = 𝑐𝑜𝑠𝜃
𝑠𝑖𝑛 −𝜃 = −𝑠𝑖𝑛𝜃
Further complex numbers
You can express a complex number
in the form z = reiθ
Express the following in the form
r(cosθ + isinθ), where –π < θ ≤ π
𝑧=
23𝜋
2𝑒 5 𝑖
𝑧=
23𝜋
2𝑒 5 𝑖
You can see
from the form
that r = 2
𝑟=2
You can see
from the form
that θ = 23π/5
𝜃=
23𝜋
5
𝜃=
13𝜋
5
𝜃=
3𝜋
5
The value of θ is not
in the range we want.
We can keep
subtracting 2π until
it is!
Subtract
2π
Subtract
2π
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
Replace r and θ
𝑧 = 2 𝑐𝑜𝑠
3𝜋
3𝜋
+ 𝑖𝑠𝑖𝑛
5
5
3A
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 = 𝑟𝑒 𝑖𝜃
𝑠𝑖𝑛 −𝜃 = −𝑠𝑖𝑛𝜃
𝑐𝑜𝑠 −𝜃 = 𝑐𝑜𝑠𝜃
Further complex numbers
You can express a complex number
in the form z = reiθ
𝑒 𝑖𝜃 = 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑒
Use:
𝑒
𝑖𝜃
𝑐𝑜𝑠𝜃 =
1 𝑖𝜃
𝑒 + 𝑒 −𝑖𝜃
2
= 𝑐𝑜𝑠 −𝜃 + 𝑖𝑠𝑖𝑛 −𝜃
𝑒 −𝑖𝜃 = 𝑐𝑜𝑠𝜃 − 𝑖𝑠𝑖𝑛𝜃
= 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
To show that:
𝑖(−𝜃)
1)
𝑒 𝑖𝜃 = 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
2)
𝑒 −𝑖𝜃 = 𝑐𝑜𝑠𝜃 − 𝑖𝑠𝑖𝑛𝜃
Let θ = -θ
Use the relationships
above to rewrite
Add 1 and 2
𝑒 −𝑖𝜃 + 𝑒 𝑖𝜃 = 2𝑐𝑜𝑠𝜃
1 −𝑖𝜃
𝑒
+ 𝑒 𝑖𝜃 = 𝑐𝑜𝑠𝜃
2
Divide by 2
3A
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠 𝜃1 + 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 + 𝜃2
Further complex numbers
You need to know how multiplying and
dividing affects both the modulus and
argument of the resulting complex
number
To be able to do this you need to be
able to use the identities for sine and
cosine of two angles added or
subtracted – you will have seen these in
Core 3
𝑠𝑖𝑛 𝜃1 ± 𝜃2 = 𝑠𝑖𝑛𝜃1 𝑐𝑜𝑠𝜃2 ± 𝑐𝑜𝑠𝜃1 𝑠𝑖𝑛𝜃2
𝑐𝑜𝑠 𝜃1 ± 𝜃2 = 𝑐𝑜𝑠𝜃1 𝑐𝑜𝑠𝜃2 ∓ 𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛𝜃2
𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1
Multiplying a complex number z1 by another complex
number z2, both in the modulus-argument form
𝑧1 = 𝑟1 𝑐𝑜𝑠𝜃1 + 𝑖𝑠𝑖𝑛𝜃1
𝑧2 = 𝑟2 𝑐𝑜𝑠𝜃2 + 𝑖𝑠𝑖𝑛𝜃2
𝑧1 𝑧2 = 𝑟1 𝑐𝑜𝑠𝜃1 + 𝑖𝑠𝑖𝑛𝜃1 × 𝑟2 𝑐𝑜𝑠𝜃2 + 𝑖𝑠𝑖𝑛𝜃2
Rewrite
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠𝜃1 + 𝑖𝑠𝑖𝑛𝜃1 𝑐𝑜𝑠𝜃2 + 𝑖𝑠𝑖𝑛𝜃2
Now you can expand the double bracket as you would with a quadratic
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠𝜃1 𝑐𝑜𝑠𝜃2 + 𝑖𝑐𝑜𝑠𝜃1 𝑠𝑖𝑛𝜃2 + 𝑖𝑠𝑖𝑛𝜃1 𝑐𝑜𝑠𝜃2 + 𝑖 2 𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛𝜃2
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠𝜃1 𝑐𝑜𝑠𝜃2 + 𝑖𝑐𝑜𝑠𝜃1 𝑠𝑖𝑛𝜃2 + 𝑖𝑠𝑖𝑛𝜃1 𝑐𝑜𝑠𝜃2 − 𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛𝜃2
Group terms using the identities to the left
 You can also factorise the ‘i’ out
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠 𝜃1 + 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 + 𝜃2
So when multiplying two complex numbers in the modulusargument form:
 Multiply the moduli
 Add the arguments together
 The form of the answer is the same
3B
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠 𝜃1 + 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 + 𝜃2
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑒 𝑖(𝜃1 +𝜃2 )
Further complex numbers
You need to know how multiplying and
dividing affects both the modulus and
argument of the resulting complex
number
Multiplying a complex number z1 by another complex
number z2, both in the exponential form
To be able to do this you need to be
able to use the identities for sine and
cosine of two angles added or
subtracted – you will have seen these in
Core 3
𝑧1 𝑧2 =
𝑧1 = 𝑟1 𝑒 𝑖𝜃1
𝑟1 𝑒 𝑖𝜃1
𝑟2 𝑒 𝑖𝜃2
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑒 𝑖𝜃1 +𝑖𝜃2
𝑧2 = 𝑟2 𝑒 𝑖𝜃2
Rewrite
 Remember you add the
powers in this situation
You can factorise the
power
𝑠𝑖𝑛 𝜃1 ± 𝜃2 = 𝑠𝑖𝑛𝜃1 𝑐𝑜𝑠𝜃2 ± 𝑐𝑜𝑠𝜃1 𝑠𝑖𝑛𝜃2
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑒 𝑖(𝜃1 +𝜃2)
𝑐𝑜𝑠 𝜃1 ± 𝜃2 = 𝑐𝑜𝑠𝜃1 𝑐𝑜𝑠𝜃2 ∓ 𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛𝜃2
You can see that in this form the process is
essentially the same as for the modulus-argument
form:
𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1
 Multiply the moduli together
 Add the arguments together
 The answer is in the same form
3B
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠 𝜃1 + 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 + 𝜃2
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑒 𝑖(𝜃1 +𝜃2 )
𝑧1
𝑟1
=
𝑐𝑜𝑠 𝜃1 − 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 − 𝜃2
𝑧2
𝑟2
Further complex numbers
You need to know how multiplying and
dividing affects both the modulus and
argument of the resulting complex
number
To be able to do this you need to be
able to use the identities for sine and
cosine of two angles added or
subtracted – you will have seen these in
Core 3
𝑠𝑖𝑛 𝜃1 ± 𝜃2 = 𝑠𝑖𝑛𝜃1 𝑐𝑜𝑠𝜃2 ± 𝑐𝑜𝑠𝜃1 𝑠𝑖𝑛𝜃2
𝑐𝑜𝑠 𝜃1 ± 𝜃2 = 𝑐𝑜𝑠𝜃1 𝑐𝑜𝑠𝜃2 ∓ 𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛𝜃2
2
2
𝑐𝑜𝑠 𝜃 + 𝑠𝑖𝑛 𝜃 = 1
So when dividing two complex numbers in
the modulus-argument form:
 Divide the moduli
 Subtract the arguments
 The form of the answer is the same
Dividing a complex number z1 by another complex
number z2, both in the modulus-argument form
𝑧1 = 𝑟1 𝑐𝑜𝑠𝜃1 + 𝑖𝑠𝑖𝑛𝜃1
𝑧2 = 𝑟2 𝑐𝑜𝑠𝜃2 + 𝑖𝑠𝑖𝑛𝜃2
𝑟1 𝑐𝑜𝑠𝜃1 + 𝑖𝑠𝑖𝑛𝜃1
𝑧1
=
𝑟2 𝑐𝑜𝑠𝜃2 + 𝑖𝑠𝑖𝑛𝜃2
𝑧2
𝑐𝑜𝑠𝜃2 − 𝑖𝑠𝑖𝑛𝜃2
𝑟1 𝑐𝑜𝑠𝜃1 + 𝑖𝑠𝑖𝑛𝜃1
𝑧1
×
=
𝑐𝑜𝑠𝜃2 − 𝑖𝑠𝑖𝑛𝜃2
𝑟2 𝑐𝑜𝑠𝜃2 + 𝑖𝑠𝑖𝑛𝜃2
𝑧2
Multiply to cancel terms
on the denominator
𝑧1
𝑟1 𝑐𝑜𝑠𝜃1 𝑐𝑜𝑠𝜃2 − 𝑖𝑐𝑜𝑠𝜃1 𝑠𝑖𝑛𝜃2 + 𝑖𝑠𝑖𝑛𝜃1 𝑐𝑜𝑠𝜃2 − 𝑖 2 𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛𝜃2
=
𝑧2
𝑟2 𝑐𝑜𝑠𝜃2 𝑐𝑜𝑠𝜃2 − 𝑖𝑐𝑜𝑠𝜃2 𝑠𝑖𝑛𝜃2 + 𝑖𝑠𝑖𝑛𝜃2 𝑐𝑜𝑠𝜃2 − 𝑖 2 𝑠𝑖𝑛𝜃2 𝑠𝑖𝑛𝜃2
𝑧1
𝑟1 𝑐𝑜𝑠𝜃1 𝑐𝑜𝑠𝜃2 − 𝑖𝑐𝑜𝑠𝜃1 𝑠𝑖𝑛𝜃2 + 𝑖𝑠𝑖𝑛𝜃1 𝑐𝑜𝑠𝜃2 + 𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛𝜃2
=
𝑧2
𝑟2 𝑐𝑜𝑠𝜃2 𝑐𝑜𝑠𝜃2 − 𝑖𝑐𝑜𝑠𝜃2 𝑠𝑖𝑛𝜃2 + 𝑖𝑠𝑖𝑛𝜃2 𝑐𝑜𝑠𝜃2 + 𝑠𝑖𝑛𝜃2 𝑠𝑖𝑛𝜃2
𝑟1 𝑐𝑜𝑠𝜃1 𝑐𝑜𝑠𝜃2 + 𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛𝜃2 + 𝑖 𝑠𝑖𝑛𝜃1 𝑐𝑜𝑠𝜃2 − 𝑐𝑜𝑠𝜃1 𝑠𝑖𝑛𝜃2
𝑧1
=
𝑟2 𝑐𝑜𝑠 2 𝜃2 + 𝑠𝑖𝑛2 𝜃2
𝑧2
𝑟1 𝑐𝑜𝑠 𝜃1 − 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 − 𝜃2
𝑧1
=
𝑟2
𝑧2
𝑧1
𝑟1
=
𝑐𝑜𝑠 𝜃1 − 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 − 𝜃2
𝑧2
𝑟2
Multiply
out
Remove
i2
Group
real and
complex
Rewrite
terms
Rewrite
(again!)
3B
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠 𝜃1 + 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 + 𝜃2
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑒 𝑖(𝜃1 +𝜃2 )
𝑧1
𝑟1
=
𝑐𝑜𝑠 𝜃1 − 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 − 𝜃2
𝑧2
𝑟2
𝑧1
𝑟1
= 𝑒 𝑖(𝜃1 −𝜃2 )
𝑧2
𝑟2
Further complex numbers
You need to know how multiplying and
dividing affects both the modulus and
argument of the resulting complex
number
To be able to do this you need to be
able to use the identities for sine and
cosine of two angles added or
subtracted – you will have seen these in
Core 3
𝑠𝑖𝑛 𝜃1 ± 𝜃2 = 𝑠𝑖𝑛𝜃1 𝑐𝑜𝑠𝜃2 ± 𝑐𝑜𝑠𝜃1 𝑠𝑖𝑛𝜃2
𝑐𝑜𝑠 𝜃1 ± 𝜃2 = 𝑐𝑜𝑠𝜃1 𝑐𝑜𝑠𝜃2 ∓ 𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛𝜃2
2
2
𝑐𝑜𝑠 𝜃 + 𝑠𝑖𝑛 𝜃 = 1
Dividing a complex number z1 by another complex
number z2, both in the exponential form
𝑧1 = 𝑟1 𝑒 𝑖𝜃1
𝑟1 𝑒 𝑖𝜃1
𝑧1
=
𝑧2
𝑟2 𝑒 𝑖𝜃2
𝑟1
𝑧1
= 𝑒 𝑖𝜃1 𝑒 −𝑖𝜃2
𝑧2 𝑟2
𝑟1
𝑧1
= 𝑒 𝑖𝜃1 −𝑖𝜃2
𝑧2 𝑟2
𝑟1
𝑧1
= 𝑒 𝑖(𝜃1−𝜃2)
𝑧2 𝑟2
𝑧2 = 𝑟2 𝑒 𝑖𝜃2
Rewrite terms
 The denominator can be
written with a negative power
Multiplying so add the powers
Factorise the power
You can see that in this form the process is essentially the
same as for the modulus-argument form:
 Divide the moduli
 Subtract the arguments
 The answer is in the same form
3B
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠 𝜃1 + 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 + 𝜃2
𝑧1
𝑟1
=
𝑐𝑜𝑠 𝜃1 − 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 − 𝜃2
𝑧2
𝑟2
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑒 𝑖(𝜃1 +𝜃2 )
𝑧1
𝑟1
= 𝑒 𝑖(𝜃1 −𝜃2 )
𝑧2
𝑟2
Further complex numbers
You need to know how multiplying and
dividing affects both the modulus and
argument of the resulting complex
number
3 𝑐𝑜𝑠
5𝜋
5𝜋
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
× 4 𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛
12
12
12
12
3(4) 𝑐𝑜𝑠
Express the following calculation in the
form x + iy:
3 𝑐𝑜𝑠
5𝜋
5𝜋
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
× 4 𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛
12
12
12
12
12 𝑐𝑜𝑠
5𝜋 𝜋
5𝜋 𝜋
+
+ 𝑖𝑠𝑖𝑛
+
12 12
12 12
Combine using one of
the rules above
 Multiply the moduli
 Add the arguments
Simplify
terms
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
2
2
Calculate the cos and sin parts (in
terms of i where needed)
12 0 + 𝑖(1)
Multiply out
= 12𝑖
3B
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠 𝜃1 + 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 + 𝜃2
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑒 𝑖(𝜃1 +𝜃2 )
𝑧1
𝑟1
=
𝑐𝑜𝑠 𝜃1 − 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 − 𝜃2
𝑧2
𝑟2
𝑧1
𝑟1
= 𝑒 𝑖(𝜃1 −𝜃2 )
𝑧2
𝑟2
Further complex numbers
You need to know how multiplying and
dividing affects both the modulus and
argument of the resulting complex
number
Express the following calculation in the
form x + iy:
2 𝑐𝑜𝑠
2 𝑐𝑜𝑠
𝜋
𝜋
2𝜋
2𝜋
+ 𝑖𝑠𝑖𝑛
× 3 𝑐𝑜𝑠
− 𝑖𝑠𝑖𝑛
15
15
5
5
2 𝑐𝑜𝑠
𝜋
𝜋
2𝜋
2𝜋
+ 𝑖𝑠𝑖𝑛
× 3 𝑐𝑜𝑠 −
+ 𝑖𝑠𝑖𝑛 −
15
15
5
5
2(3) 𝑐𝑜𝑠
𝜋
𝜋
2𝜋
2𝜋
+ 𝑖𝑠𝑖𝑛
× 3 𝑐𝑜𝑠
− 𝑖𝑠𝑖𝑛
15
15
5
5
Combine using a
rule from above
𝜋 2𝜋
𝜋 2𝜋
−
+ 𝑖𝑠𝑖𝑛
−
15 5
15
5
Simplify
6 𝑐𝑜𝑠 −
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛 −
3
3
cos(-θ) = cosθ
sin(-θ) = -sinθ
The cos and sin
terms must be added
for this to work!
 Rewrite using the
rules you saw in 3A
6
1
3
+𝑖 −
2
2
Calculate the cos and
sin parts
Multiply out
= 3 − 3 3𝑖
3B
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑐𝑜𝑠 𝜃1 + 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 + 𝜃2
𝑧1 𝑧2 = 𝑟1 𝑟2 𝑒 𝑖(𝜃1 +𝜃2 )
𝑧1
𝑟1
=
𝑐𝑜𝑠 𝜃1 − 𝜃2 + 𝑖𝑠𝑖𝑛 𝜃1 − 𝜃2
𝑧2
𝑟2
𝑧1
𝑟1
= 𝑒 𝑖(𝜃1 −𝜃2 )
𝑧2
𝑟2
Further complex numbers
You need to know how multiplying and
dividing affects both the modulus and
argument of the resulting complex
number
Express the following calculation in the
form x + iy:
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
12
12
5𝜋
5𝜋
2 𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛
6
6
2 𝑐𝑜𝑠
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
12
12
5𝜋
5𝜋
2 𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛
6
6
2 𝑐𝑜𝑠
Combine using one of the
rules above
 Divide the moduli
 Subtract the arguments
𝜋 5𝜋
𝜋 5𝜋
2
𝑐𝑜𝑠
−
+ 𝑖𝑠𝑖𝑛
−
12 6
12 6
2
Simplify
3𝜋
3𝜋
2
𝑐𝑜𝑠 −
+ 𝑖𝑠𝑖𝑛 −
4
4
2
You can work out the
sin and cos parts
1
1
2
−
+𝑖 −
2
2
2
Multiply out
1 1
=− − 𝑖
2 2
3B
Further complex numbers
You need to be able to prove that
[r(cosθ + isinθ)]n = rn(cos(nθ +
isinnθ) for any integer n
Let:
z = r(cosθ + isinθ)
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 2 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
1
2
= 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
= 𝑟 2 (𝑐𝑜𝑠2𝜃 + 𝑖𝑠𝑖𝑛2𝜃)
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧2 = 𝑧 × 𝑧
𝑧 2 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃) × 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
2
𝑧 = 𝑟 (𝑐𝑜𝑠2𝜃 + 𝑖𝑠𝑖𝑛2𝜃)
3
= 𝑟 3 (𝑐𝑜𝑠3𝜃 + 𝑖𝑠𝑖𝑛3𝜃)
𝑧 4 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
4
= 𝑟 4 (𝑐𝑜𝑠4𝜃 + 𝑖𝑠𝑖𝑛4𝜃)
𝑛
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
𝑛
= 𝑟 (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
𝑧 3 = 𝑟 2 (𝑐𝑜𝑠2𝜃 + 𝑖𝑠𝑖𝑛2𝜃) × 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
3
𝑧 = 𝑟 (𝑐𝑜𝑠3𝜃 + 𝑖𝑠𝑖𝑛3𝜃)
𝑧4 = 𝑧3 × 𝑧
𝑧 4 = 𝑟 3 (𝑐𝑜𝑠3𝜃 + 𝑖𝑠𝑖𝑛3𝜃) × 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 4 = 𝑟 4 (𝑐𝑜𝑠4𝜃 + 𝑖𝑠𝑖𝑛4𝜃)
This is De Moivre’s Theorem
 You need to be able to prove this
Multiply the moduli,
add the arguments
𝑧3 = 𝑧2 × 𝑧
3
𝑧 3 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
2
Use the modulusargument form
Use the modulusargument form
Multiply the moduli,
add the arguments
Use the modulusargument form
Multiply the moduli,
add the arguments
De Moivre = ‘De Mwavre’ (pronunciation)
3C
𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
Further complex numbers
You need to be able to prove that
[r(cosθ + isinθ)]n = rn(cos(nθ +
isinnθ) for any integer n
De Moivre’s theorem can be proved
using the method of proof by induction
from FP1
Basis – show the statement is true for
n=1
Assumption – assume the statement is
true for n = k
Inductive – show that if true for n = k,
then the statement is also true for n =
k+1
Conclusion – because the statement is
true for n = 1 and also true if any value
is, then the statement is true for all
values of n
Proving that:
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
is true for all positive integers
BASIS
 Show that the statement is true for n = 1
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
1
= 𝑟1 𝑐𝑜𝑠1𝜃 + 𝑖𝑠𝑖𝑛1𝜃
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃 = 𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
Sub in
n=1
Simplify
each side
ASSUMPTION
 Assume that the statement is true for n = k
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑘
= 𝑟 𝑘 𝑐𝑜𝑠𝑘𝜃 + 𝑖𝑠𝑖𝑛𝑘𝜃
Replace n
with k
3C
𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
Further complex numbers
You need to be able to prove that
[r(cosθ + isinθ)]n = rn(cos(nθ +
isinnθ) for any integer n
De Moivre’s theorem can be proved
using the method of proof by induction
from FP1
Basis – show the statement is true for
n=1
Assumption – assume the statement is
true for n = k
Inductive – show that if true for n = k,
then the statement is also true for n =
k+1
Conclusion – because the statement is
true for n = 1 and also true if any value
is, then the statement is true for all
values of n
Proving that:
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
is true for all positive integers
INDUCTIVE
 Show that if true for n = k, the statement is also
true for n = k + 1
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑛
Sub in n = k + 1
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑘+1
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑘
× 𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
You can write this as
two separate parts as
the powers are added
together
1
= 𝑟 𝑘 𝑐𝑜𝑠𝑘𝜃 + 𝑖𝑠𝑖𝑛𝑘𝜃 × 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
= 𝑟 𝑘+1 cos(𝑘 + 1)𝜃 + 𝑖𝑠𝑖𝑛(𝑘 + 1)𝜃
We can rewrite the first
part based on the
assumption step, and the
second based on the
basis step
Using the multiplication
rules from 3B
 Multiply the moduli,
add the arguments
3C
𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
Further complex numbers
You need to be able to prove that
[r(cosθ + isinθ)]n = rn(cos(nθ +
isinnθ) for any integer n
De Moivre’s theorem can be proved
using the method of proof by induction
from FP1
Basis – show the statement is true for
n=1
Proving that:
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
is true for all positive integers
CONCLUSION
 Explain why this shows it is true…
 We showed the statement is true for n = 1
We then assumed the following:
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑘
= 𝑟 𝑘 𝑐𝑜𝑠𝑘𝜃 + 𝑖𝑠𝑖𝑛𝑘𝜃
Assumption – assume the statement is
true for n = k
Using the assumption, we showed that:
Inductive – show that if true for n = k,
then the statement is also true for n =
k+1
As all the ‘k’ terms have become ‘k + 1’ terms, if the statement
is true for one term, it must be true for the next, and so on…
Conclusion – because the statement is
true for n = 1 and also true if any value
is, then the statement is true for all
values of n
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑘+1
= 𝑟 𝑘+1 cos 𝑘 + 1 𝜃 + 𝑖𝑠𝑖𝑛(𝑘 + 1)𝜃
 The statement was true for 1, so must be true for 2, and
therefore 3, and so on…
 We have therefore proven the statement for all positive
integers!
3C
𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
Further complex numbers
You need to be able to prove that
[r(cosθ + isinθ)]n = rn(cos(nθ +
isinnθ) for any integer n
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
=
We have just proved the theorem for
n = k where k is a positive integer
=
 Now we need to show it is also
true for any negative integer…
=
1
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
−𝑚
Write using a positive
power instead
𝑚
𝑟𝑚
1
𝑐𝑜𝑠𝑚𝜃 + 𝑖𝑠𝑖𝑛𝑚𝜃
𝑟𝑚
1
𝑐𝑜𝑠𝑚𝜃 − 𝑖𝑠𝑖𝑛𝑚𝜃
×
𝑐𝑜𝑠𝑚𝜃 + 𝑖𝑠𝑖𝑛𝑚𝜃
𝑐𝑜𝑠𝑚𝜃 − 𝑖𝑠𝑖𝑛𝑚𝜃
𝑐𝑜𝑠𝑚𝜃 − 𝑖𝑠𝑖𝑛𝑚𝜃
 If n is a negative integer, it can be =
𝑟 𝑚 𝑐𝑜𝑠 2 𝑚𝜃 − 𝑖 2 𝑠𝑖𝑛2 𝑚𝜃
written as ‘-m’, where m is a
positive integer
𝑐𝑜𝑠𝑚𝜃 − 𝑖𝑠𝑖𝑛𝑚𝜃
= 𝑚
You can see that
𝑟 𝑐𝑜𝑠 2 𝑚𝜃 + 𝑠𝑖𝑛2 𝑚𝜃
the answer has
followed the same
pattern as De
Moivre’s theorem!
=
Use De Moivre’s theorem for a positive
number (which we have proved)
1
𝑟𝑚
Multiply to change
some terms in the
fraction
Multiply out like
quadratics – the bottom
is the difference of two
squares
i2 = -1
You can cancel the
denominator as it is equal to 1
𝑐𝑜𝑠𝑚𝜃 − 𝑖𝑠𝑖𝑛𝑚𝜃
= 𝑟 −𝑚 cos −𝑚𝜃 + 𝑖𝑠𝑖𝑛(−𝑚𝜃)
Use cos(-θ) = cos(θ)
and sin(-θ) = -sinθ
3C
𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
Further complex numbers
You need to be able to prove that
[r(cosθ + isinθ)]n = rn(cos(nθ +
isinnθ) for any integer n
Having now proved that De Moivre’s
theorem works for both positive and
negative integers, there is only one
left
 We need to prove it is true for 0!
 This is straightforward. As it is
just a single value, we can
substitute it in to see what
happens…
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑛
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
0
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
Sub in n = 0
= 𝑟 0 𝑐𝑜𝑠0 + 𝑖𝑠𝑖𝑛0
1 = 1(1 + 0)
Left side = 1 as anything to
the power 0 is 1
 You can find cos0 and
sin 0 as well
‘Calculate’
1= 1
So we have shown that De Moivre’s Theorem is true for
all positive integers, all negative integers and 0’
 It is therefore true for all integers!
3C
𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
𝑧 𝑛 = 𝑟𝑒 𝑖𝜃
𝑛
= 𝑟 𝑛 𝑒 𝑖𝑛𝜃
Further complex numbers
You need to be able to prove that
[r(cosθ + isinθ)]n = rn(cos(nθ +
isinnθ) for any integer n
It is important to note that De
Moivre’s theorem can also be used in
exponential form.
𝑟𝑒 𝑖𝜃
𝑛
= 𝑟 𝑛 𝑒 𝑖𝑛𝜃
Both parts will be raised to
the power ‘n’
You can remove the bracket!
= 𝑟 𝑛 𝑒 𝑖𝑛𝜃
This is De Moivre’s theorem in exponential form!
3C
𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
𝑧 𝑛 = 𝑟𝑒 𝑖𝜃
𝑛
= 𝑟 𝑛 𝑒 𝑖𝑛𝜃
Further complex numbers
9𝜋
9𝜋
𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛
17
17
2𝜋
2𝜋
𝑐𝑜𝑠
− 𝑖𝑠𝑖𝑛
17
17
You need to be able to prove that
[r(cosθ + isinθ)]n = rn(cos(nθ +
isinnθ) for any integer n
Simplify the following:
9𝜋
9𝜋
𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛
17
17
2𝜋
2𝜋
𝑐𝑜𝑠
− 𝑖𝑠𝑖𝑛
17
17
5
𝑐𝑜𝑠
9𝜋
9𝜋
+ 𝑖𝑠𝑖𝑛
17
17
5
3
3
45𝜋
45𝜋
+ 𝑖𝑠𝑖𝑛
17
17
6𝜋
6𝜋
𝑐𝑜𝑠 −
+ 𝑖𝑠𝑖𝑛 −
17
17
𝑐𝑜𝑠
 Apply cos(-θ) = cosθ and sin(-θ)
= -sinθ
5
2𝜋
2𝜋
𝑐𝑜𝑠 −
+ 𝑖𝑠𝑖𝑛 −
17
17
𝑐𝑜𝑠
The denominator has to have the
‘+’ sign in the middle
3
Apply De Moivre’s theorem (there is no
modulus value to worry about here!)
 Just multiply the arguments by the
power
45𝜋
6𝜋
45𝜋
6𝜋
−−
+ 𝑖𝑠𝑖𝑛
−−
17
17
17
17
𝑐𝑜𝑠3𝜋 + 𝑖𝑠𝑖𝑛3𝜋
= −1 + 0𝑖
Apply the rules from 3B for the
division of complex numbers
 Divide the moduli and
subtract the arguments
Simplify the sin and cos terms
Calculate the sin
and cos terms
Simplify
= −1
3C
𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
𝑧 𝑛 = 𝑟𝑒 𝑖𝜃
𝑛
= 𝑟 𝑛 𝑒 𝑖𝑛𝜃
Further complex numbers
y
You need to be able to prove that
[r(cosθ + isinθ)]n = rn(cos(nθ +
isinnθ) for any integer n
Express the following in the form
x + iy where x Є R and y Є R
1 + 3𝑖
7
 You need to write this in one of
the forms above, and you can then
use De Moivre’s theorem
 This is easier than raising a
bracket to the power 7!
 Start with an argand diagram to
help find the modulus and
argument of the part in the
bracket
Pay attention to the directions
 The ‘x’ part is positive so
will go in the positive
direction horizontally
r
√3
θ
 The ‘y’ part is positive so
will go upwards
𝑟=
1
2
+
3
x
1
2
Calculate
𝑟=2
𝑇𝑎𝑛𝜃 =
𝜃=
3
1
𝜋
3
Inverse
Tan
𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
Sub in r and θ
2 𝑐𝑜𝑠
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
3
3
3C
𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
𝑧 𝑛 = 𝑟𝑒 𝑖𝜃
𝑛
= 𝑟 𝑛 𝑒 𝑖𝑛𝜃
Further complex numbers
You need to be able to prove that
[r(cosθ + isinθ)]n = rn(cos(nθ +
isinnθ) for any integer n
Express the following in the form
x + iy where x Є R and y Є R
1 + 3𝑖
7
 You need to write this in one of
the forms above, and you can then
use De Moivre’s theorem
 This is easier than raising a
bracket to the power 7!
1 + 3𝑖
7
𝜋
𝜋
2 𝑐𝑜𝑠 + 𝑖𝑠𝑖𝑛
3
3
27 𝑐𝑜𝑠
7
7𝜋
7𝜋
+ 𝑖𝑠𝑖𝑛
3
3
Rewrite using the
different form we
worked out before
Use De Moivre’s
Theorem as above
Calculate the cos
and sin parts
= 128
1
+𝑖
2
3
2
Multiply out and simplify
= 64 + 64 3𝑖
 Start with an argand diagram to
help find the modulus and
argument of the part in the
bracket
3C
𝑎+𝑏
𝑛
= 𝑎𝑛 + 𝑛𝐶1 𝑎𝑛−1 𝑏 + 𝑛𝐶2 𝑎𝑛−2 𝑏 2 + 𝑛𝐶3 𝑎𝑛−3 𝑏3 + … … … … + 𝑏 𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
This involves changing expressions
involving a function of θ into one
without.
 For example changing a cos6θ
into powers of cosθ
 You will need to use the binomial
expansion for C2 in this section
𝑎+𝑏
𝑛
𝑎𝑛 + 𝑛𝐶1 𝑎𝑛−1 𝑏 + 𝑛𝐶2 𝑎𝑛−2 𝑏 2 + 𝑛𝐶3 𝑎𝑛−3 𝑏3 + … … … … + 𝑏 𝑛
Remember nCr is a function you can find on your calculator
 The first term has the full power of n
 As you move across you slowly swap the powers over to
the second term until it has the full power of n
For example:
𝑎+𝑏
𝑎4 +
𝐶1 𝑎3 𝑏 +
4
4
𝐶2 𝑎2 𝑏2 +
4
Follow the
pattern above
𝐶3 𝑎𝑏 3 + 𝑏 4
4
𝑎4 + 4𝑎3 𝑏 + 6𝑎2 𝑏2 + 4𝑎𝑏 3 + 𝑏 4
You can work
out the nCr
parts
3D
𝑎+𝑏
𝑛
= 𝑎𝑛 + 𝑛𝐶1 𝑎𝑛−1 𝑏 + 𝑛𝐶2 𝑎𝑛−2 𝑏 2 + 𝑛𝐶3 𝑎𝑛−3 𝑏3 + … … … … + 𝑏 𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
3
Express cos3θ using powers of cosθ.
 This type of question involves
making a comparison between
two processes
 One which will give you a ‘cos3θ’
term – you will use De Moivre’s
Theorem for this
 If we apply De Moivre’s theorem to this, we will end up
with a ‘cos3θ’ term
 If we apply the binomial expansion to it, we will end up
with some terms with cosθ in
 So this expression is a good starting point!
 One which will give you an
expression in terms of cosθ –
you will use the binomial
expansion for this
 You have to think logically and
decide where to start
3D
𝑎+𝑏
𝑛
= 𝑎𝑛 + 𝑛𝐶1 𝑎𝑛−1 𝑏 + 𝑛𝐶2 𝑎𝑛−2 𝑏 2 + 𝑛𝐶3 𝑎𝑛−3 𝑏3 + … … … … + 𝑏 𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Express cos3θ using powers of cosθ.
 This type of question involves
making a comparison between
two processes
𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
Apply De Moivre’s theorem
𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
3
Follow the rules you know
= 𝑐𝑜𝑠3𝜃 + 𝑖𝑠𝑖𝑛3𝜃
Apply the Binomial expansion
𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
 One which will give you a ‘cos3θ’
term – you will use De Moivre’s
Theorem for this
3
= 𝑐𝑜𝑠𝜃
3
3
3
3
+ 𝐶1 𝑐𝑜𝑠𝜃
2
2
3
𝑖𝑠𝑖𝑛𝜃 + 𝐶2 𝑐𝑜𝑠𝜃 𝑖𝑠𝑖𝑛𝜃
2
2
2
3
+ 𝑖𝑠𝑖𝑛𝜃
3
= 𝑐𝑜𝑠 𝜃 + 3𝑖𝑐𝑜𝑠 𝜃𝑠𝑖𝑛𝜃 + 3𝑖 𝑐𝑜𝑠𝜃𝑠𝑖𝑛 𝜃 + 𝑖 𝑠𝑖𝑛 𝜃
 One which will give you an
expression in terms of cosθ –
you will use the binomial
expansion for this
 You have to think logically and
decide where to start
= 𝑐𝑜𝑠 3 𝜃 + 3𝑖𝑐𝑜𝑠 2 𝜃𝑠𝑖𝑛𝜃 − 3𝑐𝑜𝑠𝜃𝑠𝑖𝑛2 𝜃 − 𝑖𝑠𝑖𝑛3 𝜃
3
Write
out
‘Tidy
up’
Replace i2
parts with -1
The two expressions we have made must be equal
 Therefore the real parts in each and the imaginary parts in each must
be the same
 Equate the real parts
𝑐𝑜𝑠3𝜃 = 𝑐𝑜𝑠 3 𝜃 − 3𝑐𝑜𝑠𝜃𝑠𝑖𝑛2 𝜃
3D
𝑎+𝑏
𝑛
= 𝑎𝑛 + 𝑛𝐶1 𝑎𝑛−1 𝑏 + 𝑛𝐶2 𝑎𝑛−2 𝑏 2 + 𝑛𝐶3 𝑎𝑛−3 𝑏3 + … … … … + 𝑏 𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Express cos3θ using powers of cosθ.
 This type of question involves
making a comparison between
two processes
 One which will give you a ‘cos3θ’
term – you will use De Moivre’s
Theorem for this
𝑐𝑜𝑠3𝜃 = 𝑐𝑜𝑠 3 𝜃 − 3𝑐𝑜𝑠𝜃𝑠𝑖𝑛2 𝜃
𝑐𝑜𝑠3𝜃 = 𝑐𝑜𝑠 3 𝜃 − 3𝑐𝑜𝑠𝜃 1 − 𝑐𝑜𝑠 2 𝜃
Replace sin2θ with an
expression in cos2θ
Expand the bracket
𝑐𝑜𝑠3𝜃 = 𝑐𝑜𝑠 3 𝜃 − 3𝑐𝑜𝑠𝜃 + 3𝑐𝑜𝑠 3 𝜃
Simplify
3
𝑐𝑜𝑠3𝜃 = 4𝑐𝑜𝑠 𝜃 − 3𝑐𝑜𝑠𝜃
We have successfully expressed cos3θ
as posers of cosθ!
 One which will give you an
expression in terms of cosθ –
you will use the binomial
expansion for this
 You have to think logically and
decide where to start
3D
𝑎+𝑏
𝑛
= 𝑎𝑛 + 𝑛𝐶1 𝑎𝑛−1 𝑏 + 𝑛𝐶2 𝑎𝑛−2 𝑏 2 + 𝑛𝐶3 𝑎𝑛−3 𝑏3 + … … … … + 𝑏 𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Express the following as powers of
cosθ:
𝑠𝑖𝑛6𝜃
𝑠𝑖𝑛𝜃
 So we need something that will
give us sin6θ using De Moivre’s
theorem
 It also needs to give us terms of
cosθ from the binomial
expansion
𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
6
 If we apply De Moivre’s theorem to this, we will end up
with a ‘sin6θ’ term
 If we apply the binomial expansion to it, we will end up
with some terms with cosθ in
 So this expression is a good starting point!
(and yes you will have to do some expansions larger than
powers of 3 or 4!)
3D
𝑎+𝑏
𝑛
= 𝑎𝑛 + 𝑛𝐶1 𝑎𝑛−1 𝑏 + 𝑛𝐶2 𝑎𝑛−2 𝑏 2 + 𝑛𝐶3 𝑎𝑛−3 𝑏3 + … … … … + 𝑏 𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
6
𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
Apply De Moivre’s theorem
𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
Express the following as powers of
cosθ:
6
Follow the rules you know
= 𝑐𝑜𝑠6𝜃 + 𝑖𝑠𝑖𝑛6𝜃
𝑠𝑖𝑛6𝜃
𝑠𝑖𝑛𝜃
Apply the Binomial expansion
𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
= 𝑐𝑜𝑠𝜃
6
6
+ 6𝐶1 𝑐𝑜𝑠𝜃 5 (𝑖𝑠𝑖𝑛𝜃) + 6𝐶2 𝑐𝑜𝑠𝜃
4
𝑖𝑠𝑖𝑛𝜃
2
+ 6𝐶3 𝑐𝑜𝑠𝜃
3
𝑖𝑠𝑖𝑛𝜃
3
+ 6𝐶4 𝑐𝑜𝑠𝜃
2
𝑖𝑠𝑖𝑛𝜃
4
+ 6𝐶5 𝑐𝑜𝑠𝜃 𝑖𝑠𝑖𝑛𝜃
= 𝑐𝑜𝑠 6 𝜃 + 6𝑖𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃 + 15𝑖 2 𝑐𝑜𝑠 4 𝜃𝑠𝑖𝑛2 𝜃 + 20𝑖 3 𝑐𝑜𝑠 3 𝜃𝑠𝑖𝑛3 𝜃 + 15𝑖 4 𝑐𝑜𝑠 2 𝜃𝑠𝑖𝑛4 𝜃 + 6𝑖 5 𝑐𝑜𝑠𝜃𝑠𝑖𝑛5 𝜃 + 𝑖 6 𝑠𝑖𝑛6 𝜃
= 𝑐𝑜𝑠 6 𝜃 + 6𝑖𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃 − 15𝑐𝑜𝑠 4 𝜃𝑠𝑖𝑛2 𝜃 − 20𝑖𝑐𝑜𝑠 3 𝜃𝑠𝑖𝑛3 𝜃 + 15𝑐𝑜𝑠 2 𝜃𝑠𝑖𝑛4 𝜃 + 6𝑖𝑐𝑜𝑠𝜃𝑠𝑖𝑛5 𝜃 − 𝑠𝑖𝑛6 𝜃
5
+ 𝑖𝑠𝑖𝑛𝜃
6
Replace terms:
i2 and i6 = -1
i4 = 1
So the two expressions created from De Moivre and the Binomial Expansion must be equal
 The real parts will be the same, as will the imaginary parts
 This time we have to equate the imaginary parts as this has sin6θ in
𝑖𝑠𝑖𝑛6𝜃 = 6𝑖𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃 − 20𝑖𝑐𝑜𝑠 3 𝜃𝑠𝑖𝑛3 𝜃 + 6𝑖𝑐𝑜𝑠𝜃𝑠𝑖𝑛5 𝜃
5
3
3
5
Divide all by i
𝑠𝑖𝑛6𝜃 = 6𝑐𝑜𝑠 𝜃𝑠𝑖𝑛𝜃 − 20𝑐𝑜𝑠 𝜃𝑠𝑖𝑛 𝜃 + 6𝑐𝑜𝑠𝜃𝑠𝑖𝑛 𝜃
3D
𝑎+𝑏
𝑛
= 𝑎𝑛 + 𝑛𝐶1 𝑎𝑛−1 𝑏 + 𝑛𝐶2 𝑎𝑛−2 𝑏 2 + 𝑛𝐶3 𝑎𝑛−3 𝑏3 + … … … … + 𝑏 𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑠𝑖𝑛6𝜃 = 6𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃 − 20𝑐𝑜𝑠 3 𝜃𝑠𝑖𝑛3 𝜃 + 6𝑐𝑜𝑠𝜃𝑠𝑖𝑛5 𝜃
Express the following as powers of
cosθ:
𝑠𝑖𝑛6𝜃
𝑠𝑖𝑛𝜃
5
2
3
6
So we have changed the expression
we were given into powers of cosθ!
4
3
𝑠𝑖𝑛6𝜃
6𝑐𝑜𝑠 𝜃𝑠𝑖𝑛𝜃 − 20𝑐𝑜𝑠 𝜃𝑠𝑖𝑛 𝜃 + 6𝑐𝑜𝑠𝜃𝑠𝑖𝑛5 𝜃
=
𝑠𝑖𝑛𝜃
𝑠𝑖𝑛𝜃
Divide all terms by sinθ
= 6𝑐𝑜𝑠 5 𝜃 − 20𝑐𝑜𝑠 3 𝜃𝑠𝑖𝑛2 𝜃 + 6𝑐𝑜𝑠𝜃𝑠𝑖𝑛4 𝜃
Replace sin2θ terms with (1 – cos2θ) terms
= 6𝑐𝑜𝑠 5 𝜃 − 20𝑐𝑜𝑠 3 𝜃 1 − 𝑐𝑜𝑠 2 𝜃 + 6𝑐𝑜𝑠𝜃 1 − 𝑐𝑜𝑠 2 𝜃
2
= 6𝑐𝑜𝑠 5 𝜃 − 20𝑐𝑜𝑠 3 𝜃 + 20𝑐𝑜𝑠 5 𝜃 + 6𝑐𝑜𝑠𝜃 1 − 2𝑐𝑜𝑠 2 𝜃 + 𝑐𝑜𝑠 4 𝜃
5
3
Expand the first bracket,
square the second
Expand the second bracket
5
= 6𝑐𝑜𝑠 𝜃 − 20𝑐𝑜𝑠 3 𝜃 + 20𝑐𝑜𝑠 5 𝜃 + 6𝑐𝑜𝑠𝜃 − 12𝑐𝑜𝑠 𝜃 + 6𝑐𝑜𝑠 𝜃
= 32𝑐𝑜𝑠 5 𝜃 − 32𝑐𝑜𝑠 3 𝜃 + 6𝑐𝑜𝑠𝜃
Group together the like terms
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Let: 𝑧 = 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
1
= 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑧
You also need to be able to work this
type of question in a different way:
1
= cos(−𝜃) + 𝑖𝑠𝑖𝑛(−𝜃)
𝑧
For example, you might have a power
or cos or sin and need to express it
using several linear terms instead
1
= cos 𝜃 − 𝑖𝑠𝑖𝑛𝜃
𝑧
Eg) Changing sin6θ to sinaθ + sinbθ
where a and b are integers
 To do this we need to know some
other patterns first!
−1
Write as ‘1 over’ 
or with a power of -1
Use De Moivre’s
theorem
Use cos(-θ) = cosθ
and sin(-θ) = -sinθ
We can add our two results together:
𝑧+
1
= 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃 + 𝑐𝑜𝑠𝜃 − 𝑖𝑠𝑖𝑛𝜃
𝑧
1
𝑧 + = 2𝑐𝑜𝑠𝜃
𝑧
Simplify
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
𝑧−
1
= 2𝑖𝑠𝑖𝑛𝜃
𝑧
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Let: 𝑧 = 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
1
= 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑧
You also need to be able to work this
type of question in a different way:
1
= cos(−𝜃) + 𝑖𝑠𝑖𝑛(−𝜃)
𝑧
For example, you might have a power
or cos or sin and need to express it
using several linear terms instead
1
= cos 𝜃 − 𝑖𝑠𝑖𝑛𝜃
𝑧
Eg) Changing sin6θ to sinaθ + sinbθ
where a and b are integers
 To do this we need to know some
other patterns first!
−1
Write as ‘1 over’ 
or with a power of -1
Use De Moivre’s
theorem
Use cos(-θ) = cosθ
and sin(-θ) = -sinθ
We could also subtract our two results:
𝑧−
1
= 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠𝜃 − 𝑖𝑠𝑖𝑛𝜃
𝑧
1
𝑧 − = 2𝑖𝑠𝑖𝑛𝜃
𝑧
Simplify
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
𝑧−
1
= 2𝑖𝑠𝑖𝑛𝜃
𝑧
𝑧𝑛 +
1
= 2𝑐𝑜𝑠𝑛𝜃
𝑧𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Let: 𝑧 𝑛 = 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
1
= 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
You also need to be able to work this
type of question in a different way:
1
= cos(−𝑛𝜃) + 𝑖𝑠𝑖𝑛(−𝑛𝜃)
𝑧𝑛
For example, you might have a power
or cos or sin and need to express it
using several linear terms instead
1
= cos 𝑛𝜃 − 𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
Eg) Changing sin6θ to sinaθ + sinbθ
where a and b are integers
 To do this we need to know some
other patterns first!
 You can also apply the rules we
just saw to powers of z
−1
Write as ‘1 over’ 
or with a power of -1
Use De Moivre’s
theorem
Use cos(-θ) = cosθ
and sin(-θ) = -sinθ
We could add our two results together:
𝑧𝑛 +
1
= 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
+ 𝑐𝑜𝑠𝑛𝜃 − 𝑖𝑠𝑖𝑛𝑛𝜃
Simplify
1
𝑧 𝑛 + 𝑛 = 2𝑐𝑜𝑠𝑛𝜃
𝑧
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
𝑧−
1
= 2𝑖𝑠𝑖𝑛𝜃
𝑧
𝑧𝑛 +
1
= 2𝑐𝑜𝑠𝑛𝜃
𝑧𝑛
𝑧𝑛 −
1
= 2𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Let: 𝑧 𝑛 = 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
1
= 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
You also need to be able to work this
type of question in a different way:
1
= cos(−𝑛𝜃) + 𝑖𝑠𝑖𝑛(−𝑛𝜃)
𝑧𝑛
For example, you might have a power
or cos or sin and need to express it
using several linear terms instead
1
= cos 𝑛𝜃 − 𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
Eg) Changing sin6θ to sinaθ + sinbθ
where a and b are integers
 To do this we need to know some
other patterns first!
 You can also apply the rules we
just saw to powers of z
−1
Write as ‘1 over’ 
or with a power of -1
Use De Moivre’s
theorem
Use cos(-θ) = cosθ
and sin(-θ) = -sinθ
We could also subtract our two results:
𝑧𝑛 −
1
= 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
− 𝑐𝑜𝑠𝑛𝜃 − 𝑖𝑠𝑖𝑛𝑛𝜃
Simplify
1
𝑧 𝑛 − 𝑛 = 2𝑖𝑠𝑖𝑛𝑛𝜃
𝑧
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
𝑧−
1
= 2𝑖𝑠𝑖𝑛𝜃
𝑧
𝑧𝑛 +
1
= 2𝑐𝑜𝑠𝑛𝜃
𝑧𝑛
𝑧𝑛 −
1
= 2𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Let’s now see how we can use these
‘patterns’ in solving problems:
Express cos5θ in the form
acos5θ + bcos3θ + ccosθ
 When working this way round you
need to use the identities above to
express both cos5θ and terms
with cos5θ, cos3θ and cosθ.
 You can then set the expressions
equal to each other
5
1
𝑧+
𝑧
= 2𝑐𝑜𝑠𝜃
5
= 32𝑐𝑜𝑠 5 𝜃
Creating the other cos terms – use the Binomial expansion!
1
𝑧+
𝑧
5
Where a, b and c are constants to be
found.
Using the
Identity above
Creating a cos5θ term
5
= 𝑧 + 5𝑧
4
1
1
+ 10𝑧 3
𝑧
𝑧
2
+ 10𝑧
2
1
𝑧
3
1
+ 5𝑧
𝑧
4
1
+
𝑧
5
Use the
B.E.
Cancel
some z
1
1
1
terms
+5 3 + 5
= 𝑧 5 + 5𝑧 3 + 10𝑧 + 10
𝑧
𝑧
𝑧
Group up terms
with the same
1
1
1
power
= 𝑧 5 + 5 + 5 𝑧 3 + 3 + 10 𝑧 +
𝑧
𝑧
𝑧
Rewrite using an
identity above
= 2𝑐𝑜𝑠5𝜃 + 5(2𝑐𝑜𝑠3𝜃) + 10(2𝑐𝑜𝑠𝜃)
Simplify
= 2𝑐𝑜𝑠5𝜃 + 10𝑐𝑜𝑠3𝜃 + 20𝑐𝑜𝑠𝜃
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
𝑧−
1
= 2𝑖𝑠𝑖𝑛𝜃
𝑧
𝑧𝑛 +
1
= 2𝑐𝑜𝑠𝑛𝜃
𝑧𝑛
𝑧𝑛 −
1
= 2𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Let’s now see how we can use these
‘patterns’ in solving problems:
Using the two expressions
1
𝑧+
𝑧
5
1
𝑧+
𝑧
5
= 32𝑐𝑜𝑠 5 𝜃
These two
expressions must
be equal to each
other
= 2𝑐𝑜𝑠5𝜃 + 10𝑐𝑜𝑠3𝜃 + 20𝑐𝑜𝑠𝜃
Express cos5θ in the form
acos5θ + bcos3θ + ccosθ
Where a, b and c are constants to be
found.
 When working this way round you
need to use the identities above to
express both cos5θ and terms
with cos5θ, cos3θ and cosθ.
32𝑐𝑜𝑠 5 𝜃 = 2𝑐𝑜𝑠5𝜃 + 10𝑐𝑜𝑠3𝜃 + 20𝑐𝑜𝑠𝜃
𝑐𝑜𝑠 5 𝜃 =
1
5
5
𝑐𝑜𝑠5𝜃 + 𝑐𝑜𝑠3𝜃 + 𝑐𝑜𝑠𝜃
16
16
8
Divide both
sides by 32
So we have written cos5θ using
cos5θ, cos3θ and cosθ
 You can then set the expressions
equal to each other
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
𝑧−
1
= 2𝑖𝑠𝑖𝑛𝜃
𝑧
𝑧𝑛 +
1
= 2𝑐𝑜𝑠𝑛𝜃
𝑧𝑛
𝑧𝑛 −
1
= 2𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Show that:
1
3
𝑠𝑖𝑛3 𝜃 = − 𝑠𝑖𝑛3𝜃 + 𝑠𝑖𝑛𝜃
4
4
 This is similar to the previous
question. You need to use rules above
to find a way to create a sin3θ
expression, and an expression
containing sin3θ and sinθ
Using the
Identity above
Creating a sin3θ term
3
1
𝑧−
𝑧
= 2𝑖𝑠𝑖𝑛𝜃
3
= 8𝑖 3 𝑠𝑖𝑛3 𝜃 = −8𝑖𝑠𝑖𝑛3 𝜃
Creating the other sin terms – use the Binomial expansion!
1
𝑧−
𝑧
3
3
= 𝑧 + 3𝑧
2
1
1
+ 3𝑧 −
−
𝑧
𝑧
= 𝑧 3 − 3𝑧 + 3
= 𝑧3 −
1
𝑧
−
1
1
−
3
𝑧
−
𝑧
𝑧3
= 2𝑖𝑠𝑖𝑛3𝜃 − 3(2𝑖𝑠𝑖𝑛𝜃)
1
𝑧3
2
1
+ −
𝑧
3
Use the
B.E.
Cancel
some z
terms
Group up terms with
the same power
Rewrite using an
identity above
Simplify
= 2𝑖𝑠𝑖𝑛3𝜃 − 6𝑖𝑠𝑖𝑛𝜃
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
𝑧−
1
= 2𝑖𝑠𝑖𝑛𝜃
𝑧
𝑧𝑛 +
1
= 2𝑐𝑜𝑠𝑛𝜃
𝑧𝑛
𝑧𝑛 −
1
= 2𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
Show that:
1
3
𝑠𝑖𝑛3 𝜃 = − 𝑠𝑖𝑛3𝜃 + 𝑠𝑖𝑛𝜃
4
4
 This is similar to the previous
question. You need to use rules above
to find a way to create a sin3θ
expression, and an expression
containing sin3θ and sinθ
Using the two expressions
1
𝑧−
𝑧
3
1
𝑧−
𝑧
3
= −8𝑖𝑠𝑖𝑛3 𝜃
These two
expressions must
be equal to each
other
= 2𝑖𝑠𝑖𝑛3𝜃 − 6𝑖𝑠𝑖𝑛𝜃
−8𝑖𝑠𝑖𝑛3 𝜃 = 2𝑖𝑠𝑖𝑛3𝜃 − 6𝑖𝑠𝑖𝑛𝜃
−8𝑠𝑖𝑛3 𝜃 = 2𝑠𝑖𝑛3𝜃 − 6𝑠𝑖𝑛𝜃
1
3
𝑠𝑖𝑛3 𝜃 = − 𝑠𝑖𝑛3𝜃 + 𝑠𝑖𝑛𝜃
4
4
Divide both
sides by i
Divide both
sides by -8
So we have written sin3θ using
sin3θ and sinθ!
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
𝑧−
1
= 2𝑖𝑠𝑖𝑛𝜃
𝑧
𝑧𝑛 +
1
= 2𝑐𝑜𝑠𝑛𝜃
𝑧𝑛
𝑧𝑛 −
1
= 2𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
a) Express sin4θ in the form:
𝑑𝑐𝑜𝑠4𝜃 + 𝑒𝑐𝑜𝑠2𝜃 + 𝑓
Where d, e and f are constants to be
found.
4
1
𝑧−
𝑧
= 2𝑖𝑠𝑖𝑛𝜃
𝜋
2
4
= 16𝑖 4 𝑠𝑖𝑛4 𝜃 = 16𝑠𝑖𝑛4 𝜃
Creating the cos terms – use the Binomial expansion!
1
𝑧−
𝑧
4
4
b) Hence, find the exact value of the
following integral:
Using the
Identity above
Creating a sin4θ term
= 𝑧 + 4𝑧
3
1
1
+ 6𝑧 2 −
−
𝑧
𝑧
= 𝑧 4 − 4𝑧 2 + 6 − 4
1
𝑧2
2
+
1
+ 4𝑧 −
𝑧
3
1
𝑧4
𝑠𝑖𝑛 𝜃 𝑑𝜃
 Start exactly as with the previous
questions, by finding an expression
with sin4θ and one with cos4θ, cos2θ
and a number
= 𝑧4 +
1
1
− 4 𝑧2 + 2 + 6
4
𝑧
𝑧
4
Cancel some
z terms
Group up terms with the same
power (use positive values in the
brackets so we get cos terms)
4
0
1
+ −
𝑧
Use the
B.E.
Replace using an
identity above
= 2𝑐𝑜𝑠4𝜃 − 4(2𝑐𝑜𝑠2𝜃) + 6
Simplify
= 2𝑐𝑜𝑠4𝜃 − 8𝑐𝑜𝑠2𝜃 + 6
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
𝑧−
1
= 2𝑖𝑠𝑖𝑛𝜃
𝑧
𝑧𝑛 +
1
= 2𝑐𝑜𝑠𝑛𝜃
𝑧𝑛
𝑧𝑛 −
1
= 2𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
Further complex numbers
You can apply De Moivre’s theorem
to trigonometric identities
a) Express sin4θ in the form:
𝑑𝑐𝑜𝑠4𝜃 + 𝑒𝑐𝑜𝑠2𝜃 + 𝑓
Using the two expressions
1
𝑧−
𝑧
4
1
𝑧−
𝑧
4
= 16𝑠𝑖𝑛4 𝜃
These two
expressions must
be equal to each
other
= 2𝑐𝑜𝑠4𝜃 − 8𝑐𝑜𝑠2𝜃 + 6
Where d, e and f are constants to be
found.
b) Hence, find the exact value of the
following integral:
𝜋
2
𝑠𝑖𝑛4 𝜃 𝑑𝜃
16𝑠𝑖𝑛4 𝜃 = 2𝑐𝑜𝑠4𝜃 − 8𝑐𝑜𝑠2𝜃 + 6
1
1
3
𝑠𝑖𝑛4 𝜃 = 𝑐𝑜𝑠4𝜃 − 𝑐𝑜𝑠2𝜃 +
8
2
8
Divide both
sides by 16
0
 Start exactly as with the previous
questions, by finding an expression
with sin4θ and one with cos4θ, cos2θ
and a number
So we have written sin4θ using
cos4θ and cos2θ!
3D
𝑧+
1
= 2𝑐𝑜𝑠𝜃
𝑧
𝑧−
1
= 2𝑖𝑠𝑖𝑛𝜃
𝑧
𝑧𝑛 +
1
= 2𝑐𝑜𝑠𝑛𝜃
𝑧𝑛
𝑧𝑛 −
1
= 2𝑖𝑠𝑖𝑛𝑛𝜃
𝑧𝑛
Further complex numbers
1
1
3
𝑠𝑖𝑛4 𝜃 = 𝑐𝑜𝑠4𝜃 − 𝑐𝑜𝑠2𝜃 +
8
2
8
You can apply De Moivre’s theorem
to trigonometric identities
𝜋
2
a) Express sin4θ in the form:
0
𝑑𝑐𝑜𝑠4𝜃 + 𝑒𝑐𝑜𝑠2𝜃 + 𝑓
𝜋
2
Where d, e and f are constants to be
found.
b) Hence, find the exact value of the
following integral:
𝜋
2
𝑠𝑖𝑛4 𝜃 𝑑𝜃
0
 Start exactly as with the previous
questions, by finding an expression
with sin4θ and one with cos4θ, cos2θ
and a number
𝑠𝑖𝑛4 𝜃 𝑑𝜃
0
1
1
3
𝑐𝑜𝑠4𝜃 − 𝑐𝑜𝑠2𝜃 +
𝑑𝜃
8
2
8
1
1
3
=
𝑠𝑖𝑛4𝜃 − 𝑠𝑖𝑛2𝜃 + 𝜃
32
4
8
=
=
1
𝜋
1
𝜋
3 𝜋
𝑠𝑖𝑛4
− 𝑠𝑖𝑛2
+
32
2
4
2
8 2
𝜋
2
Cosine Integrals
(in C4)
𝑐𝑜𝑠4𝜃 =
1
𝑠𝑖𝑛4𝜃
4
𝑐𝑜𝑠2𝜃 =
1
𝑠𝑖𝑛2𝜃
2
Replace with
an equivalent
expression
Integrate each term with
respect to θ, using
knowledge from C4
0
−
1
1
3
𝑠𝑖𝑛4 0 − 𝑠𝑖𝑛2 0 + 0
32
4
8
Sub in
limits
Work
out
3
𝜋
16
3D
𝐼𝑓: 𝑧 = 𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑇ℎ𝑒𝑛: 𝑧 = 𝑟 cos(𝜃 + 2𝑘𝜋) + 𝑖𝑠𝑖𝑛(𝜃 + 2𝑘𝜋)
Further complex numbers
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
You can use De Moivre’s theorem to find
the nth roots of a complex number
You already know how to find real roots of a
number, but now we need to find both real
roots and imaginary roots!
We need to apply the following results:
1)
If: 𝑧 = 𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
 Then: 𝑧 = 𝑟 cos(𝜃 + 2𝑘𝜋) + 𝑖𝑠𝑖𝑛(𝜃 + 2𝑘𝜋)
where k is an integer
This is because we can add multiples of 2π
to the argument as it will end up in the same
place (2π = 360º)
2) De Moivre’s theorem
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
3E
𝐼𝑓: 𝑧 = 𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑇ℎ𝑒𝑛: 𝑧 = 𝑟 cos(𝜃 + 2𝑘𝜋) + 𝑖𝑠𝑖𝑛(𝜃 + 2𝑘𝜋)
Further complex numbers
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
You can use De Moivre’s theorem to find
the nth roots of a complex number
In this case the
modulus and
argument are simple
to find!
Solve the equation z3 = 1 and represent your
solutions on an Argand diagram.
𝑟=1
 Now we know r and θ we can set
equal
to this expression, when written in the
modulus-argument form
1
x
𝜃=0
 First you need to express z in the
modulus-argument form. Use an Argand
diagram.
z3
y
𝑧 3 = 1 𝑐𝑜𝑠0 + 𝑖𝑠𝑖𝑛0
Apply the rule
above
𝑧 3 = cos(0 + 2𝑘𝜋) + 𝑖𝑠𝑖𝑛(0 + 2𝑘𝜋)
 We can then find an expression for z in
terms of k
𝑧 = cos 0 + 2𝑘𝜋 + 𝑖𝑠𝑖𝑛(0 + 2𝑘𝜋)
 We can then solve this to find the roots
of the equation above
𝑧 = cos
0 + 2𝑘𝜋
0 + 2𝑘𝜋
+ 𝑖𝑠𝑖𝑛
3
3
1
3
Cube root (use a
relevant power)
Apply De
Moivre’s
theorem
3E
𝐼𝑓: 𝑧 = 𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑇ℎ𝑒𝑛: 𝑧 = 𝑟 cos(𝜃 + 2𝑘𝜋) + 𝑖𝑠𝑖𝑛(𝜃 + 2𝑘𝜋)
Further complex numbers
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
You can use De Moivre’s theorem to find
the nth roots of a complex number
Solve the equation z3 = 1 and represent your
solutions on an Argand diagram.
𝑧 = 𝑐𝑜𝑠
0 + 2𝑘𝜋
0 + 2𝑘𝜋
+ 𝑖𝑠𝑖𝑛
3
3
k=0
𝑧 = 𝑐𝑜𝑠 0 + 𝑖𝑠𝑖𝑛 0
𝑧=1
 We now just need to choose different
values for k until we have found all the
roots
 The values of k you choose should keep
the argument within the range:
-π < θ ≤ π
So the roots of z3 = 1 are:
1
3
1
3
𝑧 = 1, − + 𝑖
and − − 𝑖
2
2
2
2
Sub k = 0 in and
calculate the cosine
and sine parts
k=1
𝑧 = 𝑐𝑜𝑠
2𝜋
2𝜋
+ 𝑖𝑠𝑖𝑛
3
3
1
3
𝑧 =− +𝑖
2
2
k = -1
2𝜋
2𝜋
𝑧 = 𝑐𝑜𝑠 −
+ 𝑖𝑠𝑖𝑛 −
3
3
1
3
𝑧 =− −𝑖
2
2
Sub k = 1 in and
calculate the cosine
and sine parts
Sub k = -1 in and
calculate the cosine
and sine parts
 (k = 2 would cause
the argument to be
outside the range)
3E
𝐼𝑓: 𝑧 = 𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑇ℎ𝑒𝑛: 𝑧 = 𝑟 cos(𝜃 + 2𝑘𝜋) + 𝑖𝑠𝑖𝑛(𝜃 + 2𝑘𝜋)
Further complex numbers
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
y
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
You can use De Moivre’s theorem to find
the nth roots of a complex number
Solve the equation z3 = 1 and represent your
solutions on an Argand diagram.
𝟏
𝟑
− +𝒊
𝟐
𝟐
2
𝜋
3
 We now just need to choose different
values for k until we have found all the
roots
 The values of k you choose should keep
the argument within the range:
-π < θ ≤ π
So the roots of z3 = 1 are:
1
3
1
3
𝑧 = 1, − + 𝑖
and − − 𝑖
2
2
2
2
𝟏
2
𝜋
3
x
2
𝜋
3
𝟏
𝟑
− −𝒊
𝟐
𝟐
 The solutions will all the same distance from the origin
 The angles between them will also be the same
 The sum of the roots is always equal to 0
3E
𝐼𝑓: 𝑧 = 𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑇ℎ𝑒𝑛: 𝑧 = 𝑟 cos(𝜃 + 2𝑘𝜋) + 𝑖𝑠𝑖𝑛(𝜃 + 2𝑘𝜋)
Further complex numbers
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
You can use De Moivre’s theorem to find
the nth roots of a complex number
Solve the equation
z4
- 2√3i = 2
Give your answers in both the modulusargument and exponential forms.
By rearranging…
z4 = 2 + 2√3i
 As before, use an argand diagram to
express the equation in the modulusargument form
 Then choose values of k until you have
found all the solutions
y
Find the modulus
and argument
𝑟=
2
𝜃 = 𝑡𝑎𝑛
2
+ 2 3
−1
2 3
2
2
r
𝑟=4
θ
𝜋
𝜃=
3
𝑧 4 = 4 𝑐𝑜𝑠
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
3
3
𝑧 4 = 4 𝑐𝑜𝑠
𝜋
𝜋
+ 2𝑘𝜋 + 𝑖𝑠𝑖𝑛 + 2𝑘𝜋
3
3
𝜋
𝜋
𝑧 = 4 𝑐𝑜𝑠
+ 2𝑘𝜋 + 𝑖𝑠𝑖𝑛 + 2𝑘𝜋
3
3
𝜋
𝜋
+ 2𝑘𝜋
+ 2𝑘𝜋
1
𝑧 = 44 𝑐𝑜𝑠 3
+ 𝑖𝑠𝑖𝑛 3
4
4
𝜋
𝜋
+ 2𝑘𝜋
+ 2𝑘𝜋
𝑧 = 2 𝑐𝑜𝑠 3
+ 𝑖𝑠𝑖𝑛 3
4
4
2√3
2
x
Apply the rule
above
1
4
Take the 4th
root of each
side
De Moivre’s
Theorem
Work out the
power at the
front
3E
𝐼𝑓: 𝑧 = 𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑇ℎ𝑒𝑛: 𝑧 = 𝑟 cos(𝜃 + 2𝑘𝜋) + 𝑖𝑠𝑖𝑛(𝜃 + 2𝑘𝜋)
Further complex numbers
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
You can use De Moivre’s theorem to find
the nth roots of a complex number
k=0
𝑧 = 2 𝑐𝑜𝑠
Solve the equation z4 - 2√3i = 2
Give your answers in both the modulusargument and exponential forms.
k=1
 Then choose values of k until you have
found all the solutions
𝜋
𝜋
+ 2𝑘𝜋
+ 2𝑘𝜋
3
𝑧 = 2 𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛 3
4
4
k = -1
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
12
12
7𝜋
7𝜋
+ 𝑖𝑠𝑖𝑛
12
12
𝜋
𝜋
−
2𝜋
− 2𝜋
𝑧 = 2 𝑐𝑜𝑠 3
+ 𝑖𝑠𝑖𝑛 3
4
4
𝑧 = 2 𝑐𝑜𝑠 −
k = -2
Sub k = 0 in and
simplify (you can
leave in this form)
𝜋
𝜋
+
2𝜋
+ 2𝜋
𝑧 = 2 𝑐𝑜𝑠 3
+ 𝑖𝑠𝑖𝑛 3
4
4
𝑧 = 2 𝑐𝑜𝑠
By rearranging…
z4 = 2 + 2√3i
 As before, use an argand diagram to
express the equation in the modulusargument form
𝜋
𝜋
𝑧 = 2 𝑐𝑜𝑠 3 + 𝑖𝑠𝑖𝑛 3
4
4
5𝜋
5𝜋
+ 𝑖𝑠𝑖𝑛 −
12
12
Choose
values of k
that keep
the argument
between –π
and π
𝜋
𝜋
− 2𝜋
− 2𝜋
𝑧 = 2 𝑐𝑜𝑠 3
+ 𝑖𝑠𝑖𝑛 3
4
4
𝑧 = 2 𝑐𝑜𝑠 −
11𝜋
11𝜋
+ 𝑖𝑠𝑖𝑛 −
12
12
3E
𝐼𝑓: 𝑧 = 𝑟 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝑇ℎ𝑒𝑛: 𝑧 = 𝑟 cos(𝜃 + 2𝑘𝜋) + 𝑖𝑠𝑖𝑛(𝜃 + 2𝑘𝜋)
Further complex numbers
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑛
= 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃
You can use De Moivre’s theorem to find
the nth roots of a complex number
Solve the equation z4 - 2√3i = 2
Give your answers in both the modulusargument and exponential forms.
By rearranging…
z4 = 2 + 2√3i
 As before, use an argand diagram to
express the equation in the modulusargument form
Solutions in the modulus-argument form
𝑧 = 2 𝑐𝑜𝑠
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛
12
12
𝑧 = 2 𝑐𝑜𝑠
7𝜋
7𝜋
+ 𝑖𝑠𝑖𝑛
12
12
𝑧 = 2 𝑐𝑜𝑠 −
𝑧 = 2 𝑐𝑜𝑠 −
5𝜋
5𝜋
+ 𝑖𝑠𝑖𝑛 −
12
12
11𝜋
11𝜋
+ 𝑖𝑠𝑖𝑛 −
12
12
Solutions in the exponential form
𝑧=
𝜋
12
2𝑒 𝑖
𝑧=
7𝜋
2𝑒 12 𝑖
𝑧=
5𝜋
− 𝑖
2𝑒 12
𝑧=
11𝜋
− 12 𝑖
2𝑒
 Then choose values of k until you have
found all the solutions
𝜋
𝜋
+ 2𝑘𝜋
+ 2𝑘𝜋
3
𝑧 = 2 𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛 3
4
4
3E
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
A locus a set of points which obey a
rule
x
The locus of points a
given distance from a
point O is a circle
x
O
x
 You will need to be able to
understand Loci based on Argand
diagrams
A
The locus of points
equidistant from two
fixed points A and B is
the perpendicular
bisector of line AB
B
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
z = x + iy represents a variable point
P(x,y) on an Argand diagram
P(x,y)
z
z1 = x1 + iy1 represents a fixed point
A(x1,y1) on an Argand diagram
What is represented by:
z - z1
A(x1,y1)
z1
x
𝑧 − 𝑧1
 It represents the distance between
If we want to get from the fixed point A to the variable point
the fixed point A(x1,y1) and the
P, we need to travel back along z1 and then out along z
variable point P(x,y)
(-z1 + z)
 This can be written as a vector, z – z1
 So |z – z1| represents the distance between the fixed point
and the variable point!
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
P(x,y)
If:
A(5,3)
𝑧 − 5 − 3𝑖 = 3
x
Sketch the locus of P(x,y) which is
represented by z on an Argand
diagram
𝑧 − 5 − 3𝑖 = 3
𝑧 − (5 + 3𝑖) = 3
Leave z as it is
– this is the
variable point
Put this part in a
bracket
- This is the fixed
point
So we want the locus
where the distance
between the variable
point z and the fixed
point (5,3) is equal to 3
This will be a circle
of radius 3 units,
centre (5,3)
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
𝐼𝑓: 𝑧 = 𝑥 + 𝑖𝑦
You can use complex numbers to
represent a locus of points on an
Argand diagram
𝑇ℎ𝑒𝑛:
𝑧 = 𝑥 + 𝑖𝑦 =
𝑥2 + 𝑦2
y
P(x,y)
If:
|z|
𝑧 − 5 − 3𝑖 = 3
Use an algebraic method to find a
Cartesian equation of the locus of z
x
y
x
 So you have to do this without
using the graph you drew
 We will quickly remind ourselves of
something that will be useful for
this!
If: 𝑧 = 𝑥 + 𝑖𝑦
Then: 𝑧 = 𝑥 + 𝑖𝑦 =
(By Pythagoras’ Theorem)
𝑥 2 + 𝑦2
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
𝐼𝑓: 𝑧 = 𝑥 + 𝑖𝑦
You can use complex numbers to
represent a locus of points on an
Argand diagram
If:
𝑇ℎ𝑒𝑛:
𝑧 − 5 − 3𝑖 = 3
 Now we can find the equation of
the locus algebraically…
𝑥2 + 𝑦2
Replace z with
‘x + iy’
𝑥 + 𝑖𝑦 − 5 − 3𝑖 = 3
𝑧 − 5 − 3𝑖 = 3
Use an algebraic method to find a
Cartesian equation of the locus of z
𝑧 = 𝑥 + 𝑖𝑦 =
𝑥 − 5 + 𝑖(𝑦 − 3) = 3
𝑥−5
2
+ 𝑦−3
2
Group the real and
imaginary terms
Use the rule above to
remove the modulus
=3
Square both sides
2
(𝑥 − 5) + 𝑦 − 3
2
=9
You (hopefully) recognise that this is
the equation of a circle, radius 3 and
with centre (5,3)!
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
As a general rule, the locus of:
𝑧 − 𝑧1 = 𝑟
is a circle of radius r and centre (x1,y1) where z1 = x1 + iy1
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
𝑧 − 𝑧1 = 𝑟
 Circle radius r and centre (x1,y1)
You can use complex numbers to
represent a locus of points on an
Argand diagram
Give a geometrical interpretation of
each of the following loci of z:
a)
d)
𝑧 − 3𝑖 = 4
 Circle, centre (0,3) radius 4
2 − 5𝑖 − 𝑧 = 3
(−1)(−2 + 5𝑖 + 𝑧) = 3
−1 𝑧 − 2 + 5𝑖 = 3
b)
𝑧 − (2 + 3𝑖) = 5
𝑧 − (2 − 5𝑖) = 3
 Circle, centre (2,3) radius 5
c)
𝑧 + 3 − 5𝑖 = 2
𝑧 − (−3 + 5𝑖) = 2
Put the
‘fixed’ part
in a bracket
 Circle, centre (-3,5) radius 2
‘Factorise’ the part
inside the modulus
You can write this as
2 moduli multiplied
|-1| = 1, put the ‘fixed’ part
in a bracket
 Circle, centre (2,-5) radius 3
Effectively for d), you just swap the signs of
everything in the modulus, its value will not change
 |10 - 8| = |-10 + 8|
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
Sketch the locus of P(x,y) which is
represented by z on an Argand
diagram, if:
𝑧 = 𝑧 − 6𝑖
𝑧
𝑧 − 6𝑖
𝑧 − (6𝑖)
This is the distance of P(x,y)
from the origin (0,0)
This is the distance of
P(x,y) from (0,6)
y
(0,6)
 We therefore need the set of
points that are the same distance
from (0,0) and (0,6)
 This will be the bisector of the
line joining the two co-ordinates
y=3
(0,0)
x
 You can see that it is the line with
equation y = 3
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
𝐼𝑓: 𝑧 = 𝑥 + 𝑖𝑦
You can use complex numbers to
represent a locus of points on an
Argand diagram
𝑇ℎ𝑒𝑛:
𝑧 = 𝑥 + 𝑖𝑦 =
𝑥2 + 𝑦2
𝑧 = 𝑧 − 6𝑖
Replace z with x + iy
Sketch the locus of P(x,y) which is
represented by z on an Argand
diagram, if:
𝑥 + 𝑖𝑦 = 𝑥 + 𝑖𝑦 − 6𝑖
Factorise the ‘i’ terms
on the right side
𝑥 + 𝑖𝑦 = 𝑥 + 𝑖(𝑦 − 6)
𝑧 = 𝑧 − 6𝑖
 Show that the locus is y = 3 using
an algebraic method
𝑥2 + 𝑦2 =
𝑥2 + 𝑦 − 6
2
Use the rule above to remove
the moduli
Square both sides
𝑥2 + 𝑦2 = 𝑥2 + 𝑦 − 6
2
Expand the bracket
𝑥 2 + 𝑦 2 = 𝑥 2 + 𝑦 2 − 12𝑦 + 36
Cancel terms on
each side
0 = −12𝑦 + 36
Add 12y
12𝑦 = 36
Divide by 12
𝑦=3
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
𝐼𝑓: 𝑧 = 𝑥 + 𝑖𝑦
You can use complex numbers to
represent a locus of points on an
Argand diagram
𝑇ℎ𝑒𝑛:
𝑧 = 𝑥 + 𝑖𝑦 =
𝑥2 + 𝑦2
𝑧−3 = 𝑧+𝑖
Replace z with x + iy
a)
Use an algebraic method to find
the Cartesian equation of the
locus of z if:
𝑧−3 = 𝑧+𝑖
𝒚 = −𝟑𝒙 + 𝟒
b) Represent the locus of z on an
Argand diagram
𝑥 + 𝑖𝑦 − 3 = 𝑥 + 𝑖𝑦 + 𝑖
Group real and
imaginary parts
𝑥 − 3 + 𝑖𝑦 = 𝑥 + 𝑖(𝑦 + 1)
𝑥−3
2
𝑥−3
2
+ 𝑦2 =
𝑥2 + 𝑦 + 1
2
Use the rule above to
remove the moduli
Square both sides
2
2
+𝑦 =𝑥 + 𝑦+1
2
Expand brackets
𝑥 2 − 6𝑥 + 9 + 𝑦 2 = 𝑥 2 + 𝑦 2 + 2𝑦 + 1
Cancel terms
−6𝑥 + 9 = 2𝑦 + 1
Subtract 1
−6𝑥 + 8 = 2𝑦
Divide by 2
−3𝑥 + 4 = 𝑦
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
y
(0,4)
a)
Use an algebraic method to find
the Cartesian equation of the
locus of z if:
𝑧−3 = 𝑧+𝑖
x
𝒚 = −𝟑𝒙 + 𝟒
b) Represent the locus of z on an
Argand diagram
(3,0)
(0,-1)
𝑧−3 = 𝑧+𝑖
y = -3x + 4
Distance
from (3,0)
Distance
from (0,-1)
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
𝐼𝑓: 𝑧 = 𝑥 + 𝑖𝑦
You can use complex numbers to
represent a locus of points on an
Argand diagram
𝑇ℎ𝑒𝑛:
𝑧 − 6 = 2 𝑧 + 6 − 9𝑖
Replace z with ‘x + iy’
𝑥 + 𝑖𝑦 − 6 = 2 𝑥 + 𝑖𝑦 + 6 − 9𝑖
If:
𝑧 − 6 = 2 𝑧 + 6 − 9𝑖
a)
Use algebra to show that the locus
of z is a circle, stating its centre
and radius
𝒙 + 𝟏𝟎
𝟐
+ 𝒚 − 𝟏𝟐
𝟐
= 𝟏𝟎𝟎
Circle, centre (-10,12) and radius 10
b) Sketch the locus of z on an Argand
diagram
𝑥2 + 𝑦2
𝑧 = 𝑥 + 𝑖𝑦 =
Group real and
imaginary parts
𝑥 − 6 + 𝑖𝑦 = 2 𝑥 + 6 + 𝑖(𝑦 − 9)
𝑥−6
2
+ 𝑦2 = 2
𝑥+6
2
+ 𝑦−9
2
𝑥−6
2
+ 𝑦2 = 4 𝑥 + 6
2
+ 𝑦−9
2
Replace the moduli
using the rule above
Square both sides
(remember the ‘2’)
Expand some
brackets
𝑥 2 − 12𝑥 + 36 + 𝑦 2 = 4 𝑥 2 + 12𝑥 + 36 + 𝑦 2 − 18𝑦 + 81
Expand another
bracket
Group all terms
on one side
𝑥 2 − 12𝑥 + 36 + 𝑦 2 = 4𝑥 2 + 48𝑥 + 144 + 4𝑦 2 − 72𝑦 + 324
0 = 3𝑥 2 + 60𝑥 + 3𝑦 2 − 72𝑦 + 432
0 = 𝑥 2 + 20𝑥 + 𝑦 2 − 24𝑦 + 144
0 = 𝑥 + 10
2
− 10
2
0 = 𝑥 + 10
2
+ 𝑦 − 12
2
100 = 𝑥 + 10
2
+ 𝑦 − 12
2
+ 𝑦 − 12
− 100
2
Divide by 3
− 12
2
+ 144
Completing
the square
Simplify
Add 100
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
(-10,12)
If:
The circle shows the set
of points that are twice
as far from (6,0) as they
are from (-6,9)!
(-6,9)
𝑧 − 6 = 2 𝑧 + 6 − 9𝑖
a)
Use algebra to show that the locus
of z is a circle, stating its centre
and radius
𝒙 + 𝟏𝟎
𝟐
+ 𝒚 − 𝟏𝟐
𝟐
P(x,y)
(6,0)
x
= 𝟏𝟎𝟎
Circle, centre (-10,12) and radius 10
b) Sketch the locus of z on an Argand
diagram
𝑧 − 6 = 2 𝑧 + 6 − 9𝑖
The distance
from (6,0)
Is equal Twice the distance
to
from (-6,9)
3F
𝑧 − 𝑧1
Is the distance between the variable point z and the fixed
point z1 when they are represented on a Argand diagram
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
Therefore:
When 𝑧 − 𝑧1 = 𝑎 𝑧 − 𝑧2
 An Algebraic method will most likely be the best
way to find the equation of the locus of z
If:
𝑧 − 6 = 2 𝑧 + 6 − 9𝑖
a)
Use algebra to show that the locus
of z is a circle, stating its centre
and radius
𝒙 + 𝟏𝟎
𝟐
+ 𝒚 − 𝟏𝟐
𝟐
 You will probably have to use completing the
square (sometimes with fractions as well!)
= 𝟏𝟎𝟎
Circle, centre (-10,12) and radius 10
b) Sketch the locus of z on an Argand
diagram
3F
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
If:
𝑎𝑟𝑔𝑧 =
𝜋
4
Sketch the locus of P(x,y) which is
represented by z on an Argand
diagram. Then find the Cartesian
equation of this locus algebraically.
The line is not extended
back downwards
 It is known as a ‘halfline’
𝝅
𝟒
x
 The locus will be the set of points
which start at (0,0) and make an
argument of π/4 with the positive xaxis
3F
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
y
𝝅
𝟒
If:
𝑎𝑟𝑔𝑧 =
x
𝜋
4
Sketch the locus of P(x,y) which is
represented by z on an Argand
diagram. Then find the Cartesian
equation of this locus algebraically.
 The locus will be the set of points
which start at (0,0) and make an
argument of π/4 with the positive xaxis
𝑎𝑟𝑔𝑧 =
𝜋
4
𝜋
𝑎𝑟𝑔(𝑥 + 𝑖𝑦) =
4
𝑇𝑎𝑛−1
x
𝑦
𝜋
=
𝑥
4
Replace z with ‘x + iy’
The value of the argument
is tan-1(opposite/adjacent)
𝑦
𝜋
= 𝑇𝑎𝑛
𝑥
4
𝑦
=1
𝑥
𝑦=𝑥
‘Normal tan’
Calculate the tan part
Multiply by x
(x > 0)
3F
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
If:
arg(𝑧 − 2) =
(2,0)
𝝅
𝟑
x
𝜋
3
Sketch the locus of P(x,y) which is
represented by z on an Argand
diagram. Then find the Cartesian
equation of this locus algebraically.
 The locus will be the set of values
that, when we subtract 2 from
them, make an angle of π/3 with
the origin
 The locus must therefore start at
(2,0) rather than (0,0)!
3F
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
y
(2,0)
If:
x-2
𝜋
arg(𝑧 − 2) =
3
Sketch the locus of P(x,y) which is
represented by z on an Argand
diagram. Then find the Cartesian
equation of this locus algebraically.
 The locus will be the set of values
that, when we subtract 2 from
them, make an angle of π/3 with
the origin
 The locus must therefore start at
(2,0) rather than (0,0)!
𝝅
𝟑
arg(𝑧 − 2) =
𝜋
3
𝜋
𝑎𝑟𝑔(𝑥 + 𝑖𝑦 − 2) =
3
𝑇𝑎𝑛−1
x
𝑦
𝜋
=
𝑥−2
3
Replace z with ‘x + iy’
The value of the argument
is tan-1(opposite/adjacent)
𝑦
𝜋
= 𝑇𝑎𝑛
𝑥−2
3
𝑦
= 3
𝑥−2
𝑦 = 3𝑥 − 2 3
‘Normal tan’
Calculate the tan part
Multiply by (x – 2)
(x > 2)
3F
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
If:
𝑎𝑟𝑔(𝑧 + 3 + 2𝑖) =
3𝜋
4
𝟑𝝅
𝟒
x
(-3,-2)
Sketch the locus of z on an Argand
diagram and use an algebraic method
to find the equation of the line.
 When we add 3 and 2i to z, the
argument from (0,0) and the
positive x-axis will be 3π/4
 So the line will have to start at
(-3,-2)
3F
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
y+2
If:
𝑎𝑟𝑔(𝑧 + 3 + 2𝑖) =
3𝜋
4
Sketch the locus of z on an Argand
diagram and use an algebraic method
to find the equation of the line.
 When we add 3 and 2i to z, the
argument from (0,0) and the
positive x-axis will be 3π/4
 So the line will have to start at
(-3,-2)
x
𝟑𝝅
𝟒
x + 3 (-3,-2)
𝑎𝑟𝑔(𝑧 + 3 + 2𝑖) =
3𝜋
4
3𝜋
𝑎𝑟𝑔(𝑥 + 𝑖𝑦 + 3 + 2𝑖) =
4
𝑇𝑎𝑛−1
𝑦+2
3𝜋
=
𝑥+3
4
Replace z with ‘x + iy’
The value of the argument
is tan-1(opposite/adjacent)
𝑦+2
3𝜋
= 𝑇𝑎𝑛
𝑥+3
4
𝑦+2
= −1
𝑥+3
𝑦 + 2 = −𝑥 − 3
𝑦 = −𝑥 −5
‘Normal tan’
Calculate the tan part
Multiply by (x + 3)
Subtract 2
(x < -3)
3F
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
So therefore:
𝑎𝑟𝑔 𝑧 − 𝑧1 = 𝜃
Is represented by a half line starting
at z1 and making an angle of θ with a
line parallel to the x-axis
3F
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
Joining the ends of a chord to different points on the
circumference will always create the same angle, if the
points are in the same sector
“Angles in the same sector are equal”
For the next set of Loci, you need to remember some rules relating to circles
Major arc – θ is acute
Minor arc – θ is obtuse
θ
θ
A
θ
Semi-circle – θ is 90°
θ
B
θ
x
θ
B
A
2x
A
A
B
If they are joined to a
point on the major arc
If they are joined to a point
on the minor arc
If the chord is the
diameter of the circle
 The angle will be acute
 The angle will be obtuse
 The angle will be 90°
B
 The angle at the
centre is twice the angle
at the circumference
3F
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
If:
𝑎𝑟𝑔
a)
𝑧−6
𝜋
=
𝑧−2
4
Sketch the locus of P(x,y), which
is represented by z on an Argand
diagram
 The argument above can be
rewritten using this rule:
𝑎𝑟𝑔
𝑎𝑟𝑔
𝑧−6
𝜋
=
𝑧−2
4
𝑧1
= 𝑎𝑟𝑔 𝑧1 − 𝑎𝑟𝑔 𝑧2
𝑧2
𝑎𝑟𝑔 𝑧 − 6 − 𝑎𝑟𝑔 𝑧 − 2 =
So what we are doing is drawing the locus of
points where the difference between these
arguments is π/4
𝜋
4
3F
Further complex numbers
y
arg(z – 6) = θ1
You can use complex numbers to
represent a locus of points on an
Argand diagram
arg(z – 2) = θ2
If:
𝑎𝑟𝑔
a)
θ2
𝑧−6
𝜋
=
𝑧−2
4
Sketch the locus of P(x,y), which
is represented by z on an Argand
diagram
𝑎𝑟𝑔 𝑧 − 6 − 𝑎𝑟𝑔 𝑧 − 2 =
This angle must
therefore be θ1 – θ2,
the difference
between the
arguments!
θ1
𝜋
4
So what we are doing is drawing the locus
of points where the difference between
these arguments is π/4
However, there are more
points that satisfy this rule!
θ1
θ2
(2,0)
(6,0)
x
 Imagine drawing both arguments – we will use θ1 and θ2 to represent
their values
 Using alternate angles, we can show the angle between the
arguments is their difference
 We want this difference to be π/4
3F
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
If:
π/
𝑧−6
𝜋
𝑎𝑟𝑔
=
𝑧−2
4
a)
π/
Sketch the locus of P(x,y), which
is represented by z on an Argand
diagram
𝑎𝑟𝑔 𝑧 − 6 − 𝑎𝑟𝑔 𝑧 − 2 =
𝜋
4
So what we are doing is drawing the locus
of points where the difference between
these arguments is π/4
Geogebra
Example
4
4
θ2
(2,0)
θ1
θ2
θ1
(6,0)
x
 If we move the point where the lines cross along the major arc of a
circle, then the value of π/4 will remain the same
 The arguments will change but this doesn’t matter, it is the
difference that matters!
 So the locus of a difference between arguments is always given by
an arc of a circle
3F
Further complex numbers
y
“The angle at the centre
is twice the angle at the
circumference”
You can use complex numbers to
represent a locus of points on an
Argand diagram
π/
4
If:
𝑎𝑟𝑔
a)
𝑧−6
𝜋
=
𝑧−2
4
π/
Sketch the locus of P(x,y), which
is represented by z on an Argand
diagram
b) Find the Cartesian equation of this
locus
We need the centre of the ‘circle’ and
its radius
 We need to use another of the
rules we saw:
(2,0)
2
(6,0)
x
We can use this isosceles triangle to find the
information we need…
Centre
Radius
Radius
3F
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
Centre
Radius
Radius
If:
𝑎𝑟𝑔
a)
(4,2)
𝑧−6
𝜋
=
𝑧−2
4
Sketch the locus of P(x,y), which
is represented by z on an Argand
diagram
b) Find the Cartesian equation of this
locus
We need the centre of the ‘circle’ and
its radius
Centre (4,2)
Radius 2√2
π/
4
2
π/
4
(2,0)
2
(4,0)
(6,0)
 Split the triangle in the middle, the smaller angles will both be
π/ (45ᵒ) (because the top angle was π/ )
4
2
 The middle of the base will be (4,0), and you can work out the
side lengths from this
 The top will therefore be at (4,2)
 Use Pythagoras’ Theorem to find the diagonal (the radius)
3F
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
π/
4
If:
𝑎𝑟𝑔
a)
𝑧−6
𝜋
=
𝑧−2
4
Sketch the locus of P(x,y), which
is represented by z on an Argand
diagram
b) Find the Cartesian equation of this
locus
We need the centre of the ‘circle’ and
its radius
Centre (4,2)
Radius 2√2
(6,0)
(2,0)
x
The locus is therefore the arc of a circle with the
following equation:
𝑥−4
2
+ 𝑦−2
2
=8
𝑦>0
3F
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
Generally, for this type of question,
you need to follow 3 steps:
Step 1: Mark on the Argand diagram
the two points where the arguments
start
Step 2: Decide whether the arc is
going to be major, minor, or a semicircle, by considering the angle
Step 3: Draw the arc between the
points. You always draw from the
numerator point to the denominator
point
 Anti-clockwise if θ is positive
 Clockwise if θ is negative
If the value we want is positive, then θ1 > θ2
If the value we want is negative, then θ2 > θ1
Drawing in the direction indicated in step 3 means
you will ensure the arguments are correct to give a
positive or negative answer
 As we do some examples we will refer to this!
3F
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
Sketch the locus of P(x,y) on an Argand diagram if:
𝑧
𝜋
𝑎𝑟𝑔
=
𝑧 − 4𝑖
2
y
Generally, for this type of question,
you need to follow 3 steps:
Step 1: Mark on the Argand diagram
the two points where the arguments
start
Step 2: Decide whether the arc is
going to be major, minor, or a semicircle, by considering the angle
Step 3: Draw the arc between the
points. You always draw from the
numerator point to the denominator
point
 Anti-clockwise if θ is positive
 Clockwise if θ is negative
(0,4)
(0,0) and (0,4)
The angle to
make is π/2
 A semi-circle
(0,0)
x
Θ is positive, so draw
anti-clockwise from (0,0)
(numerator point) to
(0,4) (denominator point)
3F
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
 In step 3 we had to choose whether to draw the diagram clockwise or
anti-clockwise from the numerator point to the denominator point
 Lets show why this is correct!
y
y
(0,4)
(0,4)
θ2
θ2
θ2
θ1
θ2
θ1
θ1
θ1
x
(0,0)
 We drew the angle anticlockwise from (0,0) to
(0,4)
 However, as θ2 is
actually negative, the
sum is really θ1 + (-θ2)
 Using the alternate
angles, the angle between
the arguments is θ1 + θ2
= θ1 – θ2
 This angle is therefore
what we were wanting!
Basically, always use the rule in step 3!
x
(0,0)
 If we drew the arc
the other way clockwise from (0.0)
to (0,4)
 Using the alternate
angles, the on the
outside is θ1 + θ2
 However, as θ2 is
actually negative, the sum
is really θ1 + (-θ2)
= θ1 – θ2
 But of course it is on
the wrong side of the
arc so we do not want
this part of the circle!
3F
Further complex numbers
You can use complex numbers to
represent a locus of points on an
Argand diagram
Sketch the locus of P(x,y) on an Argand diagram if:
𝑎𝑟𝑔
𝑧 + 3𝑖
𝜋
=
𝑧−2
3
y
Generally, for this type of question,
you need to follow 3 steps:
Step 1: Mark on the Argand diagram
the two points where the arguments
start
Step 2: Decide whether the arc is
going to be major, minor, or a semicircle, by considering the angle
Step 3: Draw the arc between the
points. You always draw from the
numerator point to the denominator
point
 Anti-clockwise if θ is positive
 Clockwise if θ is negative
(0,-3) and (0,2)
(0,2)
The angle to
make is π/3
 A major arc
Θ is positive, so draw anticlockwise from (0,-3)
(numerator point) to (0,2)
(denominator point)
x
(0,-3)
3F
Further complex numbers
y
You can use complex numbers to
represent a locus of points on an
Argand diagram
3
Given that the complex number
z = x + iy satisfies the equation:
3
(12,5)
𝑧 − 12 − 5𝑖 = 3
Find the minimum and maximum values
of |z|
 Start by drawing this on an
Argand diagram
 It is a circle, centre (12,5) radius
3 units
13
x
 The smallest and largest values for |z| will be on the same straight line
through the circle’s centre
 You can mark the size of the radius on the diagram
 Find the distance from (0,0) to (12,5), then add/subtract 3 to find the
largest and smallest values
52 + 122 = 13
 So the largest value of |z| will be 16 and the smallest will be 10
3F
Further complex numbers
y
You can use complex numbers to
represent regions on a Argand diagram
This is very similar to what you have
been doing with loci
(4,2)
The only extra part is that once you
have drawn the locus representing the
point, you need to indicate the area
required
x
Shade on an Argand diagram the region
indicated by:
𝑧 − 4 − 2𝑖 ≤ 2
 Start with a circle, centre (4,2) and
radius 2 units (as 2 is the ‘limit’)
The region we want is where the absolute value of z is
less than 2
 This will be the region inside the circle
3G
Further complex numbers
y
You can use complex numbers to
represent regions on a Argand diagram
This is very similar to what you have
been doing with loci
The only extra part is that once you
have drawn the locus representing the
point, you need to indicate the area
required
(4,0)
(6,0)
x
Shade on an Argand diagram the region
indicated by:
𝑧−4 < 𝑧−6
 Start with the perpendicular bisector
between (4,0) and (6,0) as this is the
‘limit’
𝑧−4 < 𝑧−6
The distance to |z – 4| must be less than the
distance to |z – 6|
 Shade the region closest to (4,0)
3G
Further complex numbers
y
You can use complex numbers to
represent regions on a Argand diagram
This is very similar to what you have
been doing with loci
𝝅
𝟒
(2,2)
The only extra part is that once you
have drawn the locus representing the
point, you need to indicate the area
required
Shade on an Argand diagram the region
indicated by:
𝜋
0 ≤ 𝑎𝑟𝑔 𝑧 − 2 − 2𝑖 ≤
4
 Start by drawing the limits of the
argument from the point (2,2)
x
The argument must be between these two values
 Shade the region between the two arguments
3G
Further complex numbers
y
You can use complex numbers to
represent regions on a Argand diagram
This is very similar to what you have
been doing with loci
The only extra part is that once you
have drawn the locus representing the
point, you need to indicate the area
required
x
Shade on an Argand diagram the region
indicated by:
𝑧 − 4 − 2𝑖 ≤ 2
Imagine all the regions were on the same diagram
𝑧−4 < 𝑧−6
and
0 ≤ 𝑎𝑟𝑔 𝑧 − 2 − 2𝑖 ≤
𝜋
4
 The region we want will have to satisfy all of
these at the same time!
3G
Further complex numbers
y
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
The z-plane
(uses x and y)
x
Effectively we take a set of points in
the complex plane, transform them all
and map them on a new complex plane
 You will need to use Algebraic
methods a lot for this as visualising the
transformations can be very difficult!
Transformation
from one plane
to the next!
v
The w-plane
(uses u and v)
u
3H
Further complex numbers
y
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
The point P represents the complex
number z on an Argand diagram where
|z| = 2.
T1 represents a transformation from the
z plane, where z = x + iy, to the w-plane
where w = u + iv.
v
x
The z-plane
u
The w-plane
Circle centre
(0,0), radius 2
Circle centre (-2,4),
radius 2
 To start with, make z the subject
𝑤 = 𝑧 − 2 + 4𝑖
Add 2, subtract 4i
Describe the locus of P under the
transformation T1, when T1 is given by:
𝑧 = 𝑤 + 2 − 4𝑖
𝑇1 : 𝑤 = 𝑧 − 2 + 4𝑖
𝑧 = 𝑤 + 2 − 4𝑖
 We will work out the new set of
points algebraically…
2 = 𝑤 + 2 − 4𝑖
The modulus of each
side must be the same
We know |z| from
the question
Circle, centre (-2,4), radius 2
3H
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
The point P represents the complex
number z on an Argand diagram where
|z| = 2.
T2 represents a transformation from
the z plane, where z = x + iy, to the wplane where w = u + iv.
Describe the locus of P under the
transformation T2, when T2 is given by:
𝑇2 : 𝑤 = 3𝑧
 We will work out the new set of
points algebraically…
The z-plane
y
The w-plane
v
x
Circle centre
(0,0), radius 2
u
Circle centre (0,0),
radius 6
 To start with, make z the subject
𝑤 = 3𝑧
𝑤
=𝑧
3
Divide by 3
𝑤
= 𝑧
3
𝑤
=2
3
|z|= 2
Split the modulus up
Modulus of
both sides
𝑧1
𝑧1
=
𝑧2
𝑧2
|3|=3 so multiply by 3
𝑤 =6
3H
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
The point P represents the complex
number z on an Argand diagram where
|z| = 2.
T2 represents a transformation from
the z plane, where z = x + iy, to the wplane where w = u + iv.
Describe the locus of P under the
transformation T3, when T3 is given by:
1
𝑇3 : 𝑤 = 𝑧 + 𝑖
2
 We will work out the new set of
points algebraically…
The z-plane
y
The w-plane
v
x
Circle centre
(0,0), radius 2
u
Circle centre (0,1),
radius 1
 To start with, make z the subject
1
𝑧+𝑖
2
1
𝑤−𝑖 = 𝑧
2
𝑤=
1
𝑤−𝑖 = 𝑧
2
1
𝑤−𝑖 =
𝑧
2
1
𝑤−𝑖 =
2
2
Subtract i
Leaving z like this can make
the problem easier! (rather
than rearranging
completely)
Modulus of both sides
You can split the
modulus on the right
𝑧1 𝑧2 = 𝑧1 𝑧2
|z| = 2
Simplify the right side
𝑤−𝑖 =1
3H
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
For the transformation w = z2, where
z = x + iy and w = u + iv, find the locus of
w when z lies on a circle with equation
x2 + y2 = 16
 It is very important for this topic
that you draw information on z or
|z| from the question
The z-plane
y
The w-plane
x
Circle centre
(0,0), radius 4
u
Circle centre (0,0),
radius 16
𝑤 = 𝑧2
𝑤 = 𝑧2
 The equation x2 + y2 = 16 is a circle,
centre (0,0) and radius 4
𝑤 = 𝑧 𝑧
 Therefore |z| = 4
𝑤 =4×4
 We now proceed as before, by
writing the equation linking w and z
in such a way that |z| can be
replaced
v
Modulus of both sides
Split the modulus up
Replace |z| with 4
Calculate
𝑤 = 16
Circle, centre (0,0) and radius 16
3H
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
The transformation T from the z-plane,
where z = x + iy, to the w-plane, where
w = u + iv, is given by:
𝑤=
5𝑖𝑧 + 𝑖
𝑧+1
Show that the image, under T, of the
circle |z| = 1 in the z-plane, is a line l in
the w-plane. Sketch l on an Argand
diagram.
 Make z the subject!
 Now eliminate z using what we
know…
𝑤=
5𝑖𝑧 + 𝑖
𝑧+1
Multiply by (z + 1)
𝑤(𝑧 + 1) = 5𝑖𝑧 + 𝑖
Expand the bracket
𝑤𝑧 + 𝑤 = 5𝑖𝑧 + 𝑖
𝑤𝑧 − 5𝑖𝑧 = 𝑖 − 𝑤
Subtract 5iz and
subtract w
Factorise the left side
𝑧(𝑤 − 5𝑖) = 𝑖 − 𝑤
𝑖−𝑤
𝑧=
𝑤 − 5𝑖
𝑧 =
1=
𝑖−𝑤
𝑤 − 5𝑖
𝑖−𝑤
𝑤 − 5𝑖
Divide by (w – 5i)
Modulus of both sides
|z| = 1
Multiply by |w – 5i|
𝑤 − 5𝑖 = 𝑖 − 𝑤
𝑤 − 5𝑖 = 𝑤 − 𝑖
|i - w| = |w – i|
3H
Further complex numbers
Transformation T
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
The transformation T from the z-plane,
where z = x + iy, to the w-plane, where
w = u + iv, is given by:
𝑤=
5𝑖𝑧 + 𝑖
𝑧+1
Show that the image, under T, of the
circle |z| = 1 in the z-plane, is a line l in
the w-plane. Sketch l on an Argand
diagram.
𝑤 − 5𝑖 = 𝑤 − 𝑖
𝑧 =1
Circle centre
(0,0), radius 1
The z-plane
Perpendicular bisector
between (0,1) and (0,5)
 The line v = 3
y
The w-plane
v
x
u
So a circle can be transformed into a
straight line!
3H
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
The transformation T from the z-plane,
where z = x + iy, to the w-plane, where
w = u + iv, is given by:
𝑤=
3𝑧 − 2
𝑧+1
𝑤 𝑧 + 1 = 3𝑧 − 2
Expand the bracket
𝑤𝑧 + 𝑤 = 3𝑧 − 2
Subtract wz and add 2
𝑤 + 2 = 3𝑧 − 𝑤𝑧
Factorise the right side
𝑤 + 2 = 𝑧(3 − 𝑤)
3𝑧 − 2
𝑤=
𝑧+1
𝑤+2
=𝑧
3−𝑤
Show that the image, under T, of the
circle with equation x2 + y2 = 4 in the
z-plane, is a different circle C in the
w-plane.
𝑤+2
= 𝑧
3−𝑤
State the centre and radius of C.
Multiply by (z + 1)
𝑤+2
=2
3−𝑤
Divide by (3 – w)
Modulus of each side
Split up the modulus
|z| = 2
Multiply by |3 - w|
𝑤+2 =2 3−𝑤
|3 - w| = |w - 3|
 Remember that x2 + y2 = 4 is the same
as |z| = 2
𝑤+2 =2 𝑤−3
We now need to find what the equation of this will be!
3H
We will find the equation as we did in the early part of section 3F!
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
The transformation T from the z-plane,
where z = x + iy, to the w-plane, where
w = u + iv, is given by:
𝑤=
3𝑧 − 2
𝑧+1
Show that the image, under T, of the
circle with equation x2 + y2 = 4 in the
z-plane, is a different circle C in the
w-plane.
State the centre and radius of C.
x2
y2
 Remember that
+
= 4 is the same
as |z| = 2
𝑤+2 =2 𝑤−3
Replace w with ‘u + iv’
𝑢 + 𝑖𝑣 + 2 = 2 𝑢 + 𝑖𝑣 − 3
Group real/imaginary
terms
𝑢 + 2 + 𝑖𝑣 = 2 (𝑢 − 3) + 𝑖𝑣
Remove the modulus
𝑢+2
2
2
+𝑣 = 4 𝑢−3
2
+𝑣
2
Expand brackets
𝑢2 + 4𝑢 + 4 + 𝑣 2 = 4 𝑢2 − 6𝑢 + 9 + 𝑣 2
𝑢2 + 4𝑢 + 4 + 𝑣 2 = 4𝑢2 − 24𝑢 + 36 + 4𝑣 2
0 = 3𝑢2 − 28𝑢 + 3𝑣 2 + 32
0 = 𝑢2 −
2
100
14
= 𝑢−
9
3
2
14
−
3
Move all to
one side
Divide by 3
28
32
𝑢 + 𝑣2 +
3
3
14
0= 𝑢−
3
Expand more
brackets!
2
+ 𝑣2 +
+ 𝑣2
32
3
Use
completing
the square
Move the
number terms
across
Circle, centre (14/3,0), radius 10/3
3H
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
The transformation T from the z-plane,
where z = x + iy, to the w-plane, where
w = u + iv, is given by:
𝑤=
3𝑧 − 2
𝑧+1
Show that the image, under T, of the
circle with equation x2 + y2 = 4 in the
z-plane, is a different circle C in the
w-plane.
𝑧 =2
2
Transformation T
14
𝑢−
3
2
𝑥 +𝑦 =4
Circle centre
(0,0), radius 2
The z-plane
𝑤+2 =2 𝑤−3
2
+ 𝑣2 =
100
9
Circle, centre (14/3,0),
radius 10/3
y
The w-plane
x
v
u
State the centre and radius of C.
3H
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
A transformation T of the z-plane to
the w-plane is given by:
𝑤=
𝑖𝑧 − 2
1−𝑧
Show that as z lies on the real axis in
the z-plane, then w lies on a line l in the
w-plane. Find the equation of l and
sketch it on an Argand diagram.
 Start by rearranging to make z the
subject ‘as usual’
𝑤=
𝑖𝑧 − 2
1−𝑧
𝑤 1 − 𝑧 = 𝑖𝑧 − 2
𝑤 − 𝑤𝑧 = 𝑖𝑧 − 2
𝑤 + 2 = 𝑖𝑧 + 𝑤𝑧
Multiply by (1 – z)
Expand the bracket
Add 2, Add wz
Factorise the right side
𝑤 + 2 = 𝑧(𝑖 + 𝑤)
𝑤+2
=𝑧
𝑖+𝑤
𝑧=
𝑤+2
𝑖+𝑤
𝑤+2
𝑧=
𝑤+𝑖
Divide by (i + w)
Write the other way round (if
you feel it is easier!)
i+w=w+i
At this point we have a problem, as we do not know anything about |z|
 However, as z lies on the ‘real’ axis, we know that y = 0
 Replace z with ‘x + iy’
𝑥 + 𝑖𝑦 =
𝑤+2
𝑤+𝑖
3H
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
A transformation T of the z-plane to
the w-plane is given by:
𝑤=
𝑖𝑧 − 2
1−𝑧
Show that as z lies on the real axis in
the z-plane, then w lies on a line l in the
w-plane. Find the equation of l and
sketch it on an Argand diagram.
 Now, you need to rewrite the right
side so you can separate all the real
and imaginary terms
 You must be extremely careful with
positives and negatives here!
𝑥 + 𝑖𝑦 =
𝑥 + 𝑖𝑦 =
𝑤+2
𝑤+𝑖
𝑢 + 𝑖𝑣 + 2
𝑢 + 𝑖𝑣 + 𝑖
Replace w with u + iv
Group real and imaginary terms
𝑥 + 𝑖𝑦 =
𝑢 + 2 + 𝑖𝑣
𝑢 + 𝑖(𝑣 + 1)
𝑥 + 𝑖𝑦 =
𝑢 + 2 + 𝑖𝑣 𝑢 − 𝑖(𝑣 + 1)
×
𝑢 + 𝑖(𝑣 + 1) 𝑢 − 𝑖(𝑣 + 1)
𝑥 + 𝑖𝑦 =
𝑢(𝑢 + 2) − 𝑖(𝑢 + 2)(𝑣 + 1) + 𝑖𝑢𝑣 − 𝑖 2 𝑣(𝑣 + 1)
𝑢2 − 𝑖𝑢(𝑣 + 1) + 𝑖𝑢(𝑣 + 1) − 𝑖 2 (𝑣 + 1)2
𝑥 + 𝑖𝑦 =
𝑥 + 𝑖𝑦 =
Multiply by the
denominator but with the
opposite sign (this will
cancel ‘i’ terms on the
bottom
𝑢(𝑢 + 2) + 𝑣(𝑣 + 1) + 𝑖𝑢𝑣 − 𝑖(𝑢 + 2)(𝑣 + 1)
𝑢2 + (𝑣 + 1)2
𝑢𝑣 − (𝑢 + 2)(𝑣 + 1)
𝑢 𝑢 + 2 + 𝑣(𝑣 + 1)
𝑖
+
𝑢2 + 𝑣 + 1 2
𝑢2 + 𝑣 + 1 2
Simplify
i2 = -1
Separate
real and
‘i’ terms
As z lies on the x-axis, we know y = 0
 Therefore, the imaginary part on the right side must
also equal 0
3H
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
A transformation T of the z-plane to
the w-plane is given by:
𝑤=
𝑖𝑧 − 2
1−𝑧
Show that as z lies on the real axis in
the z-plane, then w lies on a line l in the
w-plane. Find the equation of l and
sketch it on an Argand diagram.
𝑥 + 𝑖𝑦 =
𝑢𝑣 − (𝑢 + 2)(𝑣 + 1)
𝑢 𝑢 + 2 + 𝑣(𝑣 + 1)
𝑖
+
𝑢2 + 𝑣 + 1 2
𝑢2 + 𝑣 + 1 2
 Set the imaginary part equal to 0
𝑢𝑣 − (𝑢 + 2)(𝑣 + 1)
=0
𝑢2 + 𝑣 + 1 2
𝑢𝑣 − (𝑢 + 2)(𝑣 + 1) = 0
Multiply by (u2 + (v + 1)2)
(you will be left with the
numerator)
Multiply out the double bracket
𝑢𝑣 − (𝑢𝑣 + 2𝑣 + 𝑢 + 2) = 0
Subtract all these terms
𝑢𝑣 − 𝑢𝑣 − 2𝑣 − 𝑢 − 2 = 0
The ‘uv’ terms cancel out
−2𝑣 − 𝑢 − 2 = 0
Make v the subject
2𝑣 = −𝑢 − 2
 You can now find an equation for the
line in the w-plane
1
𝑣 =− 𝑢−1
2
Divide by 2
So the transformation has created this line in the w-plane
(remember v is essentially ‘y’ and u is essentially ‘x’)
3H
Further complex numbers
You can apply transformations that
map points on the z-plane to points on
the w-plane by applying a formula
relating to z = x + iy to w = u + iv
Transformation T
𝑦=0
(z lies on the
real axis)
1
𝑣 =− 𝑢−1
2
Straight line, gradient is
1/ and y-intercept at
2
(0,-1)
A transformation T of the z-plane to
the w-plane is given by:
𝑖𝑧 − 2
𝑤=
1−𝑧
The z-plane
y
v
The w-plane
Show that as z lies on the real axis in
the z-plane, then w lies on a line l in the
w-plane. Find the equation of l and
sketch it on an Argand diagram.
x
𝒚=𝟎
-2
u
-1
𝟏
𝒗=− 𝒖−𝟏
𝟐
3H
Summary
• You have learnt a lot in this chapter!!
• You have seen proofs of and uses of De
Moivre’s theorem
• You have found real and complex roots of
powers
• You have see how to plot Loci and perform
transformations of complex functions