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5MM001 - Calculus and Linear Algebra
Lecture 2
Complex Numbers 2: Graphical Representation
The aim of these notes is to describe complex numbers graphically.
Graphical Representation of Complex Numbers
We are all familiar with the real number line.
-1
0
1
2
3 
If we draw a perpendicular line through the real number line at 0 we introduce a second
axis that can be used to represent the imaginary part of a complex number. The real part
is obviously represented by the real number line. The result is called an Argand diagram.
The diagram below illustrates the Argand diagram with the complex number 2+4i.
Im
.
4
2  4i
1
Re
-2
-1
-1
1
2
Any complex number can be represented by a point on the Argand diagram. In this way we
can imagine complex numbers as being points in the plane.
Representation of a Complex Number as a Vector
There is an alternative way to represent complex numbers. We can also represent complex
numbers as position vectors. For example, the complex number 2  4i can be thought of as
the position vector OA where A is the point (2, 4).
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5MM001 - Calculus and Linear Algebra
Lecture 2
.
Im
4
A 2, 4
Re
O
2
We can also represent complex numbers as free vectors. The advantage of vector
representation means that addition of complex numbers is then equivalent to the addition
of the corresponding vectors.
If z1  x1  y1i and z 2  x2  y2i then z1  z 2  x1  x2    y1  y2  i and z1  z 2 will form the
diagonal OC of the parallelogram OABC, z1  z 2 will form the diagonal BA .
Im
C
z1  z 2
z1
B
z2
z2
z1  z 2
A
z1
Re
O
Example 1
Suppose
z1  5  2i and z 2  3  4i , then z1  z 2  8  6i and z1  z 2  2  2i . We
illustrate these complex numbers as free vectors on the diagram below:
2
5MM001 - Calculus and Linear Algebra
Lecture 2
Im
8  6i
z1  z 2
z1
3  4i
z2
z2
z1  z 2
5  2i
z1
Re
Modulus and Argument (or Polar) Form of a Complex Number
.
P(x, y)
y
θ
O
x
If z  x  yi then the point P(x, y) can be represented by the polar co-ordinate r ,   where
r is called the modulus and is the length OP and θ, the argument, is the angle made by OP
with the real axis.
Calculating the Modulus
The modulus, written z is calculated via Pythagoras’ formula
r  z  x2  y 2 .
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5MM001 - Calculus and Linear Algebra
Lecture 2
Calculating the Argument
The argument is written arg z  . From the above diagram
y
y
y
tan   ,   arg z   arctan  tan 1 .
x
x
x
Note: since there are an infinite set of angles whose tangent is
range       . This is called the principal value set.
y
we restrict θ to the
x
Im
2nd quadrant
1st quadrant
π
Re
-π
3rd quadrant
4th quadrant
If the line representing z is in the first quadrant arg(z) will be in the range 0  arg z  
If z is in the second quadrant arg(z) will be in the range

2

2
.
 arg  z    .
If z is in the third quadrant arg(z) will be in the range    arg  z   


2
.
 arg z   0 .
2
When finding the modulus argument form of a complex number it is essential that you plot
it on an Argand diagram.
If z is in the fourth quadrant arg(z) will be in the range 
Example
Find the modulus and argument of the following complex numbers
1)
z  2  2i
(2, 2)
r  z  22  22  8  2 2
2 
  arg z   arctan  .
2 4
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5MM001 - Calculus and Linear Algebra
2)
z  2  2i
r z 
Lecture 2
(-2, 2)
 22  2 2
 82 2
 2  3
.

2 4
  arg z   arctan 
3)
z  2  2i
(2, -2)

2
r  z  2 2 ,   arg z   arctan 
 .
4
 2 
4)
z  2  2i
(-2, -2)
r  z  2 2,
  2   3
.

4
2
  arg z   arctan 
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5MM001 - Calculus and Linear Algebra
Lecture 2
Tutorial Questions
The tutorial questions consist of three sections: Warm-up questions, section A and section
B.



The warm-up questions are designed to allow you attempt some simple questions on
the lecture material and test your own learning. They also help prepare you for the
section A and B questions.
Section A questions show the level of learning required to answer the section A
questions on the test and exam.
Section B questions show the level of learning required to answer the section B
questions on the test and exam.
Note that not every set of tutorial questions will contain section A and B questions.
Warm-up Questions
1
If z1  3  i , z2  1  4i and z3  4  i represent the following by lines on an Argand
diagram showing the direction of each with an arrow.
a)
b)
c)
z3
z1
z2
d)
e)
f)
z1  z3
z2  z3
z1  z2
g)
h)
i)
z1  z3
z2  z3
z2  z1
2
For the following complex numbers
(i)
represent each on an Argand diagram
(ii)
give the modulus and argument of each
a)
1
b)
-2
c)
d)
e)
f)
 4i
1 i
g)
h)
i)
1 i
4i
2
j)
k)
l)
i 1  i 
i1  i 
7i
4i
m)
n)
1 i
3  4i
3
3i
3  2i
1  i
i 3 1  i 
Plot the square roots obtained in Question 6 Tutorial 1 on an Argand diagram. What
to you notice?
Section A Questions
4
If z  4  2i find iz , i 2 z and i 3 z and represent them on an Argand diagram.
Choose another complex number z and find iz , i 2 z and i 3 z .
Give a geometric representation of multiplication by i, i 2 and i 3 .
5
Points 2  i and  1 2i form two vertices of a square with centre at the origin.
Using techniques from question 3 find the other two vertices.
6
5MM001 - Calculus and Linear Algebra
Lecture 2
6
If z1  1  3i and z2  4  2i find
a)
b)
c)
z3  z1  z2
z1
z2
d)
e)
z3 .
z1  z2
Plot all the points on an Argand diagram.
By considering the points z1 and z1 , z2 and z2 what can you deduce about the
geometric representation of z and its conjugate?
By considering z3 , z3 and z1  z2 what do you deduce?
7
On an Argand diagram draw the set of points that satisfy the equation |z-2+3i|=4.
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5MM001 - Calculus and Linear Algebra
Lecture 2
Solutions
1
d
h
4  3i
 3  5i
2
a
z  1, argz   0
c
z  3, arg  z  
e
z  2 , arg z  
g
i
k
m
5  3i
 2  5i .
e
i

2

4

z  2 , arg  z  
4
3
z  2 , arg z  
4
3
z  2 , arg z  
4
z  2 , arg z  

4
f
7  2i
g
b
z  2, argz   
d
z  4, arg  z  
f
z  13, argz   0.588
h
z  17 , argz   2.897
1

2
 3
4

z  2 , arg  z  
4
34
z
, arg z   1.030
2
z  2 , arg z  
j
l
n
3
For the pairs of solutions for each square root, the second solution can be obtained
by a rotation of the first solution π radians (180˚) around the origin, either clockwise
or anticlockwise.
4
iz  2  4i , i 2 z  4  2i , i 3 z  2  4i .

radians (90˚) anticlockwise
2
around the origin. Hence multiplication by i 2 is a rotation of π radians around the
3

origin and i 3 is a rotation of
radians anticlockwise (or
radians clockwise)
2
2
around the origin.
 2  i, 1  2i .
Multiplication by i is equivalent to a rotation of
5
6
a
b
c
d
1 3i
4  2i
5i
e
5i
The conjugate of z is obtained by a reflection in the Re axis.
z1  z 2   z1  z 2  .
5i
Sample worked solutions
7
If |z-2+3i|=4, then |z-(2-3i)|=4. This means that the distance of the point z from the
point 2-3i is always 4. Therefore, the set of points is a circle with centre 2-3i radius
4.
8