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Further Concepts for Advanced Mathematics - FP1
Unit 1 Complex Numbers – Section3d The Argand Diagram
The Argand Diagram
Real numbers are usually represented as positions on a horizontal number line.
Real
-3
-2
-1
0
1
2
3
4
5
Addition, subtraction, multiplication and division with real numbers takes place on this number line.
Complex numbers also have an imaginary part so another dimension needs to be added to the number line
like this:
Imaginary
This is called an ‘Argand diagram’,
7
6
after the mathematician of that
5
name. Complex numbers can be
4
represented on the Argand diagram
3
by straight lines. Putting complex
2
numbers on an Argand diagram often
1
Real
-8 -7 -6 -5 -4 -3 -2 -1
1
2
3
4
5
6
7
8
-2
-3
-4
-5
-6
-7
1
helps give a feel for a problem.
The complex numbers u  6  2 j , v  2  7 j , w  5  4 j and z  7  6 j are drawn on the Argand
diagram below:
Imaginary
Two important properties of a
7
w
6
complex number can be visualised
5
on the Argand diagram; the modulus
4
(represented by the length of the
3
line) and the argument (represented
u
2
by the angle of the line from the
1
Real
-8 -7 -6 -5 -4 -3 -2 -1
1
2
3
4
5
6
7
horizontal axis).
8
-2
-3
-4
-5
-6
z
-7
v
Complex numbers and their conjugates appear on an Argand diagram as a reflection of each other in the
axis. For the complex numbers w  5  4 j and z  6  3 j with conjugates w*  5  4 j and
z*  6  3 j , the Argand diagram would look like this:
Imaginary
7
6
5
w
4
z*
3
2
1
Real
-8 -7 -6 -5 -4 -3 -2 -1
1
2
3
4
5
6
7
8
-2
z
-3
-4
w*
-5
-6
-7
2
x
Addition and Subtraction Using the Argand Diagram
To show the effect of addition and subtraction using an Argand diagram, arrows are put on the lines
representing the direction of the complex number from the origin.
For the complex numbers w  5  4 j and z  6  3 j , the calculation
w  z is shown on the diagram
below:
Imaginary
The result
7
6
from the origin. You can check this
z
by calculating
5
w+ z 4
w z.
5  4 j  (6  3 j )  1  7 j
3
w
2
Do you think you would get to the
1
Real
-8 -7 -6 -5 -4 -3 -2 -1
w  z is the direct route
1
2
3
4
5
6
7
8
same point if you calculated
z  w?
-2
-3
-4
-5
-6
-7
In the case of subtraction, the direction of the complex number being subtracted is reversed.
For the complex numbers u  2  3 j and v  6  3 j , the calculation
u  v is shown on the diagram
below:
Imaginary
The direction of
7
(notice the arrow).
6
5
4
u- v
-v
3
2
u
1
Real
-8 -7 -6 -5 -4 -3 -2 -1
v is reversed
1
2
3
4
5
6
7
8
-2
-3
-4
-5
-6
-7
3
The Modulus and Argument of a Complex Number
Imaginary
The diagram shows the point x 
yj on an Argand
diagram. The modulus of the complex number is the
length of the line joining the origin to the point.
x + yj
You can see that by using Pythagoras’ theorem, the
length of the line is
y
O
Real
x 2  y 2 . In algebra, this would
modulus of
x  yj is
be written
x  yj  x 2  y 2 .
x
If we define a complex variable z where
x 2  y 2 . We would say that the
z  x  yj , then we would write z  x 2  y 2 .
The argument of a complex number is the angle that the line joining the point to the origin makes with
the positive Real axis. The values for the argument are usually given as being between -180o and 180o. Any
argument given in this range is described as the principal argument. There are other possibilities that
work but by using the principal argument we can give each complex number (except 0) a unique value for
the argument. The argument of 0 is undefined.
To see how you calculate the argument, we will use two complex numbers z  2  3 j and w  3  5 j .
The diagram shows how each of these may be calculated.
The argument of
Imaginary
7
diagram) is calculated by using the
6
inverse tangent of
5
as
4
z=2 + 3j
3
-2
. This is written
arctan( 32 ) or tan 1 ( 32 ) .
notation.
1
-8 -7 -6 -5 -4 -3 -2 -1
3
2
For this module, we’ll use the arctan
2
θ
1
α
Real
2
3
4
5
6
7
8
We can calculate the argument of
(called
w
α on the diagram) using
arctan( 53 )  arctan( 53 ) . On a
-3
-4
calculator, this returns the value 59o
-5
w=-3 - 5j
z (called θ on the
(to 2 s.f.).
-6
-7
This is clearly not the argument of
actual argument of
argument of
w (which shows why the Argand diagram is important). We can find the
w by realising that the size of the angle α on the diagram is 180o-59o. Hence, the
w is -121o.
4
The correct notation is:
arg( z )  arctan( 32 )
arg( z )  56
arg( w)  121
Always use at least a rough sketch of the position of a complex number on an Argand diagram before you
calculate its argument.
Loci Using Complex Numbers
Loci using modulus
The modulus
z of the complex number z is the distance of z from the origin on an Argand diagram.
This diagram shows
z  w , the modulus of the complex number formed by subtracting w from z.
To make what is happening, the two numbers will be
Imaginary
defined as z  6  4 j and w  1  2 j .
7
z  w  6  4 j  (1  2 j )
6
5
z
4
3
2
 5 2j
-w
w
1
 52  2 2  29
z-w
Real
1
2
3
4
5
6
7
8
You can see from the diagram that the distance given by
z  w is the same as the distance between z and w.
This is in fact true for any modulus written like this. So
z  (3  2 j ) is the distance between z
and 3  2 j . Sometimes this involves a little algebraic fiddling about for example
be rewritten as
z  4  7 j would need to
z  (4  7 j ) before it is easy to see that it is the distance between z and  4  7 j .
We use this idea when we are trying to represent the set of complex numbers that fit the description
z  w  a such as z  (3  2 j )  4 .
The best thing to do is to put the idea into English language. In this case, we want to find all of the points
that are a distance of 4 from 3  2 j . Now imagine this on an Argand diagram.
What would you draw? Sketch it before you look at the answer on the next page.
5
The points for
Imaginary
7
description
6
z that fit the
z  (3  2 j )  4 form a circle
5
with centre (3,2) and radius 4 .
4
How would you show the points that fit
3
these rules?
2
1
Real
-8 -7 -6 -5 -4 -3 -2 -1
1
2
3
4
-2
5
6
7
8
4
-3
a)
z  (3  2 j )  4
b)
z  (3  2 j )  4
c)
z  (3  2 j )  4
-4
-5
-6
-7
Just out of interest……. (some extra ideas)
The common sense approach to finding the points that fit rules can be applied to other types of rules. One
good one is the points that fit the rule
it into words.
z  (1  4 j )  z  (5  j ) . This seems really difficult until you put
z  (1  4 j ) is the distance of points from (1,4) . z  (5  j ) is the distance of points
from (5,1) . The equals means that we want all the points that are the same distance from (1,4) as they
are from (5,1) . What do you draw when you want to show this?
Sketch a quick Argand diagram to show all the points that fit this rule.
What would you draw to show these?
a)
z  (1  4 j )  z  (5  j )
b)
z  (1  4 j )  z  (5  j )
c)
z  (1  4 j )  z  (5  j )
Radian measure
We are now going to use radians to measure our arguments instead of degrees. You need to learn the fact
that
 radians is 180 . Where possible, think about angles as being fractions of 180
 radians. Hence 90 

2
, 60 

3
, 150 
and therefore
5
and so on. Don’t let it confuse you. It is easy to see
6
when radians are being used. You will see a lot of  s!
Loci using argument
When using the argument it is important to remember that it is measured from the positive real axis and
the range is
     .
6

The set of points satisfying arg( z ) 
4
would be represented by the line segment shown on the Argand
diagram below:
Imaginary
The line starts at (0,0) but because
7
6
there is no argument defined there,
5
it is marked with a circle.
4
The line is inclined at an angle of
3
2
positive

4
1
-8 -7 -6 -5 -4 -3 -2 -1
1
2

4
to the real axis.
Real
3
4
5
6
7
8
-2
-3
-4
-5
-6
-7
Loci using the argument can be moved in the same way as those that use modulus.
arg( z  2 j ) 

6
would be first rewritten as
arg( z  (0  2 j )) 

6
Im
so that it is clear that the line would
start from (0,2) and be inclined at

6

6
to the real axis like this:
2
Re
arg( z  3  2 j )  

3
would first be rewritten
as arg( z  (3  2 j ))  

3
Im
so it is clear that the line
would start from (-3,2) and be inclined at 

3
to the real
-3
axis like this:
The blue dotted line shows that the argument cannot be
any lower than

  . The shaded region is there because
that is where the arguments lower than 

3
would lie.
Make sure you understand why this is correct.
7
 2
3
Re