Download Complex Numbers

In section 3.5 of the textbook complex numbers are introduced, along with the operations
of addition, subtraction, multiplication, and division of complex numbers. Here we will look
at the geometric interpretation of complex numbers, the absolute value and argument of a
complex number, and time permitting extracting the square root of a complex number,
REAL AND IMAGINARY PARTS OF A COMPLEX NUMBER.
Suppose that z is a complex number of the form a + ib, i.e. z= a + ib, where a and b are
both real numbers. The real part of z, denoted by Real(z), or Re(z), is equal to a and
imaginary part of z, denoted by Imag(z), Im(z), is equal to b. For example:
Real( 3-5i) = 3 and Imag( 3-5i ) = - 5.
GEOMETRIC REPRESENTATION OF COMPLEX NUMBERS.
Complex numbers can be shown on an the so called Argand diagram, which is similar to
the Cartesian plane; only the x axis of the Cartesian plane is identified with the real axis,
and the y axis of the Cartesian plane is identified with the imaginary axis. So that for a
given complex number, the horizontal axis represents the real part of the complex number
while the vertical axis represents the imaginary part.
4– 1
1
2
3
3
2
Example 1:
Plot the complex numbers z1 = 4 + 2i, z2 = 3 – i, z3 = -2 + 3i, z4 = -1 – 3i and z5 = i on an
Argand diagram.
imaginary axis
3
z3
z1
2
1
– 2
– 1
– 1
z5
1
2
3
4
real axis
z2
– 2
– 3
z4
Example 2: Let z = 3 – 4i. Plot z and z*= 3 + 4i an Argand diagram.
4– 1
5
1
2
3
3
2
imaginary axis
3
z*
2
1
1
2
3
4
5
real axis
– 1
5– 1
1
2
3
4
2
– 2
– 3
z
Example 3: Let z = 3
+ 4i and w = 2 – 2i.
Show z, w and z + w on an Argand diagram. Note that z + w = 5 + 2i.
Imaginary
z
4
3
z+w
2
1
o
1
2
3
4
5
Real
– 1
– 2
w
0.5
1
1.5
2.5
3
2
4
6
– 2
Notice that the points representing the origin, z, w and z + w form a parallelogram.
Example 4:
Let z = 3 + 4i, and w = 2 – 2i. Show z, w and z - w= 1 + 6i on an Argand diagram.
Imaginary
z-w
6
z
4
2
0.5
1
1.5
2
2.5
3
Real
MODULUS AND ARGUMENT OF A COMPLEX NUMBER z=a+ib.
Imaginary
z = a + ib
b
|z|
arg(z)
Real
a
The modulus of z, written |z|, is the distance of z from the origin.
| z | a 2  b 2 .
The argument of a complex number z is the angle that the line joining O to z makes with
the positive real axis. Anticlockwise rotation is positive and clockwise rotation is negative.
The argument is usually measured in radians and is chosen so that -π < arg(z) ≤ π.
Example: Find the modulus and argument of these complex numbers:
a) 2 + 7i
b) 5 – 2i
c) -4 + 3i
d) -2 – 3i
e) -5
f) 3i
Solution: It helps to sketch an Argand diagram in each case:
a)
| 2 + 7i | =
22  7 2  53
arg(2 + 7i) is angle θ in the diagram:
tan  
i.e.
b)
Imaginary
7
7
2

θ = 1.29 radians
| 5 - 2i | = 52  (2)2  29
Real
2
Imaginary
arg(5 - 2i) is shown by angle θ in the diagram.
5

-2
Real
tan  
i.e.
c)
2
   0.381
5
θ = -0.381 radians (it is negative as the number is below the axis).
| - 4 + 3i | =
(4)2  32  5
Imaginary
3
arg(-4 + 3i) is the angle shown on the diagram.
To find it, it is easiest to first find angle:
tan( ) 
3
4

  0.644
arg(z)

Real
-4
Therefore, arg(-4 + 3i) = π – 0.644 = 2.50 radians.
Imaginary
d)
|-2 – 3i| =
3i is
arg(-2 – 3i) is the angle shown on the diagram. As -2below the real axis, the argument will be negative.
It is simplest to initially find angle :
(2)2  (3)2  13
tan( ) 
3
2

-2
|-5| = 5
arg(-5) = π rads
f)
|3i| = 3
arg(3i) = ½ π rads
Real
-3
  0.983
π – 0.983 = 2.16 radians.
Therefore arg(-2 – 3i) = -2.16 radians.
e)
 arg(z)
Imaginary
3
Real
-5
Situation 1: |z| = k
Since |z| denotes the distance in an Argand diagram that a complex number is from the
origin, the loci of points satisfying the relationship |z| = k is a circle centre the origin, radius
k.
Imaginary
k
k
Real
Related loci are:
* |z| < k
the inside of a circle radius k, centre O (not including the circle
circumference);
Imaginary
k
k
Real
Sketch the loci satisfying these equations:
a)
|z – 1| = 2
b)
|z – 2 + i| > 1
c)
|z – 3| = |z – 1 – i|
d)
|z| = |z + 2i|
e)
arg(z – 2) = π/3
f)
arg(z – 2 – i) = -π/4.
EXTRACTING THE SQUARE ROOT OF A COMPLEX NUMBER:
Let’s investigate the process of taking the square root of a complex number. We will begin
with the following example.
Example: Find the square roots of 6 – 8i.
Solution:
So:
(a + bi)(a + bi) = 6 – 8i
a2 + abi + abi + b2i2 = 6 – 8i
a2 – b2 + 2abi = 6 – 8i.
Comparing real and imaginary parts, we get two equations:
a2 – b2 = 6
and
ab = -4
4
a
16
a2  2  6
Substituting this into the first equation gives:
a
4
2
Now a  6a  16  0 . This is a quadratic equation in a2, and it can be factored as
From the second of these equations we see that b 
(a2 – 8)(a2 + 2) = 0
This in turn gives
a2 = 8
or
Since a is real, we must have a   8  2 2 .
4
If. a  2 2 , then b 
2 2
4
If a  2 2 , then b 
.
2 2
Therefore the square roots of 6– 8i are 2 2  i
a2 = -2.
2
2
,
and
2 2 i
2
2
Exercise: Show that the square roots of 9 – 12i are 2 3  3i and 2 3  3i .