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•
INTRODUCTION
• OPERATIONS OF COMPLEX NUMBER
• THE COMPLEX PLANE
• THE MODULUS & ARGUMENT
• THE POLAR FORM
RATIONAL
NUMBERS
(Q)
COMPLEX
NUMBERS
(C)
REAL
NUMBERS
(R)
IRRATIONAL
NUMBERS
(Q)
INTEGERS
(Z)
WHOLE
NUMBERS
(W)
NATURAL
NUMBERS
(N)
 To solve algebraic equations that do not have real
solutions.
x2  4  0
x  2
Real solution
x2  4  0
x   4
 To solve Complex number:
x    4  2i
 Since,
i 2  1  i   1
No real solution
 Example 1 :
 Solve
a) i
b) i
5
8
 Example :
 Solution
a ) i  (i ) i  (1) i  i
5
2 2
2
b) i  (i )  (1)  1
8
2 4
2
 Definition 1.1
If z is a complex number, then the standard
equation of Complex number denoted by:
z  a  bi

where a, b R
 a – Real part of z (Re z)
b – Imaginary part of z (Im z)
 Example 1.2 :
 Express in the standard form, z:
a) 2   4
b) 3   49
 Example 1.2 :
 Solution:
a ) z  2   4  z  2  2i
Re(z) = 2, Im (z) = -2
b) z  3   40  z  3  2 10i
Re(z) = 3, Im (z) = 2√10
 Definition 1.2
2 complex numbers are said equal if and only
if they have the same real and imaginary parts:
a  bi  c  di
Iff a = c and b = d
 Example 1.3 :
 Find x and y if z1 = z2:
a ) 2 x  3 yi  4  9i
b) x  5 yi  10  20i
 Definition 1.3
If z1 = a + bi and z2 = c + di, then:
i ) z1  z 2  (a  c)  (b  d )i
ii ) z1  z 2  (a  c)  (b  d )i
iii ) z1 z 2  (ac  bd )  (ad  bc)i
 Example 1.4 :
 Given z1 = 2+4i and z2= 1-2i
a ) z1  z 2
b) z1  z 2
c) z1 z 2
 Definition 1.4
The conjugate of z = a + bi can be defined as:
z  a  bi  a  bi
***the conjugate of a complex number changes the
sign of the imaginary part only!!!
 Example 1.5 :
 Find the conjugate of
a)
b)
c)
d)
z  2i
z  3  2i
z  10
z  10i
 The Properties of Conjugate Complex Numbers
i) z  z
ii ) z1  z 2  z1  z 2
iii ) z1  z 2  z1  z 2
iv) z1. z 2  z1.z 2
1 1
v)   
z z

n
vi) z  z ; n
n
zz
vii )
 Re( z )
2
zz
viii )
 Im( z )
2
 Definition 1.5
(Division of Complex Numbers)
If z1 = a + bi and z2 = c + di then:
z1 a  bi

z 2 c  di
a  bi c  di


c  di c  di
(ac  bd )  bc  ad i

2
2
c d
Multiply with the
conjugate of
denominator
 Example 1.6 :
 Simplify and write in standard form, z:
2i
a)
1 i
3  4i
b)
1  3i
 The complex number z = a + bi is plotted as a point
with coordinates (a,b).
 Re (z)
 Im (z)
x – axis
y – axis
Im(z)
b
O(0,0)
z(a,b)
a
Re(z)
 Definition 1.6
(Modulus of Complex Numbers)
The modulus of z is defined by
r  z  a b
2
2
Im(z)
z(a,b)
b
r
O(0,0)
a
Re(z)
 Example 1.7 :
 Find the modulus of z:
a) z  2  i
b) z  3  5i
 The Properties of Modulus
i)
z  z
ii ) z z  z 2
iii ) z1 z 2  z1 z 2
z1
z1
iv)

, z2  0
z2
z2
v) z  z
n
n
vi) z1  z 2  z1  z 2
 Definition 1.7
(Argument of Complex Numbers)
The argument of the complex number z = a + bi is
defined as
b
  tan  
a
1
2nd QUADRANT
1st QUADRANT
90     180
0     90
180     270
270     360
3rd QUADRANT
4th QUADRANT
 Example 1.8 :
 Find the arguments of z:
a)
b)
c)
d)
z  2i
z  3  5i
z  1 i
z  2i
Im(z)
(a,b)
r
b

Re(z)
a
 Based on figure above:
a  r cos 
b  r sin 
b
  tan  
a
1
 The polar form is defined by:
z  r cos   i sin  
@
z  r ,  
 Example 1.9:
 Represent the following complex number in polar
form:
a) z  2  i
b) z  3  5i
c) z  2i
 Answer 1.9 :
 Polar form of z:
a ) z  5 cos 333.43  i sin 333.43
b) z  34 cos120.96  i sin 120.96
c) z  2cos 90  i sin 90
 Example 1.10 :
 Express the following in standard form of complex
number:
a ) z  2(cos 45  i sin 45)
b) z  3(cos180  i sin 180)
c) z  2 cos 270  i sin 270 
 Answer 1.10 :
 Standard form:
a ) z  2  2i
b) z  3
c) z   2i
 Theorem 1:
If z1 and z2 are 2 complex numbers in polar
form where
then,
z1  r1 cos 1  i sin 1 
z 2  r2 cos  2  i sin  2 
i ) z1 z 2  r1r2 cos1   2   i sin 1   2 
ii )
z1 r1
 cos1   2   i sin 1   2 
z 2 r2
 Example 1.11 :
a) If z1 = 4(cos30+isin30) and z2 = 2(cos90+isin90) .
Find :
i ) z1 z 2
ii )
z1
z2
b) If z1 = cos45+isin45 and z2 = 3(cos135+isin135) . Find :
i ) z1 z 2
ii )
z1
z2
 Answer 1.11 :
a)
i ) z1z 2  4  4 3i
z1
ii )  1  3i
z2
b)
i ) z1z 2  3
z1
1
ii )   i
z2
3
Think of Adam and Eve like an imaginary number, like
the square root of minus one: you can never see any
concrete proof that it exists, but if you include it in your
equations, you can calculate all manner of things that
couldn't be imagined without it.
Philip Pullman
In The Golden Compass (1995, 2001), 372-373.