A REVIEW:
COMPLEX NUMBERS
At the end of this module, students will be able to,
1.Recognize a complex number and its different forms.
2. Select appropriate mathematical method in solving function
containing complex numbers.
3.Solve different functions containing complex numbers.
A complex number (z) is a number of the form
z = a + ib
i – imaginary unit
a – real part of z
b – imaginary part of z
The real and imaginary parts of a complex number
z are abbreviated as Re (z) and Im (z) respectively.
Example:
z = 4 - 9i
Re (z) = 4
Im (z) = - 9
Two complex numbers are equal if their real
and imaginary parts are equal
z1 = a1 + ib1
z2 = a2 + ib2
Re (z1) = Re (z2)
Im (z1) = Im (z2)
Different Forms of Complex Number
(a) Rectangular Form
z = a + ib
(b) Polar Form
z = r 
r = (a2 + b2)
 = Tan-1 b /a
r = modulus or absolute
value
 = argument or
amplitude in degrees
Different Forms of Complex Number
(c) Trigonometric Form
cos  = a/r
sin  = b/r
z = a + ib
z = rcos  + i r sin
z = rcis 
(d) Exponential Form
z = re i
 = argument in radians
re i = rcos  + i r sin
Arithmetic Operations
(Rectangular Form)
Addition
z1 = a1 + ib1
z2 = a2 + ib2
I. Addition : z1 + z2
z1 + z2 =
(a1 + ib1) + (a2 + ib2)
=
(a1 + a2) + i (b1 + b2)
Subtraction
z1 = a1 + ib1
z2 = a2 + ib2
II . Subtration : z1 - z2
z1 - z2 = (a1 + ib1) - (a2 + ib2)
= (a1 - a2) + i (b1 - b2)
Multiplication
z1 = a1 + ib1
z2 = a2 + ib2
III. Multiplication: z1 . z2
z1 . z 2 =
(a1 + ib1) (a2 + ib2)
=
(a1a2) + (ia1b2) + (ib1a2) + (i2 b1b2)
=
a1a2 – b1b2 + i (a1b2 + a2b1)
Division
IV. Division :
z1
z2
=
z1
z2
a1 + ib1 . ( a2 - ib2 )
a2 + ib2 ( a2 - ib2 )
Given
z1 = a1 + ib1
z2 = a2 + ib2
Successive Integral of i
i
i2
i3
i4
=
=
=
=
−1
-1
-i
1
i5
i6
i7
i8
=
=
=
=
−1
-1
-i
1
i9
. .....
i12
=
−1
=
1
Example 1: Simplify i1997 + i1999
If i has an exponent divisible by 4, it is
equal to 1 . Such as,
i
i2
i3
i4
.....
=
=
=
=
−1
-1
-i
1
i4 i8 .... i12
Since it is a repetetive sequence, we
can predict the next values.
Example: Simplify
1997
i
+
Think for a value of exponent close to 1997 and 1999 that is divisible by 4.
i1996 =
i1997 =
i1998 =
i1999 =
1
i
-1
-i
i1997 + i1999 = i + (-i) = 0
1999
i
Example 2: Simplify : (3 – i)2 - 7 (3-i) + 10
(3 – i)2 - 7 (3-i) + 10
= 9 - 6i + i2 - 21 + 7i + 10
= 9 - 6i -1 - 21 + 7i + 10
= -3 + i
= (3-i)
Commutative Law, Associative Law &
Distributive Law
Commutative Law :
z1 + z2 = z2 + z1
z1 z2 = z2 z1
Associative Law
z1 + (z2 + z3) = (z1 + z2) + z3
:
z1 (z2 z3) = (z1 z2 ) z3
Distributive Law :
z1 (z2 + z3) = z1 z2
+ z1 z3
Conjugate of a Complex Number
• Conjugate : If z is a complex
number, then the number obtained
by changing the sign of its
imaginary part is called
COMPLEX CONJUGATE or
simply the conjugate of z
Arithmetic Operations (Polar Form)
I. Multiplication :
( r1 1 ) ( r2 2 ) = r1r2 (1 + 2)
II. Division :
𝒓𝟏 1
𝒓𝟐 2
=
𝒓𝟏
𝒓𝟐
𝟏 - 𝟐
III. Exponential Function :
( r  )n = rn (n)
Example 3:
If z1 = 2 + 4i and z2 = -3 + 8i
Find (a) z1 + z2 (b) z1 z2
z1 + z2 = (2-3) + i(4 + 8)
= -1 + 12 i
z1 z2 = (2 + 4i) (-3 + 8i)
= -38 + 4i
Example 4: Find the value of (1 + i)5
a=1
b=1
 = Tan -1 (1 / 1)
= 45 o
r = (a)2 + (b)2
r = (1)2 + ( 1)2
= 1.414
(1.414 45o )5
1.4145 5 (45)
5.657  225
Change into rectangular form: -4 - 4i
a =𝑟cos  = 5.657 cos 225o = -4
b =𝑟sin  = 5.657 sin 225o = -4