Complex numbers
i or j
Complex numbers
An Imaginary Number, when squared, gives
a negative result.
2
imaginary
negative
=
Complex numbers
i = √-1
i is used in maths
But
j is used in electronics and
engineering (because i is already
used as a symbol for current)
Complex numbers
i = √-1
i2 = -1
i3 = -√-1
i4 = 1
i5 = √-1
Complex numbers
Example What is i6 ?
6
i
4
i
2
i
= ×
= 1 × -1
= -1
Adding complex numbers
•(4 +j3) + (3 + j5)
•4 +j3 + 3 + j5
•7 + j8
Adding complex numbers
•(3 +j6) + (2 – j3)
•3 +j6 + 2 – j3
•5 + j3
Subtracting complex numbers
•(6 +j8) - (2 + j3)
•6 +j8 - 2 – j3
•4 + j5
Note the
change of
sign
Multiplying complex numbers
Example 1,
6(3 +j4) =
18 + j24
Example 2
j8 + 3(3 – j2) =
j8 + 9 – j6
j2 + 9
Multiplying complex numbers
(3 + j2)(4 + j)
Use F.O.I.L.
(3x4) + (3xj) + (j2 x4) + (j2 x j)
12 + j3 + j8 + j22
j2 = -1
12 +j11 – 2
10 + j11
Multiplying complex numbers
(5 - j2)(2 + j2)
Use F.O.I.L.
(5 x 2) + (5 x j2) - (j2 x2) - (-j2 x j2)
10 + j10 – j4 - j24
j2 = -1
10 – j6 + 4
14 - j10
Multiplying complex numbers
(4 - j2)(3 - j)
Use F.O.I.L.
(4 x 3) - (4 x j) - (j2 x3) + (j2 x j)
12 – j4 – j6 + j22
j2 = -1
12 - j10 – 2
10 - j10
Multiplying a conjugate pair
(4 - j2)(4 + j2)
Use F.O.I.L.
(4 x 4) + (4 x j2) - (j2 x 4) - (j2 x j2)
16 + j8 – j8 - j24
j2 = -1
16 + 4
20
Dividing complex numbers
(2 +6j)/2j =
2/2j + 6j/2j =
1/j +3
-1
J +3
Dividing complex numbers
(6 + j3)/ (3+j2)
Multiply by the conjugate of the denominator
(6 + j3) x (3 – j2)
(3 + j2) (3 - j2)
18 – j12 +j9 –j26
9 – j6 + j6 –j24
Dividing complex numbers
18 - j3 + 6
9 – j24=
24 – j3
9+4
24 – j3
13
Argand Diagrams
Imaginary
axis
y
Z = x +yj
Real axis x
Argand Diagrams
r = √(x2 + y2)
r
Argand Diagrams
tanΦ = y/x
Φ
Argand Diagrams
Imaginary
axis
y
x = r cosΦ
r
Z = x +yj
yj = r sinΦ
Φ
Real axis x
Example
•Argand diagrams are
used to calculate
impedance in RLC
circuits
Example
The impedance of a circuit is given by
the complex number 3 +j4
Construct the Argand diagram for
3 +j4
Example
Imaginary
axis
y
3
r
Z = 3 +j4
j4
Real axis x
Example
From the Argand diagram
derive the expression for
the impedance in polar
form
Example
r = √(32 + 42)
= √(9 + 16)
√(25) = 5
Imaginary
axis
y
3
r
Z = 3 +j4
j4
Real axis x
Example
tanΦ = 4/3
Φ = 53.13
Imaginary
axis
y
3
r
Z = 3 +j4
j4
Real axis x
Example
Answer
Z = 5 53.13
Imaginary
axis
y
3
r
Z = 3 +j4
j4
Real axis x
Multiplying and dividing polar
form
6∟20° x 4∟30°
Multiply the length (modulus) and add the argument (angle)
= 24∟50°
9∟10° / 3∟40° = 9/3 ∟(10°-40°)
divide the length (modulus) and subtract the argument (angle)
= 3∟-30°
Argand diagrams as phasor
diagrams
The voltage of a circuit is given as
V = 3 + j3
and the current drawn is given as
I = 8 + j2
Find the phase difference between V
and I
Find the power (VI.cosФ)
Argand diagrams as phasor
diagrams
Voltage = √ (32 + 32) = √18 = 4.24 Volts
Current = √(82 + 22)
= √ 68
= 8.25 amps
Argand diagrams as phasor
diagrams
Voltage phase angle tanΦ = 3/3 =1,
Φ = 45o
Current phase angle tanΦ = 2/8 =0.25,
Φ = 14.0o
•
Argand diagrams as phasor
diagrams
Phase difference between V
and I = 45o - 14.0o = 31o
power = VIcosΦ
4.24 x 8.25 cos31o
4.24 x 8.25 x .86
= 30 watts