21
Complex
Numbers
Prepared by:
Richard Mitchell
Humber College
21.1 - COMPLEX NUMBERS IN
RECTANGULAR FORM
21.1-DEFINITIONS
Complex Numbers
Rectangular Form
Examples
4 + j2
-7 + j8
where j = -1
The letter i is often used for the imaginary unit. In
technical work, however, we save i for electric current.
Further, j (or i) is sometimes written before or after b.
So the number j2 may also be written 2i, 2j, or i2
5.92 - j 2.93
83 + j 0 = 83 (real number)
0 + j 27 = j 27 (pure imaginary number)
21.1-EXAMPLE 3
Add or subtract and simplify.
Remove all brackets and separately combine
the real parts. Then, combine the imaginary
parts and express the result as a + jb
Examples
j 2 + (6 - j 5)
(2 - j 5) + (-4 + j 3)
(-6 + j 2) - ( 4 - j )
= j 2 + 6 - j5
= 2 - j5 - 4 + j3
= -6 + j 2 - 4 + j
= 6 + j 2 - j5
= 2 - 4 - j5 + j3
= -6 - 4 + j 2 + j
= 6 - j3
= -2 - j 2
= -10 + j3
blank
21.1-EXAMPLE 4
Evaluate j17
-j
Rule
j1 = -1 = + j1
j 5 = j 4 j1 = (+1)(+ j1 ) = + j1
j17 = j16 j1
j 2 = -1 • -1 = -1
j 6 = j 4 j 2 = (+1)(-1) = -1
= j 4 j 4 j 4 j 4 j1
j 3 = j 2 j1 = (-1) j1 = - j1
j 7 = j 4 j 3 = (+1)(- j1 ) = - j1
= (+1)(+1)(+1)(+1) j1
j 4 = j 2 j 2 = (-1)(-1) = +1
j8 = j 4 j 4 = (+1)(+1) = +1
= j1
The first four values keep repeating.
ANS:
j
21.1-EXAMPLE 5
Multiply and simplify.
-j
Examples
j2
´
j4
Multiply as with ordinary numbers
but simplify any powers of j
3
´
j4
´
j5
´
j
( j 3) 2
= j28
= j 3 60
= j 2 32
= ( -1)8
= (- j )60
= (-1)9
= -8
= - j 60
= -9
21.1-EXAMPLE 6
Multiply and simplify.
Examples
Multiply complex numbers as you would
any algebraic expression. Then, replace j2
by -1 and give the result in the form a + jb
(3 - j 2)(-4 + j 5)
(3 - j 5) 2
= 3(-4) + 3( j 5) + (- j 2)(-4) + (- j 2)( j 5)
= (3 - j 5)(3 - j 5)
= -12 + j15 + j8 - j 210
= 9 - j15 - j15 + j 2 25
= -12 + j15 + j8 - (-1)10
= 9 - j15 - j15 + (-1)25
= -12 + j15 + j8 + 10
= 9 - j15 - j15 - 25
= -2 + j 23
= -16 - j30
blank
21.1-EXAMPLE 10 and 11
Divide and simplify.
Divide as with ordinary numbers
but simplify any powers of j
Examples
j6
÷
2
=
j6
j 31
=blank
2
j3
(4 - j 6) ÷ 2
6 ÷ j3
(4 - j 6)
=
2
6 j3
= •
j3 j3
2
3
4 j6
= 21
12
j18
= 2
j 9
2
= 2 - j3
j18
j18
=
=
= - j2
( -1) 9 -91
21.1-EXAMPLE 12
Divide (3 – j4) by (2 + j) using the conjugate method.
(3 - j 4) (2 - j )
=
•
((22 + j ) (2 - j )
blank
6 - j 3 - j8 + j 2 4
=
4 + j2 - j2 - j2
6 - j11 + (-1)4
=
4 - (-1)
Multiply the dividend and divisor
by the conjugate of the divisor (the
conjugate is obtained by changing
the sign of the imaginary part).
6 - 4 - j11
=
4 +1
2 - j11
=
5
2 11
= -j
5
5
ANS:
2
11
- j
5
5
(0.4 - j 2.2 )
21.1-EXAMPLE 13
Solve 2x2 – 5x + 9 = 0 by the quadratic formula to 3 SD’s.
2 x2 - 5x + 9 = 0
(a = 2 b = -5 c = +9)
-(-5) ± (-5) 2 - 4(2)(9)
x=
2(2)
5 ± j 6.86
x=
4
5 ± 25 - 72
x=
4
x=
5 + j 6.86
4
x = 1.25 + j1.72
x=
5 - j 6.86
4
x = 1.25 - j1.72
5 ± -47
x=
4
ANS: x = 1.25 ± j1.72
21.2 - GRAPHING COMPLEX
NUMBERS
21.2-EXAMPLE 15
Plot the complex numbers P (2 + j3), Q (-1 + j2), R (-3 – j2) and
S (1 – j3) on the complex plane.
The Complex Plane
Imaginary
Axis
jb
P (2 + j3)
Q (-1 + j2)
a
Real
Axis
R (-3 – j2)
Argand Diagram
S (1 – j3)
21.3 - COMPLEX NUMBER IN
TRIGONOMETRIC
AND POLAR FORM
21.3-EXAMPLE 16
Write the complex number (2 + j3) in Polar Form.
Rectangular Form
jbImaginary
Polar
(3.61Ð56.30 )
Rectangular
(2 + j 3)
j3 -
j2 -
Rectangular
Polar
r = 3.61
j1 -
q = 56.30
|
1
|
a
2
Conversions
RECTANGULAR
(2 + j 3)
Real
r = a 2 + b2 = 22 + 32 @ 3.61
POLAR
(3.61Ð56.30 )
b
3
q = tan -1 æç ö÷ = tan -1 æç ö÷ @ 56.30
èaø
è 2ø
ANS:
(3 .6 1 Ð 5 6 .3 0 )
P o lar F o rm
21.3-EXAMPLE 17
Write (2 + j3) in Polar Form and Trigonometric Form.
jbImaginary
j3 -
j2 -
Rectangular Form
Polar
(3.61Ð56.30 )
Rectangular
(2 + j 3)
Rectangular
r = 3.61
a + jb = rÐq = r (cosq + j sin q )
j1 -
Rectangular
0
q = 56.3
|
1
|
2
Conversions
RECTANGULAR
(2 + j 3)
Polar
Polar
Trigonometric
a
Real
r = a 2 + b2 = 22 + 32 @ 3.61
TRIGONOMETRIC
3.61(cos 56.30 + j sin 56.30 )
POLAR
(3.61Ð56.30 )
b
3
q = tan -1 æç ö÷ = tan -1 æç ö÷ @ 56.30
èaø
è 2ø
ANS:
3.61(cos 56.3 0 + j sin 56.3 0 )
Trigonometric
0
(3.61 Ð
56.3
)
Polar
21.3-EXAMPLE 18
Write 6(cos 300 + jsin 300) in Polar and Rectangular Forms.
Trigonometric Form
jbImaginary
Rectangular
(5.20 + j 3.00)
j3.00 -
r=6
Rectangular Polar
Polar
(6Ð300 )
a + jb = rÐq = r (cosq + j sin q )
Rectangular
q = 300
|
5.20
Polar
Trigonometric
a
Real
Conversions
RECTANGULAR
(5.20 + j 3.00)
a = 6 cos 300 = 5.20
POLAR
(6Ð300 )
TRIGONOMETRIC
6(cos 300 + j sin 300 )
b = 6 sin 300 = 3.00
AN S:
(5.20 + j 3.00)
(6 Ð 30 0 ) Polar
R ectangular
21.3-EXAMPLE 19
0
Multiply (5Polar
ÐForm
300 ) by (3PoÐ
20
)
lar Form
Product is 5 • 3=15
Argument is 300 + 200 = 500
Multiplication, division, and raising to a
power are best done in Polar Form. To
multiply, the absolute value of the product
of two complex numbers is the product of
their absolute values. The argument is the
sum of the individual arguments.
(5Ð300 ) • (3Ð200 ) = 15Ð500
ANS:
15Ð 500
P o lar
21.3-EXAMPLE 20
Multiply (6.27PolarÐForm
3000 ) by (2.75PolarÐForm
1250 )
Product is 6.27 × 2.75=17.2
Argument is 300 + 125 = 425
0
0
0
The angle θ < 3600 . Subtract
multiples of 3600 if necessary.
(6.27Ð300 ) × (2.75Ð125 ) = 17.2Ð425 = 17.2Ð65
0
0
0
ANS:
0
1 7 .2 Ð 6 5 0
P o lar
21.3-EXAMPLE 21
0
Divide (6Polar
ÐForm
700 ) by (2Polar
ÐF50
)
orm
Quotient is 6 ÷ 2=3
Argument is 700 - 500 = 200
To divide, the absolute value of the
quotient of two complex numbers is the
quotient of their absolute values. The
argument is the difference (numerator
minus denominator) of their arguments.
(6Ð700 )
0
=
3
Ð
2
0
(2Ð500 )
ANS:
3Ð 200
P o lar
21.3-EXAMPLE 22
Evaluate (2Ð10 )5
0
Polar Form
(2Ð100 )5
=25Ð5(10 )
0
When a complex number is raised to the
nth power, the new absolute value is
equal to the original absolute value raised
to the nth power, and the new argument
is n times the original argument.
= 32Ð500
ANS:
32Ð 500
P o lar
21.3-EXAMPLE 24
Find
3
(256 + j192)
Rectangular Form
1
p
1
p
(rÐq ) = r Ð(q + 3600 k ) / p
where k = 0, 1, 2, ... , (p-1)
3
(256 + j192)
Convert the given complex number to
Polar Form and re-write using exponents.
Rectangular Form
=(320Ð36.9 )
0
1
3
Use DeMoivre’s Theorem to solve for all
roots. In this example, we have 3 roots.
Polar Form
1
3
= (320) Ð(36.9 + 360 k ) / 3
0
0
Since there are 3 roots,
k = 0, k = 1 and k = 2.
ANS
= 6.84Ð(12.30 + 1200 k )
21.4 - COMPLEX NUMBERS IN
EXPONENTIAL FORM
21.4-EXAMPLE 25
Write 5(cos 1800 + jsin 1800) in Exponential Form.
Trigonometric Form
a + jb = rÐq = r (cosq + j sin q ) = re jq
Rectangular
Polar
Trigonometric
(Exponential Form)
Conversions
RECTANGULAR
( - 5 + j 0)
Exponential
(q is in radians)
TRIGONOMETRIC
5(cos p + j sin p )
r = a 2 + b2 = -52 + 02 = 5
POLAR
(5Ð1800 )
TRIGONOMETRIC
5(cos 1800 + j sin 1800 )
EXPONENTIAL
5e jp
Remember to convert
1800 into π radians.
ANS:
5 e jp
E x p o n en tial
21.4-EXAMPLE 26
Write 3e j2 in Trigonometric, Polar and Rectangular Forms.
Exponential Form
a + jb = rÐq = r (cosq + j sin q ) = re jq
Rectangular
Polar
Trigonometric
(Exponential Form)
Conversions
RECTANGULAR
( - 1.27 + j 2.72)
a = 3 cos1150 = -1.27
Exponential
(q is in radians)
TRIGONOMETRIC
3(cos 2 + j sin 2 )
rad
rad
TRIGONOMETRIC
3(cos 1150 + j sin 1150 )
POLAR
(3Ð1150 )
b = 3 sin1150 = 2.72
EXPONENTIAL
3e j 2
Remember to convert
2 radians into 1150.
ANS:
( - 1 .2 7 + j 2 .7 2 ) = (3 Ð 1 1 5 0 ) = 3(co s1 1 5 0 + j sin 1 1 5 0 )
R ectan g u lar
P o lar
T rig o n o m etric
21.4-EXAMPLE 27
Multiply and simplify.
Multiply as with ordinary numbers
and apply the laws of exponents.
Examples
j3
2e ! 5e
= 10e j 7
j4
-13.5e - j 3 ! 2.75e j 5
= -37.1e j 2
21.4-EXAMPLE 28
Divide and simplify.
Examples
Divide as with ordinary numbers
and apply the laws of exponents.
8e j 5
4e j 2
5.82e- j 4
9.83e- j 7
= 2e j 3
= 0.592e j 3
21.4-EXAMPLE 29
Evaluate the following.
Examples
(2e j 3 )4
= 16e
j12
Multiply as with ordinary numbers
and apply the laws of exponents.
(0.233e- j 3 )-2
e j6
=
(0.233) 2
e j6
=
0.0497
= 20.1e j 6
21.5 - VECTOR OPERATIONS USING
COMPLEX NUMBERS
21.5-DEFINITIONS
Add the vectors (2 + j3) + (3 – j1).
jbImaginary
(2 + j3)
j3 -
j2 -
The result is the same as when we
add the two vectors algebraically.
Ð
(2 + j3) + (3 – j1) = (5 + j2) or
(5.39 / 21.80)
Polar
(5.39Ð21.80 )
j1 0
Add the two vectors graphically
using the parallelogram method.
Rectangular
(5 + j 2)
|
1
|
2
|
3
|
4
|
5
a
Real
- j1 -
(3 - j1)
j operator
Multiply the complex number (2 + j 3) by the j operator.
(2 + j 3) • j1
0
= (3.61Ð56.3 ) • (1Ð90 )
Convert both vectors to Polar Form
= (2 + j 3) • (0 + j1)
0
= (3.61Ð146.30 )
The j operator increases the angle by 900.
Thus, a vector would be rotated counterclockwise by one-quarter revolution.
r Ðq • j = r Ð(q + 900 )
21.5-EXAMPLE 30
Subtract the vectors (25Ð480 ) - (18Ð1750 ).
jbImaginary
(25Ð480 ) - (18Ð1750 )
(16.7 + j18.6)
(25Ð480 )
j 20 (18Ð1750 )
|
-20
|
-10
Vector addition and subtraction are best done in Rectangular Form.
(34.6 + j17.0)
j10 0
- j10 -
|
10
|
20
|
30
(17.9 - j1.57)
a
Real
a1 = 25 cos 480 = 16.7
jb1 = 25 sin 480 = j18.6
a2 = 18 cos1750 = -17.9
jb2 = 18 sin1750 = j1.57
= (16.7 + j18.6) - (-17.9 + j1.57)
= 16.7 + j18.6 + 17.9 - j1.57
= 16.7 + 17.9 + j18.6 - j1.57
= (34.6 + j17.0)
ANS:
( 3 4 .6 + j 1 7 .0 )
21.5-EXAMPLE 31
Multiply the vectors (2Ð250 ) and (3Ð150 ).
Product is 2 • 3=6
Argument is 250 + 150 = 400
(2Ð250 ) • (3Ð150 ) = 6Ð400
ANS:
6Ð 400
21.5-EXAMPLE 32
Divide the vector (8Ð440 ) by (4Ð120 ).
Quotient is 8 ÷ 4=2
Argument is 440 - 120 = 320
(8Ð440 )
0
=
2
Ð
3
2
(4Ð120 )
ANS:
2Ð 320
21.5-EXAMPLE 33
A certain current I is represented by the complex number (1.15Ð23.50 ) amperes,
and a complex impedence Z is represented by (24.6Ð14.80 ) ohms. Multiply
Z by I to obtain the voltage V (volts).
Product is 24.6 •1.15 = 28.3
Argument is 14.80 + 23.50 = 38.30
V = ZI = (24.6Ð14.80 ) • (1.15Ð23.50 ) = 28.3Ð38.30
AN S:
volts
28.3 Ð 38.3 0
volts
21.6 - ALTERNATING CURRENT
APPLICATIONS
21.6-DEFINITIONS
Alternating Current
PHASOR: Rotating vectors may also be
represented by a complex number RÐwt
by replacing the angle θ by ωt, when ω is
the angular velocity and t is the time.
=5
=280
Example
R sin(wt + f ) = RÐf
5.00 sin(wt + 280 ) = 5.00Ð280
( I and V - rms is the implied amount read on meters)
eff
eff
21.6-EXAMPLES 36 and 37
Complex current and voltage calculations.
( I and V - rms is the implied amount read on meters)
eff
eff
Example 36
Example 37
i = 2.84sin(wt + 330 ) amperes
V = (84.2Ð - 49 ) volts
æI
ö
I = (I eff Ðf ) = ç max Ðf ÷
è 2
ø
v = V • 2 sin(wt + f ) volts
blank
æ 2.84
ö
I=ç
Ð330 ÷
è 2
ø
amperes
I = (2.01Ð33 ) amperes
0
amperes
0
(
(
eff
)
)
v = 84.2 • 2 sin(wt - 49 )
0
v = 119sin(wt - 49 ) volts
0
volts
21.6-EXAMPLE 38
A circuit has a resistance of 5Ω in series with a reactance of 7Ω.
Represent the impedance (Z) by a complex number.
Z = (8.60Ð54.50 )
Polar Form
= 7W
Z = (5 + j 7)
Rectangular Form
Z = 8.60W
f = 54.50
R = 5W
Z = 52 + 7 2 = 8.60W
blank
X = 7 - 0 = 7W
7
f = tan -1 æç ö÷ = 54.5
Z = (5 + j 7)
Z = (8.60Ð54.5 )
Rectangular Form
(arctan)
= 5W
0
è5ø
0
Polar Form
ANS : Z = (5 + j 7)
Rectangular Form
Z = (8.60 Ð 54.5 0 )
Polar Form
21.6-EXAMPLE 39
A voltage of 142 sin 200t is applied to a given circuit (Example 38).
Write a sinusoidal expression for the current i.
Z=
v = 142sin(200t + 0 ) volts
æV
ö
V = (V Ðf ) = ç
Ðf ÷
è 2
ø
max
eff
V
Z
volts
I = (11.6Ð - 54.50 ) amperes
(
)
i = (11.6 • 2 ) sin(200t - 54.5 )
i = I eff • 2 sin(wt + f ) amperes
volts
0
V = (100Ð00 ) volts
0
Z = (8.60Ð54.5 )
From Example 38
Z = (8.60Ð54.50 )
I=
V æ 100Ð00 ö
I= =ç
amperes
0 ÷
Z è 8.60Ð54.5 ø
0
æ 142 0 ö
V =ç
Ð0 ÷
è 2
ø
V
I
amperes
i = 16.4sin(200t - 54.50 ) amperes
A N S : i = 1 6 .4 s in ( 2 0 0 t - 5 4 .5 0 ) A
(con’t)
21.6-EXAMPLE 39
A voltage of 142 sin 200t is applied to a given circuit.
Write a sinusoidal expression for the current i.
Summary
v = 142sin(200t + 0 ) volts
0
V = (100Ð00 ) volts
Z = (8.60Ð
54.5 )
Example 38
0
54.50 = 0.951 rad = 200t
t = 0.951/ 200 = 4.75ms
V æ 100Ð00 ö
I= =ç
amperes
0 ÷
Z è 8.60Ð54.5 ø
I = (11.6Ð - 54.5 ) amperes
0
i = 16.4sin(200t - 54.50 ) amperes
Current lags (phase difference)
the voltage by 54.50 or 4.75 ms.
21.6-EXAMPLE
extra
I=
V
Z
Z=
V
I
Find the complex impedence (Z) for the circuit shown below.
i = 4.0sin( wt + 45 ) amperes
0
v = 10sin( wt + 0 ) volts
0
æV
ö
V = (V Ðf ) = ç
Ðf ÷
è 2
ø
max
eff
æ 10
ö
V =ç
Ð00 ÷
è 2
ø
æI
ö
I = (I Ðf ) = ç
Ðf ÷
è 2
ø
max
volts
eff
æ 4.0
ö
I=ç
Ð450 ÷
è 2
ø
volts
amperes
amperes
I = (2.828Ð45 ) amperes
V = (7.071Ð0 ) volts
0
0
V (7.071Ð0 )
Z= =
= 2.5Ð - 45
I (2.828Ð45 )
0
0
0
ANS : Z = (2.5Ð - 450 )
Copyright © 2016 John Wiley & Sons Canada, Ltd.
All rights reserved. Reproduction or translation of this work beyond that permitted by
Access Copyright (The Canadian Copyright Licensing Agency) is unlawful. Requests
for further information should be addressed to the Permissions Department, John
Wiley & Sons Canada, Ltd. The purchaser may make back-up copies for his or her
own use only and not for distribution or resale. The author and the publisher assume
no responsibility for errors, omissions, or damages caused by the use of these
programs or from the use of the information contained herein.