ENGR 1990 Engineering Mathematics
Introduction to Complex Numbers
Introduction
Y
the green
Recall that when we calculate
the roots of a quadratic equation,
96 (ft/s)
we may get real roots or we may
get a complex conjugate pair. As
3 (ft)
θ = 50o
an example, consider the golf ball
trajectory problem we discussed in
X
300 (ft)
earlier notes.
To find the times when the ball is 50 feet above the ground ( y = 50 (ft) ) , we solved the
quadratic equation 16.1t 2 − 73.54 t + 50 = 0 using the quadratic formula and found
⎧0.8312 ≈ 0.831 (s)
73.54 ± 73.542 − 4(16.1)50
t1,2 =
= 2.2839 ± 1.4527 ⇒ t1,2 = ⎨
2(16.1)
⎩ 3.7366 ≈ 3.74 (s)
The ball passes the 50-foot mark on its way up and on its way down.
To find the times when the ball is 100 feet above the ground, we solved the equation
16.1t 2 − 73.54 t + 100 = 0 and found
73.54 ± 73.542 − 4(16.1)100 73.54 ± −1031.87 73.54 ± j 1031.87
=
=
2(16.1)
32.2
32.2
73.54 ± j 32.1227
=
= 2.2839 ± j 0.9976
32.2
The result is a complex conjugate pair j = −1 . This occurs because the ball never
t1,2 =
(
)
reaches 100 feet, so no real solutions exist.
Complex Numbers and the Complex Plane
Im
Generally, complex numbers have both real and
imaginary parts. The diagram shows a complex number
A plotted in the complex plane. We can express A using
either rectangular or polar coordinates.
Rectangular form:
A= x+ jy
Polar Form:
A = r e jθ
+A
r
y
θ
x
or
Re
A = r ∠θ
We can relate the rectangular and polar forms using right-triangle trigonometry.
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Given the rectangular form A = x + j y , we can find the polar form A = r e j θ .
r = x 2 + y 2 = A …the magnitude of A
( )
θ = tan −1 y x …the phase angle of A
Given the polar form A = r e j θ , we can find the rectangular form A = x + j y .
x = r cos(θ ) …the real part of A
y = r sin(θ ) …the imaginary part of A
From these results we can identify Euler’s formula:
A = x + j y = (r cos(θ )) + j (r sin(θ )) = r (cos(θ ) + j sin(θ )) = r e j θ
or
e j θ = cos(θ ) + j sin(θ ) …Euler’s formula
Complex Conjugate
Given a complex number A = a1 + j a2 , the complex conjugate of a A is defined to be
A* = a1 − j a2 …the complex conjugate
Operations with Complex Numbers
Addition/Subtraction
Addition and subtraction of complex numbers is most easily done in rectangular form.
Given two complex numbers A = a1 + j a2 and B = b1 + j b2 , then
A + B = ( a1 + b1 ) + j ( a2 + b2 ) and A − B = ( a1 − b1 ) + j ( a2 − b2 )
If A and B are given in polar form, then it is best to convert them to rectangular form
before adding or subtracting.
Multiplication/Division (Polar Form)
Multiplication and division of complex numbers is most easily done in polar form.
A × B = ( ae jα )( be jβ ) = abe j (α + β ) and A / B = ( ae jα ) ( be jβ ) = ( a b ) e j (α − β )
If A and B are given in rectangular form, it is usually best to convert them to polar form
before multiplying or dividing.
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Multiplication/Division (Rectangular Form)
Multiplication and division of complex numbers can also be done (with a little more
work) using rectangular form.
Multiplication
Given two complex numbers A = a1 + j a2 and B = b1 + j b2 , then their product is
A × B = ( a1b1 − a2b2 ) + j ( a1b2 + a2b1 )
( j × j = −1)
Note that if B is the complex conjugate of A ( B = A* ) , then the product is a real number
equal to the square of the magnitude of A.
A × A* = (a12 + a22 ) + j (a1a2 − a2 a1 ) = a12 + a22 = A
2
Division
To compute the ratio of A and B is a little more involved. To ensure that the
imaginary parts appear only in the numerator, we must make use of the complex
conjugate.
A a1 + j a2 ⎛ a1 + j a2 ⎞ ⎛ b1 − j b2 ⎞ ( a1b1 + a2b2 ) + j ( a2b1 − a1b2 )
=
=⎜
⎟⋅⎜
⎟=
B b1 + j b2 ⎝ b1 + j b2 ⎠ ⎝ b1 − j b2 ⎠
b12 + b22
Example #1
Given: A = 5 + j 10
Solution:
Find: the polar form of A
r = A = 52 + 102 = 125 ≈ 11.2
θ = tan −1 (10 5) = 1.107 (rad) = 63.4 (deg)
Example #2
Given: A = −5 + j 10
Solution:
Find: the polar form of A
r = A = 52 + 102 = 125 ≈ 11.2
θ = tan −1 (10 −5) = −1.107 + π = 2.03 (rad) = 117 (deg)
Example #3
Given: A = 5 + j 10
Solution:
Find: A × A*
A × A* = ( 5 + j 10 ) × ( 5 − j 10 ) = 52 + 102 = 125
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Example #4
Given: A = 5 + j 10 and B = 3 − j 8
Find: A + B , A × B and A B
A + B = ( 5 + j 10 ) + ( 3 − j 8 ) = 8 − j 2
Solution:
A × B = ( 5 + j 10 ) × ( 3 − j 8 ) = ( (5 × 3) − (−8 × 10 ) + j ( (3 × 10) − (5 × 8) )
A × B = 95 − j 10
A B=
( 5 + j 10 ) = ( 5 + j 10 ) × ( 3 + j 8) = ( (15 − 80) + j (30 + 40) ) = −65 + j 70
32 + 82
73
( 3 − j 8) ( 3 − j 8) × ( 3 + j 8)
A B = −0.89 + j 0.959
Example #5
Given: A = 5 e
Solution:
( )
j π3
and B = 8 e
(
A × B = 5e
(
A B = 5e
(
)
Find: A × B and A B
)( )
)
) (8 e ( ) ) = ( 5 8) e (
( )
j π3
j π3
(
j −π 6
× 8e
(
j −π 6
j −π 6
)
= ( 5 × 8) e
( (
j π 3+ − π 6
(
j π 3− −π 6
))
))
= 40 e
( )
j π6
= 0.625 e
( )
j π2
Example #6
Given: A = 5 e
( )
j π3
and B = 8 e
(
j −π 6
)
Find: A + B
Solution: We must first convert the polar forms to rectangular forms, then add
A = 5e
B = 8e
( )
= 5 ( cos(π 3 ) + j sin(π 3 ) ) = 2.5 + j 4.33
(
)
j π3
j −π 6
= 8 ( cos( −π 6 ) + j sin( −π 6 ) ) = 6.928 − j 4
A + B = 9.43 + j 0.33
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