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Review of Complex Arithmetic
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A Review of Complex
Arithmetic
A complex value C has both a real and imaginary component:
a = Re {C }
and
b = Im {C }
so that we can express this complex value as:
C = a + jb
where j 2 = −1 .
Just as a real value can be expressed as a point on the real line,
a complex value can be expressed as a point on the complex
plane:
a
Im {C }
C = a + jb
b
Re {C }
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The values (a,b) are a Cartesian representation of a point on the
complex plane. Recall that we can alternatively denote a point
on a 2-dimensional plane using polar coordinates:
C distance from the origin to the point
∠C φc = rotation angle from the horizontal ( Re {C } ) axis
i.e.,
Im {C }
C = a + jb
C
b
φc
a
Re {C }
Using our knowledge of trigonometry, we can determine the
relationship between the Cartesian (a,b) and polar ( C ,φc )
representations.
From the Pythagorean theorem, we find that:
C = a 2 + b2
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Likewise, from the definition of sine (opposite over hypotenuse),
we find:
sin φc =
b
b
=
C
a 2 + b2
or, using the definition of cosine (adjacent over hypotenuse):
cos φc =
a
a
=
C
a 2 + b2
Combining these results, we can determine the tangent
(opposite over adjacent) of φc:
tanφc =
sinφc b
=
cosφc a
Thus, we can write the polar coordinates in terms of the
Cartesian coordinates:
C = a 2 + b2




a
b
φc = tan −1 b a = cos −1 
= sin −1 


2
2
2
2
 a +b 
 a +b 
( )
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Likewise, we can use trigonometry to write the Cartesian
coordinates in terms of the polar coordinates.
For example, we can use the definition of sine to determine b:
b = C sinφc
and the definition of cosine to determine a:
a = C cosφc
Summarizing:
a = C cosφc
b = C sinφc
Note that we can explicitly write the complex value C in terms
of its magnitude C and phase angle φc :
C = a + jb
= C cosφc + j C sinφc
= C ( cosφc + j sinφc )
Hey! we can use Euler’s equation to simplify this further!
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Recall that Euler’s equation states:
e j φ = cosφ + j sinφ
so complex value C is:
C = a + jb
= C ( cosφc + j sinφc )
= C e j φc
Now we have two ways of expressing a complex value C!
C = a + jb
and/or
C = C e j φc
Note that both representations are equally valid
mathematically—either one can be successfully used in complex
analysis and computation.
Typically, we find that the Cartesian representation is easiest
to use if we are doing arithmetic calculations (e.g., addition and
subtraction).
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For example, if:
C1 = a1 + j b1
then:
C 2 = a2 + j b2
and
C1 + C 2 = ( a1 + a2 ) + j (b1 + b2 )
C1 − C 2 = ( a1 − a2 ) + j (b1 − b2 )
Conversely, for geometric calculations (multiplication and
division), it is easier to use the polar representation:
For example, if:
C1 = C1 e j φ
then:
C2 = C2 e j φ
and
1
2
C1 C 2 = C1 e j φ C2 e j φ
1
2
= C1 C 2 e j φ1e j φ2
= C1 C2 e
j (φ1 +φ2 )
and:
jφ
C1 C1 e
=
C2 C2 e j φ
1
2
C1 e j φ e − j φ
=
C2
1
=
2
C1 j (φ −φ )
e
C2
1
2
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Note in the above calculations we have used the general facts:
x x =x
y
z
( y +z )
xy
y −z
=x( )
z
x
and
Additionally, we note that powers and roots are most easily
accomplished using the polar form of C:
C n = ( C e j φc )
n
=C
(e j φc )
n
n
n
= C e jnφc
and
n
C =(C e
1
=C
1
=C
Therefore:
(
C = C e
2
j φc
j φc
)
1
n
(e j φc )
n
n
e
)
2
1
n
j  φc n 


2
=C e
j (2φc )
and:
C =(C e
j φc
)
1
2
= C e
j  φc 2 


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Finally, we define the complex conjugate (C*) of a complex value
C:
C ∗ Complex Conjugate of C
= a − jb
= C e − j φc
A very important application of the complex conjugate is for
determining the magnitude of a complex value:
C
2
=C C ∗
Typically, the proof of this relationship is given as:
C C ∗ = ( a + jb )( a − jb )
= a ( a − jb ) + jb ( a − jb )
= a 2 + jab − jba − j 2b 2
= a 2 + b2
=C
2
However, it is more easily shown as:
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C C ∗ = ( C e j φc )( C e − j φc )
2
=C e
j (φc −φc )
2
= C ej0
=C
2
Another important relationship involving complex conjugate is:
C + C ∗ = ( a + jb ) + ( a − jb )
= ( a + a ) + j (b − b )
= 2a
Thus, the sum of a complex value and its complex conjugate is a
purely real value.
Additionally, the difference of complex value and its complex
conjugate results in a purely imaginary value:
C − C ∗ = ( a + jb ) − ( a − jb )
= ( a − a ) + j (b + b )
= j 2b
Note from these results we can derive the relationships:
a = Re {C } =
C +C ∗
2
C −C ∗
b = Im {C } =
j2