Review of Complex Arithmetic/Algebra

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Complex Numbers, Arithmetic, and Algebra
Complex Numbers. Complex numbers are numbers that are expressed using the
imaginary unit, j  1 . An example of a complex number (expressed in Cartesian
form) is z = 3+2j. In this example, 3 is referred to as the real part of z, and 2 is referred to
as the imaginary part of z. Complex numbers can also be expressed in polar form, and an
example is w = 3ej2.7. In this example, 3 is referred to as the magnitude of w and 2.7 (in
radians) is referred to as the angle (or argument) of w.
Every complex number has an associated complex conjugate, which is denoted by
placing a super asterisk on its symbol and is formed by replacing each and every j by -j.
E.g. for the z and w used in the previous paragraph, z*=3-2j, w*=3e-j2.7.
Exercises. Create five complex numbers in Cartesian form and five in polar form.
Construct their complex conjugates.
The connection between the Cartesian and Polar forms of complex numbers is through
the Euler identities
e j  cos( )  j sin( )
sin( ) 
e j  e  j
2j
e j  e  j
2
The last two can be derived from the first. To derive the second, simplify e j  (e j )* and
solve for sin( ) . To derive the third, simplify e j  (e j )* and solve for cos( ) .
cos( ) 
Exercises. Derive the second two Euler Identities from the first.
Based on the Euler identities, the relationships between the real, imaginary, magnitude
and angle (or argument) are:
z  Re( z)  j Im( z)  z e jz
Computing polar form from Cartesian form
|z|=(Re(z)2+Im(z)2).5
z =atan2(Re(z),Im(z))=tan-1(Im(z)/Re(z))?+? 
Whether or not you add  when using tan-1(), depends on the quadrant that z is in. If it is
in first or fourth, then you do not need to add  , it is in the second or third you do. This
is because tan-1() produces angles between  and  .
Computing Cartesian form from polar form
Re(z)=|z|cos( z )
Im(z)=|z|sin( z )
Exercises: Begin with Cartesian form, derive polar form expressions. Begin with polar
form, derive Cartesian form expressions.
© P. D. Olivier, 2006
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Complex N, A, & A
A complex number can be viewed as a point in a 2-D plane. The real part is the
coefficient in the horizontal direction, and the imaginary part is the coefficient in the
vertical direction. The magnitude of z, is just the magnitude of the vector, the angle (or
argument) of z is the angle measured counter clockwise from the horizontal axis.
Control engineers need to be proficient at both Complex arithmetic and complex algebra.
The rules are easy, treat j just like a variable, but when you see j raised to a power,
replace it as follows: j2=-1, j3=-j, j4=1, etc.
There should be no more than a single j in the final answer, no j2’s, no j3’s, etc.
The complex conjugate of z, z*, has three very special computational properties:
zz*=|z|2
Re(z)=(z+z*)/2
Im(z)=(z-z*)/(2j)
Exercises: Verify the three identities in the previous paragraph. Begin with z=a+bj, note
that |z|2 =a2 + b2, a = Re(z) and b = Im(z).
Exercises: Given the following complex numbers, use the identities in the previous
paragraph to compute the magnitude, the real part, and the imaginary part.
Arithmetic refers to manipulating numbers, algebra refers to manipulating functions,
symbols, etc. It is about 80% correct to view algebra as arithmetic using symbols instead
of numbers.
Complex arithmetic. As was hinted at by manipulating the Euler identities, complex
numbers can be manipulated via the standard rules of arithmetic (i.e. the associative law,
distributive law, etc.) that are familiar from middle school mathematics courses. When
dealing with complex numbers in Cartesian form, the rules of arithmetic take the form
z  a  jb  Re( z )  j Im( z ), w  c  jd  Re( w)  j Im(w)
z  w  (a  jb)  (c  jd )  (a  c)  j (b  d )
z  w  (a  jb)  (c  jd )  (a  c)  j (b  d )
zw  (a  jb)(c  jd )  ac  jad  jbc  j 2bd  (ac  bd )  j (ad  bc)
z z w* a  jb c  jd (ac  bd )  j (bc  ad )



w w w* c  jd c  jd
c2  d 2
When dealing with complex numbers in polar form, the rules of arithmetic take the form
© P. D. Olivier, 2006
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Complex N, A, & A
z  z e jz  z (cos(z )  j sin(z )),
w  w e jw  w (cos(w)  j sin(w))
z  w   z cos(z )  w cos(w)   j  z sin(z )  w sin(w) 
z  w   z cos(z )  w cos(w)   j  z sin(z )  w sin(w) 
zw  z w e j  z w
z
z
 e j  z w
w w
Exercise 1: Given the following complex numbers in Cartesian form, a) plot the number,
b) convert to polar form, c) make sure the angle is in the correct quadrant.
a) 1 + j
b) 1 – j
c) -1 + j
d) -1 – j
e) 3 + 2j
Exercise 2: Given the following complex numbers in polar form, a) plot the number, b)
convert to Cartesian form
a) 3ej.8
b) 5e-j.5
c) 2.4ej1.8
d) 6.8ej1.1
Exercise 3: Given the complex numbers z= -2.5 + 3.2j, and w =ej1.2, compute the
following, use you calculator only to convert to polar or rectangular form. Half of the
results should be in polar form and half in Cartesian form
a) z+w
b) z2
c) z/w
d) w-z
Complex Algebra. Adopting the point of view that algebra is arithmetic on symbols,
combined with an ability to substitute expressions for some of the symbols, we arrive at
the algebraic skills needed to be successful at control engineering. Consider (the fairly
typical) problem of beginning with the following equations and solving for Y(s) in terms
of R(s) and G(s), i.e. eliminating E(s)
E ( s)  R( s)  Y ( s)
Y ( s)  G ( s) E ( s)
The solution goes as follows
© P. D. Olivier, 2006
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Complex N, A, & A
Y ( s)  G ( s)[ R( s )  Y ( s)]
 G ( s ) R( s )  G ( s )Y ( s)
Y ( s)  G ( s)Y ( s)  G ( s) R( s)
[1  G ( s)]Y ( s)  G ( s) R( s)
G(s)
R( s)
1  G ( s)
Now G(s) is often a ratio of polynomials (i.e. a rational function) such
s
as G ( s )  2
. Substituting this into the expression for Y(s) produces
s  s 1
s


After substitution, there is a
 s 2  s  1 
complicated fraction, i.e. a
Y (s) 
R(s)
s


fraction of fraction.
1  2

s

s

1


Multiply the complicated
s


2
fraction by 1, in the form of
 s 2  s  1  s  s  1
 
R
(
s
)
the denominator over itself.
s

 s2  s  1
1  2
 s  s  1 
Distribute and cancel.
s
s
 2
R( s)  2
R(s)
Combine all like powers of
s  s 1 s
s  2s  1
s.
Several features of the rational function in the final answer are important.
 The rational function includes only a simple fraction (no fractions of fractions)
 The numerator (i.e. s) is a polynomial
 The denominator (i.e. s2+2s+1) is a polynomial
 In both polynomials, the powers of the variable (i.e. s) begin on the left with the
highest power, and proceed to the right with lower powers
 The entire numerator is above the division bar
 The entire denominator is below the division bar
Y ( s) 
In control engineering, we encounter many rational functions of the variable s. The
variable s represents a complex number. Hence you must perform countless similar and
slightly more complicated simplifications similar to the above process. It is imperative
that you become good and quick at such computations. A word of warning, in some cases
the rational function G(s) is defined in terms of an unknown symbolic parameter. In these
cases most (if not all) calculator based algebraic manipulation features fail. Hence it is
good to avoid using the calculator so you can practice on the simple cases and therefore
be prepared for the more complicated ones. After all, your engineering education is not
about how well a calculator perform, it is about how well you can think with the
mathematical and scientific tools you learn.
Exercises:
© P. D. Olivier, 2006
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Complex N, A, & A
4s  1
, compute G(s) at the following values of s. Put the final
s  s 1
answer in both Cartesian and polar form, remember, j can occur only once in the final
answer and it cannot occur in the denominator.
o s=3j, s=0, s=1-2j, s=j  (in this last one, omega must remain as an unspecified
parameter to be determined later)
4) Given G ( s ) 
2
5) Given Y ( s)  G ( s) E ( s), E ( s)  R( s)  H ( s)Y ( s )
o eliminate E(s) and solve for Y(s)/R(s) = T(s)
o In the result just obtained, simplify the result obtained after making the
s 1
, H (s)  s  1
following substitutions G ( s )  2
s  s3
o Evaluate T(s) at the following values of s, putting the answer in both Cartesian
and polar form: s=0, s=-4j, s=-1-j, s=j   (in this last one, omega must
remain as an unspecified parameter to be determined later)
1
. Evaluate at the following values of s, putting the answer in both
 s 1
Cartesian and polar form: (The parameter  must remain as an unspecified parameter to
be determined later.) s=0, s=-4j, s=-1-j, s=j   (in this last one, omega must remain as an
unspecified parameter to be determined later)
6) Given G ( s ) 
n2
. Evaluate at the following values of s, putting the
s 2  2n s  n2
answer in both Cartesian and polar form: (The parameters n and  must remain as
unspecified parameters to be determined later.) s=0, s=-4j, s=-1-j, s=j   (in this last
7 ) Given G( s) 
one, omega must remain as an unspecified parameter to be determined later)
It should be noted that consideration of the functions G ( s ) 
1
and
 s 1
n2
was not by accident. These functions are prototypes of functions
s 2  2n s  n2
that play a very important role in control engineering, the parameters  , n , and  have
G( s) 
very important meanings and provide very useful insight into the design of control
systems.
Extra credit opportunity for students with a D or F: Turn in these exercises worked out
and I will add up to 10 points to the total points earned on the tests, quizzes, final exam,
etc. Due at time of first test.
© P. D. Olivier, 2006
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Complex N, A, & A
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