Download Chapter 4 Complex Numbers

ENGR 1405
48
Chapter 3: Complex Numbers
§ 3.1 Introduction and Definitions
If f (z) = 0 where f (z) = z2 + 2z + 5, then in order to find the roots of this equation it is necessary
to find values for “z” such that z2 + 2z + 5 = 0. This can be accomplished by applying the
quadratic formula or by completing the square. Here, the second approach will be used. Hence,
it is necessary to determine values for “z” such that
z2 + 2z = 5
z2 + 2z +1 = 5 + 1

(z+1)2 = 4

Since we know that the square of any real number must be greater than or equal to zero, it follows
that the roots of the given equation cannot, possibly, be real numbers. It was problems such as
this one that made mathematicians, of an earlier age, realize that the real number system was not
complete. A new number system, the Complex Number System, was developed that:
(a) had the real numbers as a subset, and
(b) allowed the possibility of negative squares.
Complex Number:
The number z = a + jb = a + bj is a complex number iff j2 = 1.
The real number “a” is called the real part of the complex number, and
the real number “b” is called the imaginary part of the complex number.
Since this is a new number system, it is necessary to establish how to work with these numbers.
Hence, the following are properties are defined for any complex numbers z1 = a1 + jb1, and
z2 = a2 + jb2 .
Addition:
The sum of the complex numbers z1 and z2 is the complex number “w” where
w
=
x + jy
=
z1 + z2
= (a1 + jb1) + (a2 + jb2)
= (a1 + a2) + j (b1 + b2)
x
=
a1 + a2

and
y
=
b1 + b2
Multiplication:
The product of the complex numbers z1 and z2 is the complex number “w” where
w
=
x + jy
=
z1  z2
=
(a1 + jb1)  (a2 + jb2)
=
(a1a2  b1b2) + j (a1b2 
a2b1)
x
=

a1a2  b1b2
and
y
=
a1b2  a2b1
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Conjugate:
The complex conjugate of the complex number z1 = a1 + jb1 is the complex number
z2 = a2 + jb2 iff
z1  z2 =
(a1 + jb1)  (a2 + jb2)
= (a1a2  b1b2) + j (a1b2  a2b1)
=
(a1)2 + (b1)2
a2
=
a1

and
b2
=  b1
Hence, if z2 is the complex conjugate of z1 = a1 + jb1, then z2 = a1  jb1 and the complex
conjugate is normally expressed as z2  z1  a1  jb1 . Another notation for the complex
conjugate of z is z*.
Modulus:
The square of the modulus | w | of the complex number w = x + jy is the real number
2
r 2  w  ww  x 2  y 2 .
Division:
The quotient of the complex numbers z1 and z2 with z2  0 is the complex number “w” where
w =
x + jy
=
z1  z2
=
z1 z 2
z2 z2
=
[(a1 + jb1)  (a2  jb2)]  [(a2 + jb2)  (a2  jb2)]
=
[(a1a2  b1b2) + j (a2b1  a1b2)]  [(a2)2  (b2)2]
x
=

[a1a2  b1b2]  [(a2)2  (b2)2]
and
y
=
[a2b1  a1b2]  [(a2)2  (b2)2]
Alternate Representations:
Any complex number w = x + jy can also be expressed in two alternate forms.
Polar form: w = x + jy = r [cos( ) + j sin( )]
Euler form: w = x + jy = r ej
Here r  x 2  y 2 is the modulus of the complex number
and  = arg(w) is called the argument of the complex number where
if x > 0
 = Tan-1(y/x)
-1
or  = Tan (y/x)   if x < 0 and y  0
or  = Tan-1(y/x) +  if x < 0 and y > 0
or  =  / 2
if x = 0 and y < 0
or  = + / 2
if x = 0 and y > 0
[The argument of zero is undefined.]
If   arg(w) < +, then  = arg(w) is the principal argument of w.
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Argand Diagrams:
Complex numbers can be represented as points on a grid that appears identical to the Cartesian
co-ordinate grid (the xy-plane) except that the horizontal axis is called the real axis, and the
vertical axis is interpreted as the imaginary axis.
The axes are labeled respectively as Re{z} and Im{z}.
From the Argand diagram one can see that the argument of a complex number is not unique.
If  = arg(w) is the principal argument of w, then  + 2n (for all integers n) are also
arguments of w, [because cos( + 2n )  cos  and sin( + 2n )  sin  for all integers n].
Sample Problems
Sample Problem 1:
Determine the value of z = (2 + j5) + (7  6j)

 2  j5  7  6 j 
 2  7  j 5  6

5 j
z 
Sample Problem 2:
Determine the value of z = 7 (3 + j4)  (2 + j) (5  6j)
z 
7 3  j 4   2  j 5  6 j 

21  j 28  10  j5  12 j  6 j 2 
21  j 28  10  6  j 5  12 
21  j 28  16  j  7 
21  16  j(28  7)

5  j 35



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Sample Problem 3:
ENGR 1405

7
1  j2

3 j 3 4 j
1  j 23  4 j 
73  j 

3  j 3  j  3  4 j 3  4 j 
21  7 j   3  j6  4 j  8 j 2 
9  j3  3 j  j 2  9  12 j  j12  16 j 2 
21  7 j   3  8  j10
9  1
9  16
2.1  j(0.7)   0.2  j(0.4) 

2.3  1.1 j
z 



Sample Problem 4:
7
1  j2

3 j 3 4 j
Determine the value of z 
1  j 3
1  j 3 
1  j 3 1  j 3  1  j 3 1  j 3 
1  j 3  j 3  3 j 1  j 3  j 3  3 j 
 2  j2 3   2  j2 3 
4  4 3 j  4 3 j  12 j 
4  12  j 4 3  4 3 
 8  j8 3 
4
Evaluate z 
z 






4
2
2
2
Alternative solution:
The coefficients in the binomial expansion of (a + b)4 are the entries in the n = 4 row of
Pascal’s triangle: 1 4 6 4 1 .
Also note that j 2 = 1, j 3 = j, j 4 = 1 and j(n+4) = jn for all n.
Then

z  1 j 3

4
 
1
 
2
 1  j4 3  6  3  j4  3 3  1  3  3
 1  18  9   j 4  12  3

 
3
 
 1 4 j 3  6 j 3  4 j 3 1 j 3
  8 1 j 3

4
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§ 3.2 de Moivre’s Theorem
The last sample problem of the previous section was rather long and tedious. It would be nice if
some method existed by which the amount of work needed for this and other similar problems
could be reduced. It turns out that it is indeed possible by using some elementary real number
properties.
The tools needed are:
(a) the Pythagorean theorem
(b) elementary geometry
(c) some basic series expansions already studied
(d) the Binomial theorem for arbitrary real exponents
Given any complex number z = x + jy, it is possible to represent it on an Argand diagram
Re z
x
By means of the Pythagorean theorem and elementary geometry, we know that
r z

x

y
 z 

x2  y2
r cos 
r sin  
x  jy
 r cos   j r sin 


r cos   j sin  
In our study of series it was found that the Maclaurin series expansion for f (x) = e x was

xn
ex   n 0
 n !
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ENGR 1405
If we set x = j , and use the fact that j n+4 = j n for n  1 and j 2 = 1, j 3 = j, j 4 = 1
n
  j 
e j   n 0
n !
 j    j    j   
j  j 


1! 2 ! 3! 4! 5!
j  2
j 3  4
j 5
1





1! 2! 3! 4 ! 5!
2
 1



then

 2 k
k 0  1 2k !

k
3

4
5

 2 k 1
j k 0  1
2k  1!

k
cos   j sin(  )
Hence, by combining the three representations for complex numbers, the following is easily
shown to be true
x  j y  p
zp 
p
 r p cos   j sin  
p

r p e j  

r p e j  p 
r p cos p   j sin  p 
This result is called de Moivre’s theorem, and is used primarily for the following situations
(a) determining integer powers of complex numbers,
(b) finding rational powers of complex numbers (especially roots)
(c) establishing the various trigonometric identities.
When using de Moivre’s theorem to find a non-integer power of a complex number z, express z
in its polar or Euler form using the complete set of arguments ( + 2n ).
Sample Problem 5:
Find all three distinct cube roots of 1.
We need to solve w3 = z for w, where z = 1.
| z | = | 1 + 0j | = 1 and arg(z) = arg(1 + 0j) = 0 (principal value).
In polar form, w3 = z = 1 cos(0 + 2n ) + j 1 sin(0 + 2n ), where n = any integer.
By de Moivre’s theorem,
w = [1 cos(2n ) + j 1 sin(2n )](1/3) = 1(1/3) [cos(2n / 3) + j sin(2n / 3)]
= cos(2n / 3) + j sin(2n / 3)
This generates only three distinct complex numbers, which can be evaluated by setting n to any
three consecutive values:
n = 1  w1 = (1  j3) / 2 ,
n = 0  w1 = 1 ,
n = +1  w1 = (1 + j3) / 2 .
Further examples of each of these applications will be done in class.