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ComplexNumbersReview.nb
Review
1.1 Complex Numbers
Formally the quadratic equation
1  x2  0
(1.1)
has solutions  1 . However there is no real number whose square is 1. Such roots were
initially regarded as "impossible" or "imaginary". However in the sixteenth and seventeenth
centuries, mathematicians gradually found that algebra could be applied to these imaginary
numbers and that the introduction of these numbers could simply some algebraic calculations.
It was Euler who introduced the symbol  to represent 1 .
A general complex number is composed of the sum of a real number and an imaginary number,
z  ab
(1.2)
where a is the real part of z, a  Re#z' and b is the imaginary part of z, b  Im#z'.
Complex numbers obey the same rules as normal numbers. To add two complex numbers, one
adds the real parts and the imaginary parts separately.
z1
 a1   b1 and z2  a2   b2
z1  z2  +a1  a2 /   +b1  b2 /
(1.3)
In multiplication, one distributes the multiplication over all terms. For a real constant, c, times
a complex number, one has
z
 ab
c z  +a   b/ c  a c   b c
(1.4)
Similarly, for an imaginary constant,  d, times a complex number
z
 ab
 d z  +a   b/ d   a d  2 b d   a d  b d
(1.5)
In general, the multiplication of two complex numbers becomes
z1
 a1   b1 and z2  a2   b2
z1 z2  +a1   b1 / +a2   b2 /
z1 z2  a1 a2   b1 a2   a1 b2  2 b1 b2  ,a1 a2  b1 b2 0   +a2 b1  a1 b2 /
(1.6)
The Euler equation states that for a real angle 
   Cos#'   Sin#'
(1.7)
The Euler formula also has a wonderful geometric interpretation. Any complex number
z  a   b can be plotted as a point on a two dimensional graph in which the horizontal axis (x) is
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defined as the real axis and the vertical axis (y) is defined as the imaginary axis. The real part
of z is the projection of the z on the real axis and the imaginary part of z is the projection of z on
the imaginary axis.
Im
b
b  Im#z'
 Abs#z' Sin#'
z  43
a  Re#z'
 Abs#z' Cos#'
  0.64
a
Re
The complex number, z  a   b, can be rewritten in terms of an exponential, z   which has a
modulus or absolute value, z , and a phase, . From the graph, the modulus is just the length
of the dashed line is
a2  b2 . Expanding the exponential form
z  a   b  Abs#z'    Abs#z' +Cos#'   Sin#'/
(1.8)
From this definition, we obtain
Re#z'  a  Abs#z' Cos#' and Im#z'  b  Abs#z' Sin#'
or
Abs#z'2 ,Cos#'2  Sin#'2 0  a2  b2
(1.9)
(1.10)
which is equivalent to our geometric intuition
(1.11)
a2  b2
Abs#z' 
Taking the ratio of the imaginary and real components of z, we obtain a definition for the angle,
,
Im#z'
Re#z'

b
a

Abs#z' Sin#'
Abs#z' Cos#'
 Tan#'
(1.12)
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Because the Cosine and Sine are periodic functions, there are many angles that satisfy Eq. 1.12
´ 4   , 2   , , 2   , 4    ´ or   2 n 
(1.13)
By convention, one normally picks that value between 0 and 2  and call it the principal value of
arg z.
From Euler's equation, one can express the Sine and Cosine in exponential form.
Cos#' 
Sin#' 
    
2
(1.14)
    
2
Euler's identity can also be used to prove trigonometric identities. Consider
Cos#A  B'  Cos#A' Cos#B'  Sin#A' Sin#B'
(1.15)
Cos#A  B'  Re$ +AB/ (
(1.16)
Remembering that Cos#x'  Re$ x (
Now expanding the exponential in terms of Cosine and Sine
Cos#A  B'  Re#+Cos#A'   Sin#A'/ +Cos#B'   Sin#B'/'
 Re#Cos#A' Cos#B'   Cos#B' Sin#A'   Cos#A' Sin#B'  Sin#A' Sin#B(1.17)
 Cos#A' Cos#B'  Sin#A' Sin#B'
Note that because Sin#x'  Im$ x (, we have immediately from Eq. 1.17
Sin#A  B'  Im$ +AB/ (
 Im#Cos#A' Cos#B'   Cos#B' Sin#A'   Cos#A' Sin#B'  Sin#A' Sin#B(1.18)
 Cos#B' Sin#A'  Cos#A' Sin#B'
In differentiating and differentiating, we simply treat  as a constant
d  x
dx
d2  x
d x2
   x
(1.19)
2 x
  
x
 
The conjugate of z denoted by z is formed by changing the sign of the imaginary part of z
z  ab  ab
Note that in exponential form
(1.20)
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z    Abs#z'  Abs#z' Cos#'   Sin#'
 Abs#z' +Cos#'   Sin#'/    Abs#z'
(1.21)
Therefore
z z  Abs#z'   Abs#z'    Abs#z'2
(1.22)
Abs#z' 
(1.23)
and so
zz
and
z
1
z
(1.24)
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