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Chap 1. Complex Numbers.
1. Sums and Products.
Complex numbers can be defined as ordered pairs (x,y) of real numbers that
are to be interpreted as points in the complex plane, with rectangular
coordinates x and y, just as real numbers x are thought of as points on the
real line.
• (x,0)
real number
• (0, y) pure imaginary number
• It is customary to denote a complex number (x,y) by z , so that
z   x, y 
x: real part of z
Re z = x
y: imaginary part of z
Im z = y
1
• Sum and
product of z1   x1 , y1  , z2   x2 , y2  are defined as
z1  z2   x1 , y1    x2 , y2    x1  x2 , y1  y2 
z1  z2   x1 , y1    x2 , y2    x1 x2  y1 y2 , y1 x2  x1 y2 
z   x, y    x, 0    0, y 
 0,1 y, 0    0, y 
z   x, 0    0,1 y, 0 
let i   0,1
but

Then z  x  iy
i 2  (0,1)(0,1)  1
2
2. Algebraic Properties
z1  z2  z2  z1
commutative law
z1 z2  z2 z1
 z1  z2   z3  z1   z2  z3 
 z1 z2  z3  z1  z2 z3 
z  z1  z2   zz1  zz2
associative law
distributive law
z+0=z
0=(0,0)
additive identity
z 1=z
1=(1,0)
multiplicative identity
For each z, there is a -z
such that z+(-z)=0
additive Inverse
3
For any nonzero z=(x,y),
multiplicative Inverse
There is a z 1 such that zz 1  1
less obvious than additive inverse
 x, y  u, v   1, 0 
 xu  yv  1
x

u


x2  y 2
 yu  xv  0
v
y
x2  y 2
 x
y 
 z 1   2
,
, z0
2
2
2 
x

y
x

y


• Division by a non-zero complex number
z1
1
 z1 z2
z2
 z2  0 
if z1   x1 , y1  , z2   x2 , y2 
xx  y y xx x y 
z1
 z1 z21   1 22 1 2 2 , 1 22 1 2 2 
z2
x2  y2 
 x2  y2
 x1 x2  y1 y2   x1 x2  x1 y2 
 2
i 2
2 
2 
 x2  y2   x2  y2 
z1  x1  iy1  x2  iy2 

z2  x2  iy2  x2  iy2 
得到相同結果
4
Other Identities
 z1 z2   z1 z2
1
1
1
z1 z2  z1  z2 
   
z3 z4  z3   z4 
z1  0, z2  0
z3  0, z4  0
Example.
1
 1  1 




 2  3i  1  i   2  3i 1  i 
1
5i
5i



5  i  5  i  5  i  26

5
i
5
1


 i
26 26 26 26
5
3. Moduli and conjugates
It is natural to associate any nonzero complex number z=x+iy with the directed
line segment, or vector, from the origin to the point (x,y) that represent z in
complex plane.
In fact, we often refer to z as the point z or the vector z.
y
z1  z2   x1  x2 , y1  y2 
(x,y)
(-2,1) -2+i
x+iy
z2
Z1
Z1+Z2
Z2
z1
0
x
-Z2
Z1-Z2
-Z2
z1  z2  z1    z2 
Although the product of two complex number z1 and z 2 is itself a complex number
represented by a vector, that vector lies in the same plane as the vectors for z1 and z 2 .
This product is neither the scalar nor the vector product used in vector analysis.
6
•The modulus, or absolute value, of a complex number z=x+iy is defined as
z  x2  y 2
length of the vector z.
distance between point z and 0
the distance between two points z1  x1  iy1
z2  x2  iy2
is z  z 
1
2
z  z0  R
•
 x1  x2    y1  y2 
2
a circle
z   Re z    Im z 
2
Z0 R
2
2
2
 Re z  Re z  z
Im z  Im z  z
z
x+iy
• complex conjugate of z =x+iy is
z  x  iy
z
x-iy
7
• If
z1  x1  iy1 , z2  x2  iy2
z1  z2   x1  x2   i  y1  y2    x1  iy1    x2  iy2 
 z1  z2
z1  z2  z1  z2
z1 z2  z1 z2
 z1  z1
   , z2  0
 z2  z2
z  z  2 Re z
Re z 
zz
2
zz  z
z  z  2i Im z
Im z 
zz
2i
2
z1 z2  z1 z2
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4. Triangle Inequality
z1  z2  z1  z2
Equality holds when z1, z2 ,0 are colinear.
z1  z2
z2
z1
geometrically
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algebraically,



z1  z2   z1  z2  z1  z2   z1  z2  z1  z2
2



 z1 z1  z1 z2  z1 z2  z2 z2


 z1  2 Re z1 z2  z2
2
 z1  2 z1 z2  z2
2
2
 z1  2 z1 z2  z2
2
  z1  z2

2
2
2
 z1  z2  z1  z2
Now
z1   z1  z2     z2   z1  z2   z2  z1  z2  z2
 z1  z2  z1  z2
similarly
z1  z2  z2  z1
 z1  z2  z1  z2
10
When z2 is replaced by  z2,
z1  z2  z1  z2
z1  z2  z1  z2
Example:
z on unit circle z  1
or
z  0 1
z3  2  z3  2  z  2  3
3
2
z3  2  z  2  1
3
The triangle inequality can be generalized by mathematical induction to sums …
z1  z2  ...  zn  z1  z2  ...  zn
n  2,3,...
11
5. Polar coordinates and Euler’s Formula
Let r, and  be polar coordinates of the point (x,y) that corresponds to a non-zero
complex number z=x+iy.
since
x  r cos  ,
y  r sin  ,
z  r  cos   i sin  
if z=0, the coordinate  is undefined.
r  z the length of the radius vector for z.

has an

possible values.
Each value of

is called an argument of z.
and the set of all such values is arg z
The principal value of arg z, Arg z, is that unique
arg z  Arg z  2n
using Euler’s formula
then
,
s.t.
    
 n  0, 1, 2,...
ei  cos   i sin 
z  rei
12
• Two non-zero complex numbers
z1  r1ei1
are equal iff
z2  r2ei2
r1  r2 and 1  2  2n ,
z  Rei  z  R
n  0, 1, 2,...
z  z0  Rei  z  z0  R
 0    2 
6. Product and Quotients in Exponential From
ei1 ei2   cos 1  i sin 1  cos  2  i sin  2 

i
 cos 1   2   i sin 1   2   e
If

i 1 2 
z1  r1ei1 , z2  r2ei2
z1 z2  r1r2 e  1
i   2 
(1)
13
z1 r1 ei1 e i2 r1 i1 2 

 e
i 2  i 2
z2 r2 e e
r2
1 1 i
z   e
z r
z n  r n ein
1
if
e 
i
r  1,
 n  0, 1, 2,...
n
 ein
 cos  i sin  

Ex. Find

3 i
 64


7
3 i
n
 n  1, 2,...
 cos n  i sin n
Moivre’s formula

7

7



i 7
  2e 6   27 e 6  26 ei  2ei 6


i
3i

14
Z1Z2
argument of product
arg  z1 z2   arg z1  arg z2
Z2
(7)
1   2
Z1
1
If we know two of these, can find the third.
A. If arg z1  1
arg z2   2
B. If
1   2 From Expression (1) is a value of arg  z1 z2 
arg  z1 z2   1  2   2n
arg z1  1  2n1
If we choose
arg z2  2  2  n  n1  
(7) is satisfied.
15
C. Similarly for
Z1Z2
arg  z1 z2   1  2   2n
Z2
1   2
Z1
arg z2  2  2n2
1
Then choose arg z1  1  2  n  n2  
Finally
Ex:
arg
   arg z  arg z
Z1
Z2
1
Arg
2
arg
z1  i
z2  1
z1 z2  i
16
7. Roots of Complex Numbers
Suppose z is nth root of a nonzero number z0 .
z n  z0
or
r n e jn  r0 ei0
n   0  2k
r n  r0 and
r  n r0

k  0, 1, 2,...
 0  2k
n
0
2k
k  0, 1, 2,...
n
n
   2 k  
 z  n r0 exp i  0 
k  0, 1, 2,...

n 
 n

are the nth root of

z0
These roots are on the circle z  n r0
and are equally spaced every
2
n
17
All of the distinct roots are obtained when
k = 0,1,2,…,n-1
Let Ck  n r0 exp i  0  2k  


 n
and z0
1
n
n 
k  0,1, 2,..., n  1
denote the set of nth roots of
denote these distinct roots
z0
z0is a positive real number r0 then r0 1n denotes the entire set of roots.
(іі) if  0in (1) is the principal value of arg z0
(і) if
   0   
  
C0  n r0 exp i 0 
 n
is referred to as the principal root.
18
Ex. nth roots of unity
1  1exp i  0  2k  
k  0, 1, 2,..., n  1
1
  0 2k  
1 n  n 1 exp i  

n
n



 2k 
 exp  i
k  0,1, 2,..., n  1

 n 
n  2 : 1
n3 :
1
n=4 ：
1
19
 2 
Let n  exp  i

 n 
 2 k 
k
then n  exp  i

 n 
and 1 n  1, n , n ,..., n
1
Ex.2. Find  8i 
2
1
n 1
3
  

8i  8  i   8exp i    2k  

  2
   2k
Ck  2 exp i   
3
  6



k  0,1, 2
   
C0  2 exp i      3  i
  6 
C1  2i  C03
C2   3  i  C03
k  0, 1, 2,...
c2
c1
2
c0
2
20

• Regions in the complex Plane
closeness of points to one another
•
 -neighborhood or neighborhood
z  z0  
of a given point
z0
• Deleted neighborhood
Z

Z0
0  z  z0  
• Interior point
z0 is said to be an interior point of a set S whenever there is
some neighborhood of z0 that contains only points of S.
A point
Z0
S
• Exterior point
when there exists a neighborhood of z0 containing no points of S
21
• Boundary point
all of whose neighborhoods contain points is S and points not in S
Boundary = { all boundary points }
Ex. z  1 is the boundary of
z 1
and
z 1
• A set is open if it contains none of its boundary points
• A set is closed if it contains all of its boundary points.
• The closure of a set S is the closed set consisting of all points in S together with
the boundary of S
z  1 is open
z  1 is closed and closure of
-
z  1,
z 1
0  z  1 neither open nor closed.
22
- The set of all complex number is both open and closed since it has no
boundary points.
• An open set S is connected if each pair of points
joined by a polygonal line that lies entirely in S.
z1 and z2 in it can be
1 z  2
open, connected
Z1
Z2
1
2
• An open set that is connected is called a domain. (any neighborhood is a
domain )
• A domain together with some, none, or all of its boundary points is a region.
• A set S is bounded if every point of S lies inside some circle Z  R;
otherwise it is unbounded.
23
• A point z0 is said to be an accumulation point of a set S if each deleted
neighborhood of z0 contains at least one point of S.
- If a set S is closed, then it contain s each of its accumulation points.
pf: If an accumulation point z0 were not in S, it would be a boundary
point of S;
(can not be exterior points)
but this contradicts the fact that a closed set contains all of its boundary
points.
i
z

 n  1, 2,...
Ex: For the set n
n
the origin is the only accumulation point.
24