Post on 15-Dec-2015
Chapter 1. Complex Numbers
Weiqi Luo (骆伟祺 )School of Software
Sun Yat-Sen UniversityEmail : weiqi.luo@yahoo.com Office : # A313
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Textbook: James Ward Brown, Ruel V. Churchill, Complex Variables and
Applications (the 8th ed.), China Machine Press, 2008
Reference: 王忠仁 张静《 工程数学 - 复变函数与积分变换》高等教育出版社, 2006
2
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Numbers System
3
Refer to: http://en.wikipedia.org/wiki/Number_system
Natural Numbers Zero & Negative Numbers
Integers Fraction
Rational numbers Irrational numbers
Real numbers Imaginary numbers
Complex numbers
… More advanced number systems
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Sums and Products; Basic Algebraic Properties Further Properties; Vectors and Moduli Complex Conjugates; Exponential Form Products and Powers in Exponential Form Arguments of Products and Quotients Roots of Complex Numbers Regions in the Complex Plane
4
Chapter 1: Complex Numbers
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Definition Complex numbers can be defined as ordered pairs (x,
y) of real numbers that are to be interpreted as points in the complex plane
5
1. Sums and Products
Complex plane
(x, y)
(x, 0)
(0, y)
Real axisimaginary axis
Note: The set of complex numbers Includes the real numbers as a subset
x
y
O
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Notation It is customary to denote a complex number (x,y) by z,
6
1. Sums and Products
x
yz=(x, y)
(x, 0)
(0, y)
x = Rez (Real part); y = Imz (Imaginary part)
z1=z2 iff
1. Rez1= Rez2
2. Imz1 = Imz2
OQ: z1<z2?
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Two Basic Operations Sum
(x1, y1) + (x2, y2) = (x1+x2, y1+y2)
Product
(x1, y1) (x2, y2) = (x1x2 - y1y2, y1x2+x1y2)
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1. Sums and Products
1. when y1=0, y2=0, the above operations reduce to the usual operations of addition and multiplication for real numbers.
2. Any complex number z= (x,y) can be written z = (x,0) + (0,y)
3. Let i be the pure imaginary number (0,1), then z = x (1, 0) + y (0,1) = x + i y, x & y are real numbers
i2 =(0,1) (0,1) =(-1, 0) i2=-1
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Two Basic Operations (i2 -1) Sum
(x1, y1) + (x2, y2) = (x1+x2, y1+y2)
(x1 + iy1) + (x2+ iy2) = (x1+x2)+i(y1+y2)
Product
(x1, y1) (x2, y2) = (x1x2 - y1y2, y1x2+x1y2)
(x1 + iy1) (x2+ iy2) = (x1x2+ x1 iy2) + (iy1x2 + i2 y1y2)
= (x1x2+ x1 iy2) + (iy1x2 - y1y2)
= (x1x2 - y1y2) +i(y1x2+x1y2)8
1. Sums and Products
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Various properties of addition and multiplication of complex numbers are the same as for real numbers
Commutative Laws
z1+ z2= z2 +z1, z1z2=z2z1
Associative Laws
(z1+ z2 )+ z3 = z1+ (z2+z3)
(z1z2) z3 =z1 (z2z3)
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2. Basic Algebraic Properties
e.g. Prove that z1z2=z2z1
(x1, y1) (x2, y2) = (x1x2 - y1y2, y1x2+x1y2) = (x2x1 - y2y1, y2x1 +x2y1) = (x2, y2) (x1, y1)
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For any complex number z(x,y) z + 0 = z; z ∙ 0 = 0; z ∙ 1 = z Additive Inverse
-z = 0 – z = (-x, -y) (-x, -y) + (x, y) =(0,0)=0 Multiplicative Inverse
when z ≠ 0 , there is a number z-1 (u,v) such that
z z-1 =1 , then
(x,y) (u,v) =(1,0) xu-yv=1, yu+xv=0
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2. Basic Algebraic Properties
2 2 2 2,
x yu v
x y x y
1
2 2 2 2( , ), 0
x yz z
x y x y
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pp. 5
Ex. 1, Ex.4, Ex. 8, Ex. 9
Homework
11
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If z1z2=0, then so is at least one of the factors z1 and z2
3. Further Properties
12
Proof: Suppose that z1 ≠ 0, then z1-1 exists
z1-1 (z1z2)=( z1
-1 z1) z2 =1 z2 = z2
Associative Laws
z1-1 (z1z2)=z1
-1 0 =0
Therefore we have z2=0
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Other two operations: Subtraction and Division
Subtraction: z1-z2=z1+(-z2)
(x1, y1) - (x2, y2) = (x1, y1)+(-x2, -y2) = (x1 -x2, y1-y2)
Division:
3. Further Properties
13
111 2 2
2
( 0)z
z z zz
1 2 2 1 2 1 2 1 2 1 21 1 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2
( , )( , ) ( , )+ + + +
z x y x x y y y x x yx y
z x y x y x y x y
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An easy way to remember to computer z1/z2
3. Further Properties
14
1 1 1 1 1 2 2
2 2 2 2 2 2 2
( ) ( )( )
( ) ( )( )
z x iy x iy x iy
z x iy x iy x iy
Note that 2 2
2 2 2 2 2 2( )( )x iy x iy x y R
commonly used
For instance
4 (4 )(2 3 ) 5 14 5 14
2 3 (2 3 )(2 3 ) 13 13 13
i i i ii
i i i
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3. Further Properties
15
Binomial Formula
1 2 1 20
( ) , 1, 2,...n
n k k n kn
k
z z C z z n
Where
!, 0,1,2,...,
!( )!kn
nC k n
k n k
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pp.8
Ex. 1. Ex. 2, Ex. 3, Ex. 6
3. Further Properties
16
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Any complex number is associated a vector from the origin to the point (x, y)
4. Vectors and Moduli
17
x x
y
O
z1
z2
z1+z2
Sum of two vectors
1 1 12 2| |z x y
y
z1=(x1, y1)
O
1 2 1 2 1 2( ) ( )z z x x i y y
2 1 || | |z zz2=(x2, y2)
The moduli or absolute value of z is a nonnegative real number
2 2| |z x y
Product: refer to pp.21
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Example 1 The distance between two point z1(x1, y1) and z2(x2, y2)
is |z1-z2|.
4. Vectors and Moduli
18
x
y
O
z2z1
|z1 - z2 |
-z2
z1 - z2
Note: |z1 - z2 | is the length of the vectorrepresenting the number z1-z2 = z1 + (-z2)
2 21 2 1 2 1 2| z z | ( ) ( )x x y y
1 2 1 2 1 2z z ( ) ( )x x i y y
Therefore
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Example 2 The equation |z-1+3i|=2 represents the circle whose
center is z0 = (1, -3) and whose radius is R=2
4. Vectors and Moduli
19
x
y
O z0(1, -3)
Note: | z-1+3i | = | z-(1-3i) | = 2
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Some important inequations Since we have
Triangle inequality
4. Vectors and Moduli
20
Re | Re | | |; Im | Im | | |Z Z Z Z Z Z
2 2 2| Re | | Im | | |Z Z Z
x
y
z1=(x, y)
O
12 2| |z x y
x
y
O
z1
z2
z1+z2
1 2 1 2| | | | | |z z z z
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4. Vectors and Moduli
21
1 1 2 2| | | ( ) ( ) |z z z z
1 2 1 2| | || | | ||z z z z
Proof: when |z1| ≥ |z2|, we write
1 2 2| | | ( ) |z z z 1 2 2| | | |z z z
1 2 1 2 1 2| | | | | | || | | ||z z z z z z
Similarly when |z2| ≥ |z1|, we write
2 1 2 1| | | ( ) ( ) |z z z z 1 2 1| | | ( ) |z z z 1 2 1| | | |z z z
1 2 2 1 1 2| | | | | | || | | ||z z z z z z
Triangle inequality
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4. Vectors and Moduli
22
1 2 1 2| | | | | |z z z z
1 2 1 2| | || | | ||z z z z
1 2 1 2 1 2|| | | || | | | | | |z z z z z z
1 2 1 2| ... | | | | | ... | |n nz z z z z z
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Example 3 If a point z lies on the unit circle |z|=1 about the origin,
then we have
4. Vectors and Moduli
23
| 2 |z 2
| 2 |z
x
y
O
z
1
|| | 2 | 1z
| | 2 3z
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pp. 12
Ex. 2, Ex. 4, Ex. 5
4. Homework
24
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Complex Conjugate (conjugate)
The complex conjugate or simply the conjugate, of a complex number z=x+iy is defined as the complex number x-iy and is denoted by z
5. Complex Conjugates
25
x
y
Oz z
| | | |z zz(x,y)
z (x,-y)
Properties:
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If z1=x1+iy1 and z2=x2+iy2 , then
Similarly, we have
5. Complex Conjugates
26
1 2 1 2 1 2 1 1 2 2 1 2( ) ( ) ( ) ( )z z x x i y y x iy x iy z z
1 2 1 2z z z z
1 2 1 2z z z z
1 12
2 2
, 0z z
zz z
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If , then
5. Complex Conjugates
27
,z x iy z x iy
( ) ( ) 2 2Rez z x iy x iy x z
( ) ( ) 2 2 Imz z x iy x iy yi i z
Re , Im2 2
z z z zz z
i
2 2 2( ) ( ) | |zz x iy x iy x y Z
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Example 1
5. Complex Conjugates
28
1 3?
2
i
i
1 3 ( 1 3 )(2 )
2 (2 )(2 )
i i i
i i i
2
5 5
| 2 |
i
i
5 51
5
ii
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Example 2
5. Complex Conjugates
29
1 1
2 2
| |. | |
| |
z za
z z
221 1 1 1 1 1 1 1
22 2 2 2 22 2 2
| |:| | ( )
| |
z z z z z z z zproof
z z z z zz z z
| | | |n nz z1 2 1 2| | | || |z z z z
Refer to pp. 14
3 2| 3 2 1|z z z
| | 2z
3 2| | 3 | | 2 | | 1z z z 253 2| | | 3 | | 2 | |1|z z z
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pp. 14 – 16
Ex. 1, Ex. 2, Ex. 7, Ex. 14
5. Homework
30
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Polar Form Let r and θ be polar coordinates of the point (x,y) that
corresponds to a nonzero complex number z=x+iy, since x=rcosθ and y=rsinθ, the number z can be written in polar form as z=r(cosθ + isinθ), where r>0
6. Exponential Form
31
y
xO
z(x,y)
θr
argz: the argument of zArgz: the principal value of argz
x
y
O
z(x,y)
1
rθ
θ Θ
arg 2 , 0, 1, 2,...z ArgZ n n
ArgZ
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Example 1 The complex number -1-i, which lies in the third quadrant
has principal argument -3π/4. That is
It must be emphasized that the principal argument must be in the region of (-π, +π ]. Therefore,
However,
6. Exponential Form
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3( 1 )
4Arg i
5( 1 )
4Arg i
3arg( 1 ) 2 , 0, 1, 2,...
4i n n
5arg( 1 ) 2 , 0, 1, 2,...
4i n n
argz = α + 2nπ
Here: α can be any one of arguments of z
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The symbol eiθ , or exp(iθ)
6. Exponential Form
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cos sinie i Why? Refer to Sec. 29
1 2 3
0
2 2 1
0 0
1 1 1 1 11 ... ...
1! 2! 3! ! !
1 1
(2 )! (2 1)!
x n n
n
n n
n n
e x x x x xn n
x xn n
2 2 1 2 2 2 2 2 1
0 1 0 1
2 1 2 1
0 1
1 1 1 1( ) ( ) ( ) [ ( ) ]
(2 )! (2 1)! (2 )! (2 1)!
1 1( 1) ( ) [ ( 1) ( ) ]
(2 )! (2 1)!
i n n n n n n
n n n n
n n n n
n n
e i i i i in n n n
in n
cosθ sinθ
Let x=iθ, then we have
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Example 2 The number -1-i in Example 1 has exponential form
6. Exponential Form
34
3( )
43 3
1 2(cos( ) sin( )) 24 4
ii i e
3( 2 )
43 3
1 2(cos( ) sin( )) 2 , 0, 1, 2,...4 4
i ni i e n
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z=Reiθ where 0≤ θ ≤2 π
6. Exponential Form
35
x
y
O
Reiθ
Rθ
x
y
θ
O
z0z
z=z0 +Reiθ
Reiθ
|z-z0 |=R
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Product in exponential form
7. Products and Powers in Exponential Form
36
1 2
1 2
1 1 2 2
1 2 1 2 1 2 1 2
( )1 2 1 2
(cos sin )(cos sin )
(cos cos sin sin ) (sin cos cos sin )
cos( ) sin( )
i i
i
e e i i
i
i e
1 2 1 2( )1 2 1 2 1 2( )( )i i iz z re r e r r e
1
1 2
2
( )1 1 12
2 2 2
, 0i
ii
z re re z
z r e r
2
2
0
22 2 2
1 1 1, 0
ii
i
ee z
z r e r
1 11 1 1( ) ( ) , 0, 1, 2,...i inn n nz re r e n
1 21 1 2 2&i iz re z r e
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Example 1 In order to put in rectangular form, one need
only write
7. Products and Powers in Exponential Form
37
7( 3 )i
7 /6 7 7 7 /6 7 7 7( 3 ) (2 ) 2 2 (cos sin ) 64( 3+i)
6 6i ii e e i
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Example 2
7. Products and Powers in Exponential Form
38
de Moivre’s formula
( ) (cos sin ) cos sin , 0, 1, 2,...i n ne i n i n n
2(cos sin ) cos 2 sin 2i i
2 2 2(cos sin ) cos sin (2sin cos )i i
pp. 23, Exercise 10, 11
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8. Arguments of products and quotients
39
1 2 1 2( )1 2 1 2 1 2( )( )i i iz z re r e r r e
argz1z2= θ1 +θ2 +2(n1+n2)π = (θ1 +2n1π)+ (θ2 +2n2π) = argz1+argz2
arg(z1z2)= θ1 +θ2 +2nπ, n=0, ±1, ±2 …
θ1 is one of arguments of z1 and θ2 is one of arguments of z2 then θ1 +θ2 is one of arguments of z1z2
Here: n1 and n2 are two integers with n1+n2=n
Q: Argz1z2 = Argz1+Argz2?
1 21 1 2 2& ,i iIfz re z r e then
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Example 1
When z1=-1 and z2=i, then
Arg(z1z2)=Arg(-i) = -π/2
but
Arg(z1)+Arg(z2)=π+π/2=3π/2
8. Arguments of products and quotients
40
≠Note: Argz1z2=Argz1+Argz2 is not always true.
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Arguments of Quotients
8. Arguments of products and quotients
41
1 111 2 1 2
2
arg( ) arg( ) arg( ) arg( )z
z z z zz
1 2arg( ) arg( )z z
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Example 2 In order to find the principal argument Arg z when
observe that
since
8. Arguments of products and quotients
42
2
1 3z
i
arg arg( 2) arg(1 3 )z i
2
3
( 2)Arg
2( ) 2 2
3 3argz n n
(1 3 )3
Arg i
Argz
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pp. 22-24
Ex. 1, Ex. 6, Ex. 8, Ex. 10
8. Homework
43
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Two equal complex numbers
9. Roots of Complex Numbers
44
11 1
iz re 22 2
iz r e
1 2z z
If and only if
1 2 1 2& 2r r k
for some integer k
At the same point
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Roots of Complex Number Given a complex number , we try to find all
the number z, s.t.
Let then
thus we get
9. Roots of Complex Numbers
45
00 0
iz r e
0nz z
iz re 00( ) in i n n inz re r e r e
0 0& 2 , 0, 1, 2,...nr r n k k
00
2& , 0, 1, 2,...n
kr r k
n n
The unique positive nth root of r0
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The nth roots of z0 are
9. Roots of Complex Numbers
46
00
2exp[ ( )], 0, 1, 2,...n
kz r i k
n n
00
2exp[ ( )], 0,1,2,..., 1n
k
kc r i k n
n n
Note: 1.All roots lie on the circle |z|;2.There are exactly n distinct roots!
|z|
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9. Roots of Complex Numbers
47
00
2exp[ ( )], 0,1, 2,..., 1n
k
kc r i k n
n n
00
2exp( )exp( ), 0,1, 2,..., 1n
k
kc r i i k n
n n
2exp( )nw i
n
Let then
2exp( )
n
k kw i
n
Therefore 0c , 0,1,2,..., 1kk nc w k n
where 0 00 0 0
2 0exp( )exp( ) exp( )n nc r i i r i
n n n
Note: the number c0 can be replaced by any particular nth root of z0
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Example 1 Let us find all values of (-8i)1/3, or the three roots of the
number -8i. One need only write
To see that the desired roots are
10. Examples
48
8 8exp[ ( 2 )], 0, 1, 2,...2
i i k k
22exp[ ( )], 0,1,2
6 3k
kc i k
2i
3 i3 i
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Example 2 To determine the nth roots of unity, we start with
And find that
10. Examples
49
1 1exp[ (0 2 )], 0, 1, 2,...i k k 1 0 2 2
1 1exp[ ( )] exp( ), 0,1,2,..., 1nn k ki i k n
n n n
n=3 n=4 n=6
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Example 3 the two values ck (k=0,1) of , which are the
square roots of , are found by writing
10. Examples
50
1/2( 3 )i
3 i
0 2 exp( ) 2(cos sin )12 12 12
c i i
1 0c c
3 2exp[ ( 2 )], 0, 1, 2,...6
i i k k
2 exp[ ( )], 0,112kc i k k
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pp. 29-31
Ex. 2, Ex. 4, Ex. 5, Ex. 7, Ex. 9
10. Homework
51
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ε- neighborhood The ε- neighborhood
of a given point z0 in the complex plane as shown below
11. Regions in the Complex Plane
52
0| |z z
x
y
O
z0
ε
0| |z z
z00 | |z z
Deleted neighborhood
x
y
O
z0
ε
0| |z z
z
Neighborhood
0| |z z
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Interior Point A 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
Exterior Point
A point z0 is said to be an exterior point of a set S when there exists a neighborhood of it containing no points of S;
Boundary Point (neither interior nor exterior) A boundary point is a point all of whose neighborhoods
contain at least one point in S and at least one point not in S.
The totality of all boundary points is called the boundary of S.
11. Regions in the Complex Plane
53
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Consider the set S={z| |z|≤1}
11. Regions in the Complex Plane
54
z0
x
y
O
All points z, where |z|<1 are Interior points of S;
z0
All points z, where |z|>1 are Exterior points of S;
All points z, where |z|=1 are Boundary points of S;
z0
S={z| |z|≤1-{1,0}}
?
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Open Set
A set is open if it and only if each of its points is an interior point.
Closed Set A set is closed if it contains all of its boundary points.
Closure of a set
The closure of a set S is the closed set consisting of all points in S together with the boundary of S.
11. Regions in the Complex Plane
55
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Examples S={z| |z|<1} ?
Open Set S={z| |z|≤1} ?
Closed Set S={z| |z|≤1} – {(0,0)} ?
Neither open nor closed S= all points in complex plane ?
Both open and closed
Key: identify those boundary points of a given set
11. Regions in the Complex Plane
56
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Connected An open set S is connected if each pair of points z1 and
z2 in it can be joined by a polygonal line, consisting of a finite number of line segments joined end to end, that lies entirely in S.
11. Regions in the Complex Plane
57
The open set 1<|z|<2 is connected.
xO
The set S={z| |z|<1 U |z-(2+i)|<1} is openHowever, it is not connected.
y
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Domain A set S is called as a domain iff
1. S is open;
2. S is connected.
e.g. any neighborhood is a domain.
Region A domain together with some, none, or all of it
boundary points is referred to as a region.
11. Regions in the Complex Plane
58
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Bounded A set S is bounded if every point of S lies inside some
circle |z|=R; Otherwise, it is unbounded.
11. Regions in the Complex Plane
59
x
y
O
S R
e.g. S={z| |z|≤1} is bounded
S={z| Rez≥0} is unbounded
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Accumulation point 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 contains each of its accumulation points. Why?
A set is closed iff it contains all of its accumulation points
11. Regions in the Complex Plane
60
The relationships among the Interior, Exterior, Boundary and Accumulation Points!
e.g. the origin is the only accumulation point of the set Zn=i/n, n=1,2,…
An Interior point must be an accumulation point. An Exterior point must not be an accumulation point. A Boundary point must be an accumulation point?
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pp. 33
Ex. 1, Ex. 2, Ex. 5, Ex. 6, Ex.10
11. Homework
61