Download Chap. 1 - Sun Yat

Chapter 1. Complex Numbers
Weiqi Luo (骆伟祺)
School of Software
Sun Yat-Sen University
Email：[email protected] Office：# A313
 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
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Numbers System
Natural Numbers
Zero & Negative Numbers
Integers
Fraction
Rational numbers
Irrational numbers
Real numbers
Imaginary numbers
Complex numbers
… More advanced number systems
Refer to: http://en.wikipedia.org/wiki/Number_system
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Chapter 1: Complex Numbers







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
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1. Sums and Products
 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
y
Note: The set of complex numbers
Includes the real numbers as a subset
(x, y)
(0, y)
imaginary axis
Real axis
O
(x, 0)
x
Complex plane
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1. Sums and Products
 Notation
It is customary to denote a complex number (x,y) by z,
x = Rez (Real part);
y = Imz (Imaginary part)
y
z=(x, y)
(0, y)
z1=z2
iff
1. Rez1= Rez2
2. Imz1 = Imz2
O
(x, 0)
x
6
Q: z1<z2?
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1. Sums and Products
 Two Basic Operations
 Sum
(x1, y1) + (x2, y2) = (x1+x2, y1+y2)
 Product
(x1, y1) (x2, y2) = (x1x2 - y1y2, y1x2+x1y2)
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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1. Sums and Products
 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)
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2. Basic Algebraic Properties
 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)
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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2. Basic Algebraic Properties
 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
x
y
u 2
,v  2
2
x y
x  y2
x
y
z ( 2
, 2
), z  0
2
2
x y x y
1
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Homework
 pp. 5
Ex. 1, Ex.4, Ex. 8, Ex. 9
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3. Further Properties
 If z1z2=0, then so is at least one of the factors z1 and z2
Proof: Suppose that z1 ≠ 0, then z1-1 exists
z1-1 (z1z2)=z1-1 0 =0
z1-1 (z1z2)=( z1-1 z1) z2 =1 z2 = z2
Associative Laws
Therefore we have z2=0
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3. Further Properties
 Other two operations: Subtraction and Division
 Subtraction: z1-z2=z1+(-z2)
(x1, y1) - (x2, y2) = (x1, y1)+(-x2, -y2) = (x1 -x2, y1-y2)

z1
 z1 z2 1 ( z2  0)
Division:
z2
z1
x
y
xx  y y y x x y
 ( x1 , y1 )( 2 2 2 , 2 2 2 )  ( 1 22 12 2 , 1 22 12 2 )
z2
x2 +y2 x2 +y2
x2 +y2
x2 +y2
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3. Further Properties
 An easy way to remember to computer z1/z2
z1 ( x1  iy1 ) ( x1  iy1 )( x2  iy2 )


z2 ( x2  iy2 ) ( x2  iy2 )( x2  iy2 )
commonly used
( x2  iy2 )( x2  iy2 )  x2 2  y2 2  R
Note that
For instance
4i
(4  i)(2  3i) 5  14i 5 14


  i
2  3i (2  3i)(2  3i)
13
13 13
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3. Further Properties
Binomial Formula
n
( z1  z2 ) n   Cnk z1k z2 n  k , n  1, 2,...
k 0
Where
n!
C 
, k  0,1, 2,..., n
k !(n  k )!
k
n
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3. Further Properties
pp.8
Ex. 1. Ex. 2, Ex. 3, Ex. 6
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4. Vectors and Moduli
 Any complex number is associated a vector from the
origin to the point (x, y)
y
y
z1=(x1, y1)
| z1 | x12  y12
z2=(x2, y2)
z1+z2
| z2 || z1 |
O
z1
z2
O
x
Sum of two vectors
The moduli or absolute value of z
is a nonnegative real number
| z | x  y
2
x
z1  z2  ( x1  x2 )  i( y1  y2 )
2
Product: refer to pp.21
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4. Vectors and Moduli
 Example 1
The distance between two point z1(x1, y1) and z2(x2, y2)
is |z1-z2|.
Note: |z1 - z2 | is the length of the vector
representing the number z1-z2 = z1 + (-z2)
y
|z1 - z2 |
Therefore
z2
z1
-z2
z1  z2  ( x1  x2 )  i( y1  y2 )
z1 - z2
O
| z1  z2 | ( x1  x2 )2  ( y1  y2 )2
x
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4. Vectors and Moduli
 Example 2
The equation |z-1+3i|=2 represents the circle whose
center is z0 = (1, -3) and whose radius is R=2
y
x
O
z0(1, -3)
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Note: | z-1+3i |
= | z-(1-3i) |
=2
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4. Vectors and Moduli
 Some important inequations
 Since | Re Z |  | Im Z | | Z | we have
2
2
2
Re Z | Re Z || Z |; Im Z | Im Z || Z |
 Triangle inequality
y
z1=(x, y)
| z1 | x2  y 2
O
| z1  z2 || z1 |  | z2 |
x
y
z1+z2
z1
z2
O
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x
4. Vectors and Moduli

| z1  z2 ||| z1 |  | z2 ||
Proof: when |z1| ≥ |z2|, we write
Triangle inequality
| z1 || ( z1  z2 )  ( z2 ) | | z1  z2 |  | ( z2 ) | | z1  z2 |  | z2 |
| z1  z2 || z1 |  | z2 ||| z1 |  | z2 ||
Similarly when |z2| ≥ |z1|, we write
| z2 || ( z1  z2 )  ( z1 ) | | z1  z2 |  | ( z1 ) | | z1  z2 |  | z1 |
| z1  z2 || z2 |  | z1 ||| z1 |  | z2 ||
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4. Vectors and Moduli

| z1  z2 || z1 |  | z2 |

| z1  z2 ||| z1 |  | z2 ||

|| z1 |  | z2 ||| z1  z2 || z1 |  | z2 |
| z1  z2  ...zn || z1 |  | z2 | ... | zn |
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4. Vectors and Moduli
 Example 3
If a point z lies on the unit circle |z|=1 about the origin,
then we have
y
| z  2 | || z | 2 | 1
z
| z  2 | | z | 2  3
O
23
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2
x
4. Homework
pp. 12
Ex. 2, Ex. 4, Ex. 5
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5. Complex Conjugates
 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
y
Properties:
z(x,y)
O
| z || z |
zz
x
z (x,-y)
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5. Complex Conjugates
 If z1=x1+iy1 and z2=x2+iy2 , then
z1  z2  ( x1  x2 )  i( y1  y2 )  ( x1  iy1 )  ( x2  iy2 )  z1  z2
 Similarly, we have
z1  z2  z1  z2
z1 z2  z1  z2
z1 z1
 , z2  0
z2 z2
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5. Complex Conjugates
 If
z  x  iy, z  x  iy
, then
z  z  ( x  iy )  ( x  iy )  2 x  2 Re z
z  z  ( x  iy )  ( x  iy )  2 yi  2i Im z
zz
zz
Re z 
, Im z 
2
2i

z z  ( x  iy )  ( x  iy )  x 2  y 2 | Z |2
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5. Complex Conjugates
 Example 1
1  3i
?
2i
1  3i (1  3i)(2  i)

2i
(2  i)(2  i)
5  5i

| 2  i |2
5  5i

 1  i
5
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5. Complex Conjugates
z1 | z1 |
a. | |
z2 | z2 |
z1 2 z1 z1
z1 z1 z1 z1 | z1 |2
proof :| |   ( )   

z2
z2 z2
z2 z2 z2 z2 | z2 |2
| z1 z2 || z1 || z2 |
| z n || z |n
Refer to pp. 14
 Example 2
| z | 2
| z 3  3z 2  2 z  1| | z 3 |  | 3z 2 |  | 2 z |  |1| | z |3 3 | z |2 2 | z | 1  25
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5. Homework
pp. 14 – 16
Ex. 1, Ex. 2, Ex. 7, Ex. 14
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6. Exponential Form
 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
θ
y
y
z(x,y)
argz: the argument of z
Argz: the principal value of argz
r
θ
O
arg z  ArgZ  2n , n  0, 1, 2,...
z(x,y)
r
θ
x
O
Θ
  ArgZ  
1
x
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6. Exponential Form
 Example 1
The complex number -1-i, which lies in the third quadrant
has principal argument -3π/4. That is
3
Arg (1  i )  
4
It must be emphasized that the principal argument must be in
the region of (-π, +π ]. Therefore,
5
Arg (1  i ) 
However,
arg(1  i )  
arg(1  i ) 
4
3
 2n , n  0, 1, 2,...
4
5
 2n , n  0, 1, 2,...
4
32
argz = α + 2nπ
Here: α can be any one
of arguments of z
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6. Exponential Form
 The symbol eiθ , or exp(iθ)
ei  cos   i sin 
Why? Refer to Sec. 29

1 1 1 2 1 3
1 n
1
e  1  x  x  x  ...  x  ...   x n
1!
2!
3!
n!
n 0 n !

1 2 n 
1

x 
x 2 n 1
Let x=iθ, then we have
n  0 (2n)!
n  0 (2 n  1)!
x




1
1
1
1
2n
2 n 1
2n
2n
e 
(i )  
(i )
  (i )
 i[ (i) 2 n  2
 2 n 1 ]
(2n)!
(2n  1)!
n  0 (2n)!
n 1 (2 n  1)!
n 0
n 1
i


1
1
2n
n 1
  (1)
( ) i[ (1)
( ) 2 n 1 ]
(2n)!
(2n  1)!
n 0
n 1
n
cosθ
sinθ
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6. Exponential Form
 Example 2
The number -1-i in Example 1 has exponential form
3
i(  )
3
3
1  i  2(cos(  )  i sin(  ))  2e 4
4
4
3
i (    2 n )
3
3
1  i  2(cos(  )  i sin(  ))  2e 4
, n  0, 1, 2,...
4
4
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6. Exponential Form
 z=Reiθ where 0≤ θ ≤2 π
y
y
Reiθ θ
Reiθ
z
θ
R
O
z0
O
x
x
z=z0 +Reiθ
|z-z0 |=R
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7. Products and Powers in Exponential Form
 Product in exponential form
ei1 ei2  (cos 1  i sin 1 )(cos  2  i sin  2 )
 (cos 1 cos  2  sin 1 sin  2 )  i(sin 1 cos  2  cos 1 sin  2 )
 cos(1   2 )  i sin(1   2 )  ei (1 2 )
z1  r1ei1 & z2  r2ei2
z1 z2  (r1ei1 )(r2ei2 )  r1r2ei (1 2 )
( z1 ) n  (r1ei1 ) n  r1n ein1 , n  0, 1, 2,...
z1 r1ei1 r1 i (1 2 )
 i2  e
, z2  0
z2 r2e
r2
1 1ei 0
1 i2
 i2  e , z2  0
z2 r2e
r2
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7. Products and Powers in Exponential Form
 Example 1
In order to put
only write
( 3  i )  (2e
7
i /6 7
( 3  i)7
) 2 e
7
in rectangular form, one need
i 7 /6
7
7
 2 (cos
 i sin )  64( 3+i)
6
6
7
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7. Products and Powers in Exponential Form
 Example 2
de Moivre’s formula
(ei )n  (cos   i sin  )n  cos n  i sin n , n  0, 1, 2,...
(cos   i sin  )  cos 2  i sin 2
2
(cos   i sin  )2  cos2   sin 2   i(2sin  cos  )
pp. 23, Exercise 10, 11
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8. Arguments of products and quotients
Ifz1  r1ei1 & z2  r2ei2 , then
z1 z2  (r1ei1 )(r2ei2 )  r1r2ei (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
arg(z1z2)= θ1 +θ2 +2nπ, n=0, ±1, ±2 …
argz1z2= θ1 +θ2 +2(n1+n2)π
= (θ1 +2n1π)+ (θ2 +2n2π)
= argz1+argz2
Q: Argz1z2 = Argz1+Argz2?
Here: n1 and n2 are two integers with n1+n2=n
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8. Arguments of products and quotients
 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
Note: Argz1z2=Argz1+Argz2 is not always true.
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8. Arguments of products and quotients
 Arguments of Quotients
z1
arg( )  arg( z1 z21 )  arg( z1 )  arg( z21 )
z2
 arg( z1 )  arg( z2 )
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8. Arguments of products and quotients
 Example 2
In order to find the principal argument Arg z when
2
z
1 i 3
observe that
arg z  arg(2)  arg(1  3i)
since Arg (2)  
Arg (1  3i ) 


2
argz  (  )  2n 
 2n
3
3
Argz
42
3
2
 
 
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8. Homework
pp. 22-24
Ex. 1, Ex. 6, Ex. 8, Ex. 10
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9. Roots of Complex Numbers
 Two equal complex numbers
z1  r1ei1
z2  r2 ei2
At the same point
z1  z2
If and only if
r1  r2 & 1  2  2k
for some integer k
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9. Roots of Complex Numbers
 Roots of Complex Number
i 0
z

r
e
Given a complex number 0 0 , we try to find all
the number z, s.t.z n  z0
Let z  rei then z n  (rei )n  r n ein  r0ei
0
thus we get
r n  r0 & n   0  2k , k  0, 1, 2,...
0
2 k
r  r0 &   
, k  0, 1, 2,...
n
n
n
The unique positive nth root of r0
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9. Roots of Complex Numbers
The nth roots of z0 are
z  n r0 exp[i (
0
n

2 k
)], k  0, 1, 2,...
n
Note:
1.All roots lie on the circle |z|;
2.There are exactly n distinct roots!
ck  n r0 exp[i (
0
n

2 k
)], k  0,1, 2,..., n  1
n
|z|
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9. Roots of Complex Numbers
ck  n r0 exp[i (
ck  n r0 exp(i
Let
0
n
0
wn  exp(i
n

2 k
)], k  0,1, 2,..., n  1
n
) exp(i
2
)
n
2 k
), k  0,1, 2,..., n  1
n
wnk  exp(i
then
Therefore
ck  c0 wnk , k  0,1, 2,..., n  1
where
c0  n r0 exp(i
0
n
) exp(i
2 k
)
n

2 0
)  n r0 exp(i 0 )
n
n
Note: the number c0 can be replaced by any particular nth root of z0
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10. Examples
 Example 1
Let us find all values of (-8i)1/3, or the three roots of the
number -8i. One need only write
8i  8exp[i(

2
 2k )], k  0, 1, 2,...
To see that the desired roots are
ck  2 exp[i(

6

2i
2 k
)], k  0,1, 2
3
 3 i
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3 i
School of Software
10. Examples
 Example 2
To determine the nth roots of unity, we start with
1  1exp[i (0  2k )], k  0, 1, 2,...
And find that
n=3
0 2k
2k
1  n 1exp[i( 
)]  exp(i
), k  0,1, 2,..., n  1
n
n
n
1
n
n=4
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n=6
School of Software
10. Examples
 Example 3
the two values ck (k=0,1) of ( 3  i)1/2, which are the
square roots of3  i , are found by writing

3  i  2 exp[i (  2k )], k  0, 1, 2,...
6

ck  2 exp[i (  k )], k  0,1
12
c0  2 exp(i

12
)  2(cos

12
 i sin
c1  c0
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
12
)
10. Homework
pp. 29-31
Ex. 2, Ex. 4, Ex. 5, Ex. 7, Ex. 9
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School of Software
11. Regions in the Complex Plane
 ε- neighborhood
The ε- neighborhood
| z  z0 | 
of a given point z0 in the complex plane as shown below
| z  z0 |
y
z
| z  z0 |
y
ε
z0
O
| z  z0 | 
z
O
x
Neighborhood
ε
z0
0 | z  z0 | 
x
Deleted neighborhood
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School of Software
11. Regions in the Complex Plane
 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.
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School of Software
11. Regions in the Complex Plane
 Consider the set S={z| |z|≤1}
All points z, where |z|>1
are Exterior points of S;
y
S={z| |z|≤1-{1,0}}
z0
?
z0
O
z0
All points z, where |z|<1
are Interior points of S;
x
All points z, where |z|=1
are Boundary points of S;
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School of Software
11. Regions in the Complex Plane
 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.
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School of Software
11. Regions in the Complex Plane
 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
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School of Software
11. Regions in the Complex Plane
 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.
y
O
x
The set S={z| |z|<1 U |z-(2+i)|<1} is open
However, it is not connected.
The open set
1<|z|<2 is connected.
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School of Software
11. Regions in the Complex Plane
 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.
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School of Software
11. Regions in the Complex Plane
 Bounded
A set S is bounded if every point of S lies inside some
circle |z|=R; Otherwise, it is unbounded.
y
e.g. S={z| |z|≤1} is bounded
S
O
S={z| Rez≥0} is unbounded
R
x
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School of Software
11. Regions in the Complex Plane
 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
e.g. the origin is the only accumulation point of the set Zn=i/n, n=1,2,…
The relationships among the Interior, Exterior, Boundary and Accumulation Points!
 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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School of Software
11. Homework
pp. 33
Ex. 1, Ex. 2, Ex. 5, Ex. 6, Ex.10
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School of Software