Chapter 1. Complex Numbers

Chapter 1. Complex Numbers

Weiqi Luo (骆伟祺 )School of Software

Sun Yat-Sen UniversityEmail ： [email protected] Office ： # A313

mailto:[email protected]

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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

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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

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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

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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


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. 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


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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

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If z1z2=0, then so is at least one of the factors z1 and z2

3. Further Properties

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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:


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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


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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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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


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Any complex number is associated a vector from the origin to the point (x, y)

4. Vectors and Moduli

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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|.


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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


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


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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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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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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


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| 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

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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

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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


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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


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,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


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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


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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

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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

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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

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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

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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

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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


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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


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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

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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


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≠Note: Argz1z2=Argz1+Argz2 is not always true.

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Arguments of Quotients


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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


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

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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


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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


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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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

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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.


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Consider the set S={z| |z|≤1}


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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.


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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


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.


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.


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.


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


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

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Chapter 1. Complex Numbers