Chapter 2. Analytic Functions Weiqi Luo ( ) School of Software Sun Yat-Sen University Email...

Chapter 2. Analytic Functions

Weiqi Luo (骆伟祺 )School of Software

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

mailto:[email protected]

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Functions of a Complex Variable; Mappings Mappings by the Exponential Function Limits; Theorems on Limits Limits Involving the Point at Infinity Continuity; Derivatives; Differentiation Formulas Cauchy-Riemann Equations; Sufficient Conditions for

Differentiability; Polar Coordinates; Analytic Functions; Harmonic Functions; Uniquely

Determined Analytic Functions; Reflection Principle

2

Chapter 2: Analytic Functions

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Function of a complex variable Let s be a set complex numbers. A function f defined on

S is a rule that assigns to each z in S a complex number w.

12. Functions of a Complex Variable

3

S

Complex numbers

f

S’

Complex numbers

z

The domain of definition of f The range of f

w

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Suppose that w=u+iv is the value of a function f at z=x+iy, so that

Thus each of real number u and v depends on the real variables x and y, meaning that

Similarly if the polar coordinates r and θ, instead of x and y, are used, we get


4

( )u iv f x iy

( ) ( , ) ( , )f z u x y iv x y

( ) ( , ) ( , )f z u r iv r

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Example 2 If f(z)=z2, then

case #1:

case #2:


5

2 2 2( ) ( ) 2f z x iy x y i xy 2 2( , ) ; ( , ) 2u x y x y v x y xy

z x iy

iz re 2 2 2 2 2( ) ( ) cos 2 sin 2i if z re r e r ir

2 2( , ) cos 2 ; ( , ) sin 2u r r v r r

When v=0, f is a real-valued function.

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Example 3 A real-valued function is used to illustrate some important concepts later in this chapter is

Polynomial function

where n is zero or a positive integer and a0, a1, …an are complex constants, an is not 0; Rational function the quotients P(z)/Q(z) of polynomials


6

2 2 2( ) | | 0f z z x y i

20 1 2( ) ... n

nP z a a z a z a z

The domain of definition is the entire z plane

The domain of definition is Q(z)≠0

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Multiple-valued function A generalization of the concept of function is a rule that

assigns more than one value to a point z in the domain of definition.


7

f

S

Complex numbers

z

S’

Complex numbers

w1

w2

wn

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Example 4 Let z denote any nonzero complex number, then z1/2 has

the two values

If we just choose only the positive value of


8

1/2 exp( )2

z r i

Multiple-valued function

r

1/2 exp( ), 02

z r i r

Single-valued function

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pp. 37-38

Ex. 1, Ex. 2, Ex. 4

12. Homework

9

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Graphs of Real-value functions

13. Mappings

10

f=tan(x)

f=ex

Note that both x and f(x) are real values.

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Complex-value functions

13. Mappings

11

( ) ( ) ( , ) ( , )f z f x yi u x y iv x y

x

y

u

v

Note that here x, y, u(x,y) and v(x,y) are all real values.

z(x,y) w(u(x,y),v(x,y))

mapping

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Examples

13. Mappings

12

1 ( 1)w z x iy

x

y

z(x,y)

u

v

w(x+1,y)

w z x yi

u

v

w(x,-y)

Translation Mapping

Reflection Mapping

y

z(x,y)

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Example

13. Mappings

13

( ) exp( ( ))2

iw iz i re r i

y

rθ

z

u

v

rθ

w

2

Rotation Mapping

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Example 1

13. Mappings

14

2w z 2 2 , 2u x y v xy

Let u=c1>0 in the w plane, then x2-y2=c1 in the z plane

Let v=c2>0 in the w plane, then 2xy=c2 in the z plane

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Example 2

The domain x>0, y>0, xy<1 consists of all points lying on the upper branches of hyperbolas

13. Mappings

15

x=0,y>0

x>0,y=0

2 2;

2 2 1

u x y

v xy xy

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Example 3

13. Mappings

16

2 2 2iw z r e In polar coordinates

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The exponential function

14. Mappings by the Exponential Function

17

,z x iy x iyw e e e e z x iy

ρeiθ ρ=ex, θ=y

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Example 2


18

w=exp(z)

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Example 3


19

w=exp(z)=ex+yi

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pp. 44-45

Ex. 2, Ex. 3, Ex. 7, Ex. 8

14. Homework

20

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For a given positive value ε, there exists a positive value δ (depends on ε) such that

when 0 < |z-z0| < δ, we have |f(z)-w0|< ε

meaning the point w=f(z) can be made arbitrarily chose to w0 if we choose the point z close enough to z0 but distinct from it.

15. Limits

21

00lim ( )

z zf z w

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The uniqueness of limit If a limit of a function f(z) exists at a point z0, it is unique.

Proof: suppose that

then

when

Let , when 0<|z-z0|<δ, we have

15. Limits

22

0 00 1lim ( ) & lim ( )

z z z zf z w f z w

0 1/ 2 0, 0, 0

1 0 0 1| | | ( ( ) ) ( ( ) ) |w w f z w f z w

0 1| ( ) | | ( ) |2 2

f z w f z w

0 1min( , )

0| ( ) | / 2;f z w 0 00 | |z z

1| ( ) | / 2;f z w 0 10 | |z z

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Example 1 Show that in the open disk |z|<1, then

Proof:

15. Limits

23

( ) / 2f z iz

1lim ( )

2z

if z

| || 1| | 1|| ( ) | | |

2 2 2 2 2

i iz i i z zf z

0, 2 , . .s t

0 | 1| ( 2 )z when

| 1|0 | ( ) |

2 2

z if z

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Example 2 If then the limit does not exist.

15. Limits

24

( )z

f zz

0

lim ( )z

f z

0

0lim 1

0x

x i

x i

( ,0)z x

(0, )z y0

0lim 1

0y

iy

iy

≠

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Theorem 1

Let

and

then

if and only if

16. Theorems on Limits

25

00lim ( )

z zf z w

0 00

( , ) ( , )lim ( , )

x y x yu x y u

0 00

( , ) ( , )lim ( , )

x y x yv x y v

( ) ( , ) ( , )f z u x y iv x y z x iy

0 0 0 0 0 0;z x iy w u iv

and

(a)

(b)

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Proof: (b)(a)


26

0 00

( , ) ( , )lim ( , )

x y x yu x y u

0 00

( , ) ( , )lim ( , )

x y x yv x y v

&

00lim ( )

z zf z w

1 2/ 2 0, 0, 0 . .s t

0| ( , ) |2

u x y u

When 2 20 0 10 ( ) ( )x x y y

2 20 0 20 ( ) ( )x x y y 0| ( , ) |

2v x y v

1 2min( , ) Let 2 20 0 00 ( ) ( ) , . .0 | |x x y y i e z z When

0 0 0| ( ) | | ( ( , ) ( , )) ( ) |f z w u x y iv x y u iv

0 0| ( , ) | | ( , ) |2 2

u x y u v x y v

0 0| ( , ) ( ( , ) ) |u x y u i v x y v

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Proof: (a)(b)


27

0 00

( , ) ( , )lim ( , )

x y x yu x y u

0 00

( , ) ( , )lim ( , )

x y x yv x y v

&

00lim ( )

z zf z w

0, 0 . .s t 0| ( ) |f z w When 00 | |z z

0 0 0| ( ) | | ( , ) ( , ) ( ) |f z w u x y iv x y u iv

0 0| ( ( , ) ) ( ( , ) ) |u x y u i v x y v

Thus 0 0| ( , ) | ;| ( , ) |u x y u v x y v

0 0 0| ( , ) | | ( ( , ) ) ( ( , ) ) |u x y u u x y u i v x y v

0 0 0| ( , ) | | ( ( , ) ) ( ( , ) ) |v x y v u x y u i v x y v

When (x,y)(x0,y0)

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Theorem 2

Let and

then


28

00lim ( )

z zf z w

00lim ( )

z zF z W

00 0lim[ ( ) ( )]

z zf z F z w W

00 0lim[ ( ) ( )]

z zf z F z w W

0

00

0

( )lim[ ] , 0

( )z z

wf zW

F z W

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29

( ) ( , ) ( , ), ( ) ( , ) ( , )f z u x y iv x y F z U x y iV x y

0 0 0 0 0 0 0 0 0; ;z x iy w u iv W U iV

( ) ( ) ( ) ( )f z F z uU vV i vU uV

Let

00lim ( )

z zf z w

00lim ( )

z zF z W

When (x,y)(x0,y0); u(x,y)u0; v(x,y)v0; & U(x,y)U0; V(x,y)V0;

0 0 0 0( )u U v VRe(f(z)F(z)):

0 0 0 0( )v U u VIm(f(z)F(z)):w0W0

00 0lim[ ( ) ( )]

z zf z F z w W

00lim ( )

z zf z w

00lim ( )

z zF z W

&

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30

0

limz z

c c

0

0limz z

z z

0

0lim ( 1,2,...)n n

z zz z n

20 1 2( ) ... n

nP z a a z a z a z

It is easy to verify the limits

For the polynomial

00lim ( ) ( )

z zP z P z

We have that

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Riemannsphere & Stereographic Projection

17. Limits Involving the Point at Infinity

31

N: the north pole

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The ε Neighborhood of Infinity


32

O x

y

R1

R2

When the radius R is large enough

i.e. for each small positive number ε

R=1/ε

The region of |z|>R=1/ε is called the ε Neighborhood of Infinity(∞)

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Theorem If z0 and w0 are points in the z and w planes,

respectively, then


33

0

lim ( )z z

f z

iff0

1lim 0

( )z z f z

0lim ( )z

f z w

iff00

1lim ( )z

f wz

lim ( )z

f z

iff0

1lim 0

1( )

zf

z

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Examples


34

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Continuity A function is continuous at a point z0 if

meaning that

1. the function f has a limit at point z0 and

2. the limit is equal to the value of f(z0)

18. Continuity

35

00lim ( ) ( )

z zf z f z

For a given positive number ε, there exists a positive number δ, s.t.

0| |z z When 0| ( ) ( ) |f z f z

00 | | ?z z

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Theorem 1 A composition of continuous functions is itself continuous.

18. Continuity

36

Suppose w=f(z) is a continuous at the point z0; g=g(f(z)) is continuous at the point f(z0)

Then the composition g(f(z)) is continuous at the point z0

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Theorem 2 If a function f (z) is continuous and nonzero at a point z0,

then f (z) ≠ 0 throughout some neighborhood of that point.

18. Continuity

37

Proof 0

0lim ( ) ( ) 0z z

f z f z

0| ( ) |0, 0, . .

2

f zs t

0| |z z When

00

| ( ) || ( ) ( ) |

2

f zf z f z

r

f(z0)f(z)

If f(z)=0, then 0

0

| ( ) || ( ) |

2

f zf z

Contradiction!

0| ( ) |f z

Why?

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Theorem 3 If a function f is continuous throughout a region R that is

both closed and bounded, there exists a nonnegative real number M such that

where equality holds for at least one such z.

18. Continuity

38

| ( ) |f z M for all points z in R

2 2| ( ) | ( , ) ( , )f z u x y v x y Note:

where u(x,y) and v(x,y) are continuous real functions

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pp. 55-56

Ex. 2, Ex. 3, Ex. 6, Ex. 9, Ex. 11, Ex. 12

18. Homework

39

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Derivative

Let f be a function whose domain of definition contains a neighborhood |z-z0|<ε of a point z0. The derivative of f at z0 is the limit

And the function f is said to be differentiable at z0 when f’(z0) exists.

19. Derivatives

40

0

00

0

( ) ( )'( ) lim

z z

f z f zf z

z z

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Illustration of Derivative

19. Derivatives

41

0 00

0

( ) ( )'( ) lim

z

f z z f zf z

z

0limz

dw w

dz z

0 0( ) ( )w f z z f z

0

00

0

( ) ( )'( ) lim

z z

f z f zf z

z z

O u

v

f(z0)

f(z0+Δz)

Δw

0z z z

Any position

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Example 1 Suppose that f(z)=z2. At any point z

since 2z + Δz is a polynomial in Δz. Hence dw/dz=2z or f’(z)=2z.

19. Derivatives

42

2 2

0 0 0

( )lim lim lim(2 ) 2z z z

w z z zz z z

z z

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Example 2 If f(z)=z, then

19. Derivatives

43

w z z z z z z z

z z z z

Case #1: Δx0, Δy=0

( , ) (0,0)z x y In any direction

O Δx

Δy

Case #1

Case #2

0

0lim 1

0x

z x i

z x i

Case #2: Δx=0, Δy0

0

0lim 1

0x

z i y

z i y

Since the limit is unique, this function does not exist anywhere

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Example 3 Consider the real-valued function f(z)=|z|2. Here

19. Derivatives

44

2 2| | | | ( )( )w z z z z z z z zz zz z z

z z z z

Case #1: Δx0, Δy=0

0 0

0lim ( ) lim ( )

0x x

z x iz z z z x z z z

z x i

Case #2: Δx=0, Δy0

0 0

0lim ( ) lim ( )

0y y

z i yz z z z i y z z z

z i y

0z z z z z dw/dz can not exist when z is not 0

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Continuity & Derivative Continuity Derivative

Derivative Continuity

19. Derivatives

45

For instance, f(z)=|z|2 is continuous at each point, however, dw/dz does not exists when z is not 0

0 0 0

00 0 0

0

( ) ( )lim[ ( ) ( )] lim lim( ) '( )0 0z z z z z z

f z f zf z f z z z f z

z z

Note: The existence of the derivative of a function at a point implies the continuity of the function at that point.

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Differentiation Formulas

20. Differentiation Formulas

46

[ ( ) ( )] '( ) '( )d

f z g z f z g zdz

[ ( ) ( )] ( ) '( ) '( ) ( )d

f z g z f z g z f z g zdz

2

( ) '( ) ( ) ( ) '( )[ ]

( ) [ ( )]

d f z f z g z f z g z

dz g z g z

1[ ]n ndz nz

dz

0; 1; [ ( )] '( )d d d

c z cf z cf zdz dz dz

( ) ( ( ))F z g f z

0 0 0'( ) '( ( )) '( )F z g f z f z

dW dW dw

dz dw dz

Refer to pp.7 (13)

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Example To find the derivative of (2z2+i)5, write w=2z2+i and

W=w5. Then

20. Differentiation Formulas

47

2 5 4 2 4 2 4(2 ) (5 ) ' 5(2 ) (4 ) 20 (2 )d

z i w w z i z z z idz

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pp. 62-63

Ex. 1, Ex. 4, Ex. 8, Ex. 9

20. Homework

48

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Theorem Suppose that

and that f’(z) exists at a point z0=x0+iy0. Then the first-order partial derivatives of u and v must exist at (x0,y0), and they must satisfy the Cauchy-Riemann equations

then we have

21. Cauchy-Riemann Equations

49

( ) ( , ) ( , )f z u x y iv x y

;x y y xu v u v

0 0 0 0 0'( ) ( , ) ( , )x xf z u x y iv x y

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Proof:


50

Let 0 0 0;z x iy z x i y

0 0

0 0 0 0 0 0 0 0

( ) ( )

[ ( , ) ( , )] [ ( , ) ( , )]

w f z z f z

u x x y y u x y i v x x y y v x y

00

0 0 0 0 0 0 0 0

( , ) (0,0)

'( ) lim

( , ) ( , )] [ ( , ) ( , )lim

z

x y

wf z

zu x x y y u x y i v x x y y v x y

x i y

Note that (Δx, Δy) can be tend to (0,0) in any manner .

Consider the horizontally and vertically directions

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Horizontally direction (Δy=0)

Vertically direction (Δx=0)


51

0 0 0 0 0 0 0 00

0

( , ) ( , ) [ ( , ) ( , )]'( ) lim

0x

u x x y y u x y i v x x y y v x yf z

x i

0 0 0 0 0 0 0 0

0

( , ) ( , ) [ ( , ) ( , )]limx

u x x y u x y i v x x y v x y

x

0 0 0 0( , ) ( , )x xu x y iv x y

0 0 0 0 0 0 0 00 0

( , ) ( , ) [ ( , ) ( , )]'( ) lim

0y

u x y y u x y i v x y y v x yf z

i y

2

0 0 0 0 0 0 0 0

0

{[ ( , ) ( , )] [ ( , ) ( , )]}lim

( )y

i u x y y u x y i v x y y v x y

i i y

0 0 0 0( , ) ( , )y yv x y iu x y

;x y y xu v u v Cauchy-Riemann equations

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Example 1

is differentiable everywhere and that f’(z)=2z. To verify that the Cauchy-Riemann equations are satisfied everywhere, write


52

2 2 2( ) 2f z z x y i xy

2 2( , )u x y x y ( , ) 2v x y xy

2x yu x v 2y xu y v

'( ) 2 2 2( ) 2f z x i y x iy z

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Example 2


53

2( ) | |f z z

2 2( , )u x y x y ( , ) 0v x y

If the C-R equations are to hold at a point (x,y), then

0x y y xu u v v

0x y

Therefore, f’(z) does not exist at any nonzero point.

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Theorem

be defined throughout some ε neighborhood of a point z0 = x0 + iy0, and suppose that

a. the first-order partial derivatives of the functions u and v with respect to x and y exist everywhere in the neighborhood;

b. those partial derivatives are continuous at (x0, y0) and satisfy the Cauchy–Riemann equations

Then f ’(z0) exists, its value being f ’ (z0) = ux + ivx where the right-hand side is to be evaluated at (x0, y0).

22. Sufficient Conditions for Differentiability

54

( ) ( , ) ( , )f z u x y iv x y

;x y y xu v u v at (x0,y0)

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Proof


55

z x i y Let

0 0( ) ( )w f z z f z u i v

0 0 0 0( , ) ( , )u u x x y y u x y 0 0 0 0( , ) ( , )v v x x y y v x y

0 0 0 0 1 2( , ) ( , )x yu u x y x u x y y x y

Note : (a) and (b) assume that the first-order partial derivatives of u and v are continuous at the point (x0,y0) and exist everywhere in the neighborhood

0 0 0 0 3 4( , ) ( , )x yv v x y x v x y y x y

Where ε1, ε2, ε3 and ε4 tend to 0 as (Δx, Δy) approaches (0,0) in the Δz plane.

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56

0 0 0 0 1 2( , ) ( , )x yu u x y x u x y y x y

0 0 0 0 3 4( , ) ( , )x yv v x y x v x y y x y

Note : The assumption (b) that those partial derivatives are continuous at (x0, y0) and satisfy the Cauchy–Riemann equations

-vx(x0,y0)

ux(x0,y0)

0 0 0 0 1 2( , ) ( , )x xu u x y x v x y y x y

0 0 0 0 3 4( , ) ( , )x xv v x y x u x y y x y

0 0 0 0 1 2 0 0 0 0 3 4( , ) ( , ) ( , ) ( , )x x x x

w u i v

z zu x y x v x y y x y v x y x u x y y x y

iz z

0 0 0 0 0 0 0 0 1 3 2 4( , ) ( , ) ( , ) ( , ) ( ) ( )x x x xu x y x iu x y y v x y x iv x y y i x i yi

z z z z

i2vx(x0,y0)

0 0 0 0 1 3 2 4( , ) ( , ) ( ) ( )x x

x yu x y iv x y i i

z z

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57

0 0 0 0 1 3 2 4( , ) ( , ) ( ) ( )x x

w x yu x y iv x y i i

z z z

| | 1;| | 1x y

z z

1 3 1 3 1 3| ( ) | | | | | | |x

i iz

2 4 2 4 2 4| ( ) | | | | | | |y

i iz

0 0 0 00

lim ( , ) ( , )x xz

wu x y iv x y

z

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Example 1


58

( ) cos sinz x i y x xf z e e e e y ie y

( , ) cosxu x y e y ( , ) sinxv x y e y

Both Assumptions (a) and (b) in the theorem are satisfied.

'( ) cos sin ( )x xx xf z u iv e y ie y f z

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Example 2


59

2( ) | |f z z

2 2( , )u x y x y ( , ) 0v x y

2 0 0x yu v x x

2 0 0y xu v x y

Therefore, f has a derivative at z=0, and cannot have a derivative at any nonzero points.

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Assuming that z0≠0

23. Polar Coordinates

60

cos , sinx r y r u u x u y

r x r y r

u u x u y

x y

cos sinr x yu u u sin cosx yu u r u r

Similarly cos sinr x yv v v sin cosx yv v r v r If the partial derivatives of u and v with respect to x and y satisfy the Cauchy-Riemann equations

;x y y xu v u v

;r rru v u rv

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Theorem Let the function f(z)=u(r,θ)+iv(r,θ) be defined throughout

some ε neighborhood of a nonzero point z0=r0exp(iθ0) and suppose that

the first-order partial derivatives of the functions u and v with respect to r and θ exist everywhere in the neighborhood;

those partial derivatives are continuous at (r0, θ0) and satisfy the polar form rur = vθ, uθ = −rvr of the Cauchy-Riemann equations at (r0, θ0)

Then f’(z0) exists, its value being


61

0 0 0 0 0'( ) ( ( , ) ( , ))ir rf z e u r iv r

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Example 1 Consider the function


62

1 1 1 1( ) (cos sin ), 0i

if z e i z

z re r r

cos sin( , ) , ( , )u r v r

r r

Then

cos sin&r rru v u rv

r r

2 2 2 2 2

cos sin 1 1'( ) ( )

( )

ii i

i

ef z e i e

r r r re Z

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pp. 71-72

Ex. 1, Ex. 2, Ex. 6, Ex. 7, Ex. 8

23. Homework

63

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Analytic at a point z0

A function f of the complex variable z is analytic at a point z0 if it has a derivative at each point in some neighborhood of z0.

Analytic function

A function f is analytic in an open set if it has a derivative everywhere in that set.

24. Analytic Function

64

Note that if f is analytic at a point z0, it must be analytic at each point in some neighborhood of z0

Note that if f is analytic in a set S which is not open, it is to be understood that f is analytic in an open set containing S.

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Analytic vs. Derivative For a point

Analytic Derivative

Derivative Analytic

For all points in an open set

Analytic Derivative

Derivative Analytic


65

f is analytic in an open set D iff f is derivative in D

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Singular point (singularity) If function f fails to be analytic at a point z0 but is analytic at

some point in every neighborhood of z0, then z0 is called a singular point.

For instance, the function f(z)=1/z is analytic at every point in the finite plane except for the point of (0,0). Thus (0,0) is the singular point of function 1/z.

Entire Function

An entire function is a function that is analytic at each point in the entire finite plane.

For instance, the polynomial is entire function.


66

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Property 1 If two functions are analytic in a domain D, then their sum and product are both analytic in D their quotient is analytic in D provided the function in the

denominator does not vanish at any point in D

Property 2 From the chain rule for the derivative of a composite function, a

composition of two analytic functions is analytic.


67

( ( )) '[ ( )] '( )d

g f z g f z f zdz

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Theorem If f ’(z) = 0 everywhere in a domain D, then f (z) must be

constant throughout D.


68

( )du

gradu Uds

x=u ygradu i u j

0 & 0x y x yu u v v

'( ) 0x x y yf z u iv v iu

U is the unit vector along L

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Example 1 The quotient

is analytic throughout the z plane except for the singular points

25. Examples

69

3

2 2

4( )

( 3)( 1)

zf z

z z

3 &z z i

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Example 3 Suppose that a function and its

conjugate are both analytic in a given domain D. Show that f(z) must be constant throughout D.

25. Examples

70

( ) ( , ) ( , )f z u x y iv x y

( ) ( , ) ( , )f z u x y iv x y

( ) ( , ) ( , )f z u x y iv x y is analytic, then

,x y y xu v u v ( ) ( , ) ( , )f z u x y iv x y is analytic, then

Proof:

,x y y xu v u v

0, 0x xu v '( ) 0x xf z u iv

Based on the Theorem in pp. 74, we have that f is constant throughout D

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Example 4 Suppose that f is analytic throughout a given region D, and

the modulus |f(z)| is constant throughout D, then the function f(z) must be constant there too.

Proof:

|f(z)| = c, for all z in D

where c is real constant.

If c=0, then f(z)=0 everywhere in D.

If c ≠ 0, we have

25. Examples

71

2( ) ( )f z f z c2

( ) , ( ) 0( )

cf z f z inD

f z

Both f and it conjugate are analytic, thus f must be constant in D. (Refer to Ex. 3)

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pp. 77~78

Ex. 2, Ex. 3, Ex. 4, Ex. 6, Ex. 7

25. Homework

72

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A Harmonic Function A real-valued function H of two real variables x and y is

said to be harmonic in a given domain of the xy plane if, throughout that domain, it has continuous partial derivatives of the first and second order and satisfies the partial differential equation

Known as Laplace’s equation.

26. Harmonic Functions

73

( , ) ( , ) 0xx yyH x y H x y

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Theorem 1 If a function f (z) = u(x, y) + iv(x, y) is analytic in a domain

D, then its component functions u and v are harmonic in D.

Proof:


74

( ) ( , ) ( , )f z u x y iv x y is analytic in D

&x y y xu v u v Differentiating both sizes of these equations with respect to x and y respectively, we have

Theorem in Sec.52: a function is analytic at a point, then its real and imaginary components have continuous partial derivatives of all order at that point.

continuity

0 & 0xx yy xx yyu u v v

&xx yx yx xxu v u v

&xy yy yy xyu v u v &xx xy xy xxu v u v &xy yy yy xyu v u v

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Example 3 The function f(z)=i/z2 is analytic whenever z≠0 and since

The two functions


75

2 2 2

2 2 2 22 2

( ) 2 ( )

( )( )

i i z xy i x y

z x yz z

2 2 2

2( , )

( )

xyu x y

x y

2 2

2 2 2( , )

( )

x yv x y

x y

are harmonic throughout any domain in the xy plane that does not contain the origin.

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Harmonic conjugate

If two given function u and v are harmonic in a domain D and their first-order partial derivatives satisfy the Cauchy-Riemann equation throughout D, then v is said to be a harmonic conjugate of u.


76

If u is a harmonic conjugate of v, then

&x y y xu v u v

Is the definition symmetry for u and v?

Cauchy-Riemann equation &x y y xu v u v

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Theorem 2 A function f (z) = u(x, y) + iv(x, y) is analytic in a domain D

if and only if v is a harmonic conjugate of u.

Example 4

The function is entire function, and its real and imaginary components are

Based on the Theorem 2, v is a harmonic conjugate of u throughout the plane. However, u is not the harmonic conjugate of v, since is not an analytic function.


77

2 2( , ) & ( , ) 2u x y x y v x y xy

2( )f z z

2 2( ) 2 ( )g z xy i x y

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Example 5 Obtain a harmonic conjugate of a given function.

Suppose that v is the harmonic conjugate of the given function

Then


78

3 2( , ) 3u x y y x y

&x y y xu v u v

6x yu xy v

2 23 3y xu y x v

23 ( )v xy x

2 2 23 3 ( 3 '( ))y x y x

2 3'( ) 3 ( )=x x x x C

2 33v xy x C

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pp. 81-82

Ex. 1, Ex. 2, Ex. 3, Ex. 5

26. Homework

79

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Lemma Suppose that

a) A function f is analytic throughout a domain D;

b) f(z)=0 at each point z of a domain or line segment contained in D.

Then f (z) ≡ 0 in D; that is, f (z) is identically equal to zero throughout D.

27. Uniquely Determined Analytic Function

80

Refer to Chap. 6 for the proof.

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Theorem

A function that is analytic in a domain D is uniquely determined over D by its values in a domain, or along a line segment, contained in D.

27. Uniquely Determined Analytic Function

81

f(z)

D

g(z)

D

( ) ( )f z g z

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Theorem Suppose that a function f is analytic in some domain D

which contains a segment of the x axis and whose lower half is the reflection of the upper half with respect to that axis. Then

for each point z in the domain if and only if f (x) is real for each point x on the segment.

28. Reflection Principle

82

x

y

D

( ) ( )f z f z

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