Transcript 幻灯片 1 - Sun Yat-sen University

Chapter 4. Integrals
Weiqi Luo (骆伟祺)
School of Software
Sun Yat-Sen University
Email：[email protected] Office：# A313
Chapter 4: Integrals
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
Derivatives of Functions w(t)
Definite Integrals of Functions w(t)
Contours; Contour Integrals;
Some Examples; Example with Branch Cuts
Upper Bounds for Moduli of Contour Integrals
Anti derivatives; Proof of the Theorem
Cauchy-Goursat Theorem; Proof of the Theorem
Simply Connected Domains; Multiple Connected Domains;
Cauchy Integral Formula; An Extension of the Cauchy Integral Formula;
Some Consequences of the Extension
 Liouville’s Theorem and the Fundamental Theorem
 Maximum Modulus Principle
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37. Derivatives of Functions w(t)
 Consider derivatives of complex-valued functions w of
real variable t
w(t) = u(t) +iv(t)
where the function u and v are real-valued functions of t.
The derivative
d
w'(t), or w(t)
dt
of the function w(t) at a point t is defined as
w'(t) = u'(t)+iv'(t)
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37. Derivatives of Functions w(t)
 Properties
For any complex constant z0=x0+iy0,
u(t)
v(t)
d
[z0 w(t)] = [(x0 + iy0 )(u + iv)]' = [(x0u - y0 v)+ i(y0u + x0 v)]'
dt
= (x0u - y0 v)'+i(y0u + x0v)'
= (x0u'- y0v')+ i(y0u'+ x0v')
= (x0 + iy0 )(u'+iv') = z0 w'(t)
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37. Derivatives of Functions w(t)
 Properties
d z0 t
e = z0 e z 0 t
dt
where z0=x0+iy0. We write
( x0 iy0 )t
e e
z0t
u(t)
v(t)
 e cos y0t  ie sin y0t
x0t
x0 t
d z0t
e  (e x0t cos y0t ) ' i (e x0 t sin y0t ) '
dt
Similar rules from calculus and some simple algebra then lead us to the expression
d z0t
e  ( x0  iy0 )e( x0 iy0 )t  z0 e z0t
dt
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37. Derivatives of Functions w(t)
 Example
Suppose that the real function f(t) is continuous on an
interval a≤ t ≤b, if f’(t) exists when a<t<b, the mean value
theorem for derivatives tells us that there is a number ζ in
the interval a<ζ<b such that
f (b) - f (a)
f '(V ) =
b-a
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37. Derivatives of Functions w(t)
 Example (Cont’)
The mean value theorem no longer applies for some
complex functions. For instance, the function
w(t) = eit
on the interval 0 ≤ t ≤ 2π .
Please note that
| w'(t) |=| ieit |=1
And this means that the derivative w’(t) is never zero, while
w(2 )  w(0)
w '( ) 
 0,  （0，
2）
w(2 )  w(0)  0
2  0
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38. Definite Integrals of Functions w(t)
 When w(t) is a complex-valued function of a real variable
t and is written
w(t) = u(t) +iv(t)
where u and v are real-valued, the definite integral of
w(t) over an interval a ≤ t ≤ b is defined as
b
b
b
a
a
a
 w(t )dt   u(t )dt  i  v(t )dt
Provided the individual integrals on the right exist.
b
b
b
b
a
a
a
a
Re  w(t )dt   Re w(t )dt & Im  w(t )dt   Im w(t )dt
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38. Definite Integrals of Functions w(t)
 Example 1
1
1
1
2
0 (1  it ) dt  0 (1  t )dt  i 0 2tdt  3  i
2
2
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38. Definite Integrals of Functions w(t)
 Properties
The existence of the integrals of u and v is ensured if
those functions are piecewise continuous on the interval a
≤ t ≤ b. For instance,
b
c
b
a
a
c
 w(t )dt   w(t )dt   w(t )dt
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38. Definite Integrals of Functions w(t)
 Integral vs. Anti-derivative
Suppose that
w(t )  u(t )  iv(t ),W (t )  U (t )  iV (t )
are continuous on the interval a ≤ t ≤ b.
If W’(t)=w(t) when a ≤ t ≤ b, then U’(t)=u(t) and V’(t)=v(t).
Hence, in view of definition of the integrals of function
b
b
b
a
a
a
 w(t )dt   u(t )dt  i  v(t )dt
 U (t ) ba  iV (t )
b
a
 [U (b)  iV (b)]  [U (a)  iV (a)]
 W (b)  W (a )  W (t )
b
a
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38. Definite Integrals of Functions w(t)
 Example 2
Since
d eit
1 d it 1 it
( )
e  ie  eit
dt i
i dt
i
one can see that

it 
4
e
0 e dt  i
it
4
0
e
i

4
1


i
i
1
1

 i (1 
)
2
2
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38. Definite Integrals of Functions w(t)
 Example 3
Let w(t) be a continuous complex-valued function of t defined on an
interval a ≤ t ≤ b. In order to show that it is not necessarily true that there
is a number c in the interval a <t< b such that
b
 w(t )dt  w(c)(b  a)
a
We write a=0, b=2π and use the same function w(t)=eit (0 ≤ t ≤ 2π) as the
Example in the previous Section (pp.118). We then have that
2
2
it
e
it
2
w
(
t
)
dt

e
dt

0 0
0
0
i
However, for any number c such that 0 < c < 2π
| w(c)(b  a) || eic | 2  2
And this means that w(c)(b-a) is not zero.
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38. Homework
pp. 121
Ex. 1, Ex. 2, Ex. 4
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39. Contours
 Arc
A set of points z=(x, y) in the complex plane is said to be an
arc if
x  x(t ) & y  y(t ), a  t  b
where x(t) and y(t) are continuous functions of the real
parameter t. This definition establishes a continuous mapping
of the interval a ≤ t ≤ b in to the xy, or z, plane.
And the image points are ordered according to increasing
values of t. It is convenient to describe the points of C by
means of the equation
z (t )  x(t )  iy(t )
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39. Contours
 Simple arc (Jordan arc)
The arc C: z(t)=x(t)+iy(t) is a simple arc, if it does not
cross itself; that is, C is simple if z(t1)≠z(t2) when t1≠t2
 Simple closed curve (Jordan curve)
When the arc C is simple except for the fact that
z(b)=z(a), we say that C is simple closed curve.
Define that such a curve is positively oriented when it is in the
counterclockwise direction.
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39. Contours
 Example 1
The polygonal line defined by means of the equations
 x  ix, if 0  x  1
z
 x  i, if 1  x  2
and consisting of a line segment from 0 to 1+i followed
by one from 1+i to 2+i is a simple arc
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39. Contours
 Example 2~4
i
z

e
,(0    2 )
 The unit circle
about the origin is a simple closed curve, oriented in the
counterclockwise direction.
i
z

z

e
,(0    2 )
So is the circle
0
centered at the point z0 and with radius R.
 The set of points z  ei ,(0    2 )
This unit circle is traveled in the clockwise direction.
 The set of point z  ei 2 ,(0    2 )
This unit circle is traversed twice in the counterclockwise
Note: the same set of points can make up different arcs.
direction.
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39. Contours
 The parametric representation used for any given
arc C is not unique
To be specific, suppose that t   ( ),     
where Φ is a real-valued function mapping an interval α ≤ τ
≤ β onto a ≤ t ≤ b.
C : z(t ), a  t  b
C : Z ( )  z ( ( )),     
The same arc C
Here we assume Φ is a continuous functions with a continuous
derivative, and Φ’(τ)>0 for each τ (why?)
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39. Contours
 Differentiable arc
Suppose the arc function is z(t)=x(t)+iy(t), and the
components x’(t) and y’(t) of the derivative z(t) are continuous
on the entire interval a ≤ t ≤ b.
Then the arc is called a differentiable arc, and the real-valued
function
| z '(t ) | [ x '(t )]2  [ y '(t )]2
is integrable over the interval a ≤ t ≤ b.
In fact, according to the definition of a length in calculus, the
b
length of C is the number
L   | z '(t ) | dt   ds
a
C
Note: The value L is invariant under certain changes in the representation for C.
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39. Contours
 Smooth arc
A smooth arc z=z(t) (a ≤ t ≤ b), then it means that the derivative z’(t)
is continuous on the closed interval a ≤ t ≤ b and nonzero throughout
the open interval a < t < b.
 A Piecewise smooth arc (Contour)
Contour is an arc consisting of a finite number of smooth arcs joined
end to end. (e.g. Fig. 36)
 Simple closed contour
When only the initial and final values of z(t) are the same, a contour
C is called a simple closed contour. (e.g. the unit circle in Ex. 5 and 6)
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39. Contours
 Jordan Curve Theorem
Interior of C (bounded)
Jordan Curve Theorem asserts that
every Jordan curve divides the plane into
an "interior" region bounded by the curve
and an "exterior" region containing all of
the nearby and far away exterior points.
Jordan curve
Exterior of C (unbounded)
Refer to: http://en.wikipedia.org/wiki/Jordan_curve_theorem
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39. Homework
pp. 125-126
Ex. 1, Ex. 3, Ex.4
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40. Contour Integrals
 Consider the integrals of complex-valued function f of
the complex variable z on a given contour C, extending
from a point z=z1 to a point z=z2 in the complex plane.
 f ( z )dz
C
or
z2
 f ( z )dz
z1
When the value of the integral is independent of the choice
of the contour taken between two fixed end points.
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40. Contour Integrals
 Contour Integrals
Suppose that the equation z=z(t) (a ≤ t ≤b) represents a
contour C, extending from a point z1=z(a) to a point z2=z(b). We
assume that f(z(t)) is piecewise continuous on the interval a ≤ t
≤b, then define the contour integral of f along C in terms of the
parameter t as follows
b
 f ( z)dz   f ( z(t )) z '(t )dt
C
a
On the integral [a b] as defined previously
Contour integral
Note the value of a contour integral is invariant under a change
in the representation of its contour C.
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40. Contour Integrals
 Properties
z
0
f ( z )dz  z0  f ( z )dz
C
C
 [ f ( z )  g ( z )]dz   f ( z )dz   g( z )dz
C
C
C
 f ( z )dz   f ( z )dz
-C
C
Note that the value of the contour integrals
depends on the directions of the contour
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40. Contour Integrals
 Properties
 f ( z )dz   f ( z )dz   f ( z )dz
C
C1
C2
The contour C is called the sum of its legs C1 and C2 and is denoted by C1+C2
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41. Some Examples
 Example 1
Let us find the value of the integral I   zdz
C
when C is the right-hand half

i
z  2e , ( 
 /2
I

 /2
 zdz  4 i
C
f ( z ( ))z '( )d 
2
 /2

 

2
)
2ei (2ei ) ' d  4 i
 /2
4
C z dz  4 i
zz
C z dz  4 i
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dz
C z   i
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41. Some Examples
 Example 2
C1 denotes the polygnal line OAB, calculate the integral
I1 
 f ( z )dz


f ( z )dz 
OA
C1

f ( z )dz
AB
Where
f ( z)  y  x  i3x2 ,( z  x  iy)
The leg OA may be represented parametrically as z=0+iy, 0≤y ≤1
In this case, f(z)=yi, then we have

OA
1
f ( z )dz   yidy 
0
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41. Some Examples
 Example 2 (Cont’)
Similarly, the leg AB may be represented parametrically as z=x+i, 0≤x ≤1
In this case, f(z)=1-x-i3x2, then we have

AB
1
f ( z )dz   (1  x  i3x 2 ) 1dx 
0
1
i
2
Therefore, we get
I1 
 f ( z )dz  
OA
C1
f ( z )dz 

f ( z )dz
AB
i
1
1 i
  (  i) 
2 2
2
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41. Some Examples
 Example 2 (Cont’)
C2 denotes the polygonal line OB of the line y=x, with
parametric representation z=x+ix (0≤ x ≤1)
I2 

1
f ( z )dz   i3x 2 (1  i )dx  1  i
C2

0
f ( z )dz 
OABO

C1 -C2
-1+i
f ( z )dz =I1 -I2 =
2
A nonzero value
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41. Some Examples
 Example 3
We begin here by letting C denote an arbitrary smooth arc
z  z (t ),(a  t  b)
from a fixed point z1 to a fixed point z2. In order to calculate
the integral
b
 zdz   z(t ) z '(t )dt
C
a
Please note that
d [ z (t )]2
 z (t ) z '(t )
dt 2
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41. Some Examples
 Example 3 (Cont’)
b
[ z(t )]2
C zdz  a z(t ) z '(t )dt  2
b
a
z(b)2  z(a)2 z22  z12


2
2
The value of the integral is only dependent on the two end points z1 and z2
z2 2  z12
C zdz  z zdz  2
1
z2
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41. Some Examples
 Example 3 (Cont’)
When C is a contour that is not necessarily smooth since a
contour consists of a finite number of smooth arcs Ck
(k=1,2,…n) jointed end to end. More precisely, suppose that
each Ck extend from wk to wk+1, then
n
 zdz    zdz
k 1 Ck
C
n wk 1


k 1 wk
wk 12  wk 2
zdz  
2
k 1
n
wn12  w12 z2 2  z12


2
2
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42. Examples with Branch Cuts
 Example 1
Let C denote the semicircular path z  3ei (0     )
from the point z=3 to the point z = -3.
Although the branch
f ( z)  z
1/2
1
 exp( log z ), (| z | 0, 0  arg z  2 )
2
of the multiple-valued function z1/2 is not defined at the
initial point z=3 of the contour C, the integral
I   z1/ 2 dz
nevertheless exists.
C
Why?
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42. Examples with Branch Cuts
 Example 1 (Cont’)
Note that z ( )  3ei
3
3
f [ z ( )]z '( )  ( 3e )(3ie )  3 3 sin
 i3 3 cos
2
2
At θ=0, the real and imaginary component are 0 and 3 3
i /2
i
0  
Thus f[z(θ)]z’(θ) is continuous on the closed interval 0≤ θ ≤ π when its value at θ=0
is defined as 3 3i

I   z1/2 dz  3 3i  ei 3 /2d
C
0

36
2 i 3 /2
e
3i

0
 2 3(1  i)
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42. Examples with Branch Cuts
 Example 2
Suppose that C is the positively oriented circle
z  Rei (     )
Let a denote any nonzero real number. Using the principal branch
f ( z)  z a1  exp[(a 1) Logz]
(| z | 0,   Argz   )
of the power function za-1, let us evaluate the integral
-R
I   z a 1dz
C
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42. Examples with Branch Cuts
 Example 2 (Cont’)
when z(θ)=Reiθ, it is easy to see that
f [ z( )]z '( )  iRaeia
where the positive value of Ra is to be taken.
Thus, this function is piecewise continuous on -π ≤ θ ≤ π, the
integral exists.

eia 
2Ra
I   z dz  iR  e d  iR [ ]   i
sin a
ia
a
C

a 1
a
ia
a
 If a is a nonzero integer n, the integral becomes 0
 If a is zero, the integral becomes 2πi.
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42. Homework
pp. 135-136
Ex. 2, Ex. 5, Ex. 7, Ex. 8, Ex. 10
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43. Upper Bounds for Moduli of Contour Integrals
 Lemma
If w(t) is a piecewise continuous complex-valued function
defined on an interval a ≤ t ≤b
b
b
a
a
|  w(t )dt |  | w(t ) | dt
Proof:
b
Case #1:
b
 w(t )dt  0
|  w(t )dt |  | w(t ) | dt
a
a
a
b
b
Case #2:
b
r0   ei0 w(t )dt
i0
w
(
t
)
dt

r
e
0
0

a
a
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holds
43. Upper Bounds for Moduli of Contour Integrals
 Lemma (Cont’)
b
r0   ei0 w(t )dt
a
Note that the values in both sizes of this equation are real numbers.
b
b
a
a
r0  Re[ ei0 w(t )dt ]   Re[ei0 w(t )]dt
Re[ei0 w(t )] | ei0 w(t ) || ei0 || w(t ) || w(t ) |
b
b
a
a
r0 |  w(t )dt |  | w(t ) | dt
Why?
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43. Upper Bounds for Moduli of Contour Integrals
 Theorem
Let C denote a contour of length L, and suppose that a
function f(z) is piecewise continuous on C. If M is a
nonnegative constant such that
| f ( z) | M
For all point z on C at which f(z) is defined, then
|  f ( z ) dz | ML
C
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43. Upper Bounds for Moduli of Contour Integrals
 Theorem (Cont’)
Proof: We let z=z(t) (a ≤ t ≤ b) be a parametric
representation of C. According to the lemma, we have
b
b
a
a
|  f ( z)dz ||  f [ z(t )]z '(t )dt |  | f [ z(t )]z '(t ) | dt
C
b
b
a
a
 | f [ z(t )]z '(t ) | dt   M | z '(t ) | dt
| f ( z) | M
b
 M  | z '(t ) | dt  ML
a
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43. Upper Bounds for Moduli of Contour Integrals
 Example 1
Let C be the arc of the circle |z|=2 from z=2 to z=2i that lies
in the first quadrant. Show that
z4
6
|
C
dz |
z 1
7
3
Based on the triangle inequality,
| z  4 || z | 4  6
| z3 1||| z |3 1| 7
Then, we have
z4 6
|
3
z 1 7
And since the length of C is L=π, based on the theorem
z4
6
|  3 dz |
z 1
7
C
|
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43. Upper Bounds for Moduli of Contour Integrals
 Example 2
i
z

Re
,0    
Here CR is the semicircular path
and z1/2 denotes the branch (r>0, -π/2<θ<3π/2)
z
1/2
1
 exp( log z )  rei /2
2
Without calculating the integral, show that
z1/2
lim
dz  0
R   z 2  1
CR
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43. Upper Bounds for Moduli of Contour Integrals
 Example 2 (Cont’)
z1/2
lim  2 dz  0
R 
z 1
CR
Note that when |z|=R>1
| z1/2 || Rei /2 | R
z1/2
R
| 2
| M R  2
z 1
R 1
| z 2  1||| z |2 1| R2 1
Based on the theorem
z1/2
R
|  2 dz |  M R L  2 ( R)
z 1
R 1
CR
46
R
lim
( R)  0
R  R 2  1
School of Software
43. Homework
pp. 140-141
Ex. 3, Ex. 4, Ex. 5
47
School of Software
44. Antiderivatives
 Theorem
Suppose that a function f (z) is continuous on a domain D. If
any one of the following statements is true, then so are the others
a) f (z) has an antiderivative F(z) throughout D;
b) the integrals of f (z) along contours lying entirely in D and
extending from any fixed point z1 to any fixed point z2 all
have the same value, namely
z2

f ( z )dz  F ( z )
z2
z1
 F ( z2 )  F ( z1 )
z1
where F(z) is the antiderivative in statement (a);
c) the integrals of f (z) around closed contours lying entirely in
D all have value zero.
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44. Antiderivatives
 Example 1
The continuous function f (z) = z2 has an antiderivative
F(z) = z3/3 throughout the plane. Hence
1i

0
3
z
z 2 dz 
3
1i
0
2
 (1  i)
3
For every contour from z=0 to z=1+i
49
School of Software
44. Antiderivatives
 Example 2
The function f (z) = 1/z2, which is continuous everywhere
except at the origin, has an antiderivative F(z) = −1/z in the
domain |z| > 0, consisting of the entire plane with the origin
deleted. Consequently,
dz
C z 2  0
Where C is the positively oriented circle
50
School of Software
44. Antiderivatives
 Example 3
Let C denote the circle as previously, calculate the integral
1
I   dz
z
C
It is known that
1
For any given branch (log z ) '  , ( z  0)
z
1
I   dz  0
z
C
?
51
School of Software
44. Antiderivatives
 Example 3 (Cont’)
z  2ei , (
Let C1 denote
The principal branch

2
 

2
)
Logz  ln r  i,(r  0,      )
2i
1
1
C1 z dz  2i zdz  Logz
52
2i
2i
 Log (2i)  Log (2i)   i
School of Software
44. Antiderivatives
 Example 3 (Cont’)
z  2ei , (
Let C2 denote
Consider the function

2
 
3
)
2
logz  ln r  i ,(r  0,0    2 )
Why not Logz?
2i
1
1
C 2 zdz  2i zdz  logz
53
2i
2i
 log (2i)  log (2i)   i
School of Software
44. Antiderivatives
 Example 3 (Cont’)
The value of the integral of 1/z around the entire circle
C=C1+C2 is thus obtained:
dz
dz
dz
C z  C1 z  C2 z   i   i  2 i  0
54
School of Software
44. Antiderivatives
 Example 4
Let us use an antiderivative to evaluate the integral

z1/ 2 dz
C1
where the integrand is the branch
f ( z)  z
1/2
1
 exp( log z )  rei /2 , (r  0, 0    2 )
2
Let C1 is any contour from z=-3 to 3 that, except for
its end points, lies above the X axis.
Let C2 is any contour from z=-3 to 3 that, except for
its end points, lies below the X axis.
55
School of Software
44. Antiderivatives
 Example 4 (Cont’)
1
f ( z )  z1/2  exp( log z )  rei /2 , (r  0, 0    2 )
2
f1  rei /2 , (r  0, 

2
 
3
)
2
f1 is defined and continuous everywhere on C1
2

3
i 3 /2
F1 ( z )  r re
, (r  0,     )
3
2
2

z1/2 dz  F1 ( z )
3
3
 2 3(1  i )
C1
56
School of Software
44. Antiderivatives
 Example 4 (Cont’)
1
f ( z )  z1/2  exp( log z )  rei /2 , (r  0, 0    2 )
2
f 2  rei /2 , (r  0,

2
 
5
)
2
f2 is defined and continuous everywhere on C2
57
School of Software
45. Proof of the Theorem
Basic Idea: (a)  (b)  (c)  (a)
1. (a)  (b)
Suppose that (a) is true, i.e. f(z) has an antiderivative
F(z) on the domain D being considered.
If a contour C from z1 to z2 is a smooth are lying in D,
with parametric representation z=z(t) (a≤ t≤b), since
d
F ( z (t ))  F '[ z (t )]z '(t )  f ( z (t )) z '(t ), (a  t  b)
dt

C
b
f ( z)dz   f ( z (t )) z '(t )dt  F ( z(t ))
b
a
 F ( z(b))  F ( z(a))  F ( z2 )  F ( z1)
a
Note: C is not necessarily a smooth one, e.g. it may contain finite number of smooth arcs.
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School of Software
45. Proof of the Theorem
b) (b) (c)
Suppose the integration is independent of paths, we try
to show that the value of any integral around a closed
contour C in D is zero.
 f ( z )dz  
C1
f ( z )dz
C2
C=C1-C2 denote any integral around a closed contour C in D
 f ( z )dz   f ( z )dz  
C
C1

59
 C2
 f ( z )dz  
C1
f ( z )dz
f ( z )dz  0
C2
School of Software
45. Proof of the Theorem
c) (c)  (a)
Suppose that the integrals of f (z) around closed
contours lying entirely in D all have value zero. Then, we
can get the integration is independent of path in D (why?)
We create the following function
z
F ( z )   f ( s )ds
z0
and try to show that F’(z)=f(z) in D
i.e. (a) holds
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45. Proof of the Theorem
z
F ( z )   f ( s )ds
z0
z z
F ( z  z)  F ( z) 
z

f ( s)ds   f ( s)ds 
z0
z0
z z

f ( s)ds
z
Since the integration is independent of path in D, we consider the path
of integration in a line segment in the following. Since
z z

1
f ( z) 
z
ds  z
z
F ( z  z)  F ( z)
1
 f ( z) 
z
z
61
z z

f ( z )ds
z
z z

[ f (s)  f ( z )]ds
z
School of Software
45. Proof of the Theorem
Please note that f is continuous at the point z, thus, for each positive number ε,
a positive number δ exists such that
When
| s  z | 
| f (s)  f ( z ) | 
Consequently, if the point z+Δz is close to z so that | Δ z| <δ, then
F ( z  z)  F ( z)
1
|
 f ( z ) ||
z
z
z z

z
1
[ f (s)  f ( z )]ds |
 | z | 
| z|
F ( z  z)  F ( z)
F '( z )  lim
 f ( z)
z 0
z
62
School of Software
45. Homework
pp. 149
Ex. 2, Ex. 3, Ex. 4
63
School of Software
46. Cauchy-Goursat Theorem
 Cauchy-Goursat Theorem
 Give other conditions on a function f which ensure that
the value of the integral of f(z) around a simple closed
contour is zero.
 The theorem is central to the theory of functions of a
complex variable, some modification of it, involving
certain special types of domains, will be given in
Sections 48 and 49.
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School of Software
46. Cauchy-Goursat Theorem
Let C be a simple closed contour z=z(t) (a≤t ≤b) in the positive sense,
and f is analytic at each point. Based on the definition of the
contour integrals
b
 f ( z)dz   f ( z(t )) z '(t )dt
C
a
And if f(z)=u(x,y)+iv(x,y) and z(t)=x(t) + iy(t)
b
b
b
a
a
a
 f ( z)dz   f ( z(t )) z '(t )dt   (ux ' vy ')dt  i  (vx ' uy ')dt
C
  udx  vdy  i  vdx  udy
C
C
65
School of Software
46. Cauchy-Goursat Theorem
Based on Green’s Theorem, if the two real-valued functions P(x,y) and Q(x,y), together
with their first-order partial derivatives, are continuous throughout the closed region R
consisting of all points interior to and on the simple closed contour C, then
 Pdx  Qdy   (Q
x
C
 Py )dA
R
 f ( z )dz   udx  vdy  i  vdx  udy
C
C
 (v
x
C
 u y )dA,( P  u, Q  v)
R
 (u
x
 vy )dA,( P  v, Q  u)
R
If f(z) is analytic in R and C, then the Cauchy-Riemann equations shows that
u y  vx , ux  vy
Both become zeros
 f ( z )dz  0
C
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School of Software
46. Cauchy-Goursat Theorem
 Example
If C is any simple closed contour, in either direction,
then
3
exp(
z
) dz  0

C
This is because the composite function f(z)=exp(z3) is
analytic everywhere and its derivate f’(z)=3z2exp(z3) is
continuous everywhere.
67
School of Software
46. Cauchy-Goursat Theorem
 Two Requirements described previously
 The function f is analytic at all points interior to and on
a simple closed contour C, then
 The derivative f’ is continuous there
 Goursat was the first to prove that the condition of
continuity on f’ can be omitted.
68
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46. Cauchy-Goursat Theorem
 Cauchy-Goursat Theorem
If a function f is analytic at all points interior to and on a
simple closed contour C, then
 f ( z )dz  0
Interior of C (bounded)
C
69
School of Software
48. Simply Connected Domains
 Simple Connected domain
A simple connected domain D is a domain such that
every simple closed contour within it encloses only points
of D. For instance,
A simple
connected domain
The set of points interior to
a simple closed contour
Not a simple connected domain
70
School of Software
48. Simply Connected Domains
 Theorem 1
If a function f is analytic throughout a simply connected
domain D, then
 f ( z )dz  0
C
for every closed contour C lying in D.
Basic idea: Divide it into finite simple
closed contours. For this example,
4
 f ( z)dz    f ( z)dz  0
C
71
k 1 Ck
School of Software
48. Simply Connected Domains
 Example
If C denotes any closed contour lying in the open disk
|z|<2, then
ze z
C ( z 2  9)5 dz  0
This is because the disk is a simply
connected domain and the two
singularities z = ±3i of the integrand
are exterior to the disk.
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48. Simply Connected Domains
 Corollary
A function f that is analytic throughout a simply
connected domain D must have an antiderivative
everywhere in D.
Refer to the theorem in Section 44 (pp.142), (c)(a)
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49. Multiply Connected Domains
 Multiply Connected Domain
A domain that is not simple connected is said to be
multiply connected. For instance,
Multiply connected domain
Multiply connected domain
The following theorem is an adaptation of the Cauchy-Goursat
theorem to multiply connected domains.
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49. Multiply Connected Domains
 Theorem
Suppose that
(a) C is a simple closed contour, described in the counterclockwise direction;
(b) Ck (k = 1, 2, . . . , n) are simple closed contours interior to C, all described in
the clockwise direction, that are disjoint and whose interiors have no points
in common.
If a function f is analytic on all of these contours and throughout the multiply
connected domain consisting of the points inside C and exterior to each Ck, then
n
 f ( z)dz    f ( z)dz  0
C
k 1 Ck
Main Idea:
Multiple  Finite Simple connected domains
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49. Multiply Connected Domains
 Corollary
Let C1 and C2 denote positively oriented simple closed
contours, where C1 is interior to C2. If a function f is
analytic in the closed region consisting of those contours
and all points between them, then

C2
76
f ( z )dz 
 f ( z )dz
C1
School of Software
49. Multiply Connected Domains
 Example
When C is any positively oriented simple closed
contour surrounding the origin, the corollary can be used
to show that
dz
C z  2 i
For a positively oriented circle C0 with center at the original
dz
C z  2 i
0
 f ( z )dz  
C
pp. 136 Ex. 10
f ( z )dz  2 i
C0
77
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49. Homework
pp. 160-163
Ex. 1, Ex. 2, Ex. 3, Ex. 7
78
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50. Cauchy Integral Formula
 Theorem
Let f be analytic everywhere inside and on a simple closed contour
C, taken in the positive sense. If z0 is any point interior to C, then
1
f ( z)
f ( z0 ) 
dz Cauchy Integral Formula

2 i C z  z0
which tells us that if a function f is to be analytic within and on a
simple closed contour C, then the values of f interior to C are
completely determined by the values of f on C.
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50. Cauchy Integral Formula
f ( z0 ) 
1
f ( z)
dz

2 i C z  z0
f ( z)
C z  z0 dz  2 if ( z0 )
Proof:
Let Cρ denote a positively oriented circle |z-z0|=ρ, where ρ is small enough that Cρ
is interior to C , since the quotient f(z)/(z-z0) is analytic between and on the contours
Cρ and C, it follows from the principle of deformation of paths

C
f ( z )dz
f ( z )dz

z  z0 C z  z0
This enables us to write

C
pp. 136 Ex. 10
2πi
f ( z)dz
dz
f ( z )dz
dz
 f ( z0 ) 

 f ( z0 ) 
z  z0
z  z0 C z  z0
z  z0
C
C


C
[ f ( z)  f ( z0 )]dz
f ( z )dz
2 if ( z0 )  
z  z0
z  z0
C
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50. Cauchy Integral Formula

C
f ( z)  f ( z0 )dz
f ( z )dz
2 if ( z0 )  
z  z0
z  z0
C
Now the fact that f is analytic, and therefore continuous, at z0 ensures
that corresponding to each positive number ε, however small, there is a
positive number δ such that when |z-z0|< δ
| f ( z)  f ( z0 ) | 
|
C
f ( z )  f ( z0 )dz
f ( z )dz

2 if ( z0 ) || 
| ( )(2 )  2
z  z0
z  z0

C

C
f ( z )dz
2 if ( z0 )
z  z0
The theorem is proved.
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50. Cauchy Integral Formula
f ( z0 ) 
1
f ( z)
dz

2 i C z  z0
f ( z)
C z  z0 dz  2 if ( z0 )
This formula can be used to evaluate certain integrals along simple closed contours.
Example
Let C be the positively oriented circle |z|=2, since the function
f ( z) 
z
9  z2
is analytic within and on C and since the point z0=-i is interior to C,
the above formula tells us that
z
z /(9  z 2 )
i 
C (9  z 2 )( z  i) dz  C z-(-i) dz  2 i(10 )  5
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51. An Extension of the Cauchy Integral Formula
The Cauchy Integral formula can be extended so as to provide an
integral representation for derivatives of f at z0
f ( z)
2 i ( n )
C ( z  z0 )n1 dz  n! f ( z0 )
n  0,1, 2,...
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51. An Extension of the Cauchy Integral Formula
 Example 1
If C is the positively oriented unit circle |z|=1 and
f ( z )  exp(2 z )
then
exp(2 z )
f ( z)
2 i
8 i
C z 4 dz  C ( z  0)31 dz  3! f '''(0)  3
84
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51. An Extension of the Cauchy Integral Formula
 Example 2
Let z0 be any point interior to a positively oriented
simple closed contour C. When f(z)=1, then
dz
C z  z0  2 i
And
dz
C ( z  z0 )n1  0, n  1, 2,...
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52. Some Consequences of the Extension
 Theorem 1
If a function f is analytic at a given point, then its derivatives of all
orders are analytic there too.
f
(n)
n!
f ( z)
( z0 ) 
dz
n 1

2 i C ( z  z0 )
n  0,1, 2,...
 Corollary
If a function f (z) = u(x, y) + iv(x, y) is analytic at a point z = (x, y),
then the component functions u and v have continuous partial
derivatives of all orders at that point.
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52. Some Consequences of the Extension
 Theorem 2
Let f be continuous on a domain D. If
 f ( z )dz  0
C
For every closed contour C in D. then f is analytic
throughout D
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52. Some Consequences of the Extension
 Theorem 3
Suppose that a function f is analytic inside and on a
positively oriented circle CR, centered at z0 and with
radius R. If MR denotes the maximum value of |f (z)| on
CR, then
n!M R
(n)
| f ( z0 ) |
(n  1, 2,...)
n
R
The Cauchy’s Inequality
88
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52. Homework
pp.170-172
Ex. 2, Ex. 4, Ex. 5, Ex. 7
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