Matematika Teknik 1-3a
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Transcript of Matematika Teknik 1-3a
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Complex Integration
Engineering Mathematics 1 - 02
1
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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
Liouvilles Theorem and the Fundamental Theorem
Maximum Modulus Principle
2
Chapter 4: Integrals
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Consider derivatives of complex-valued functions w of
real variable t
where the function u and v are real-valued functions oft.
The derivative
of the function w(t) at a point tis defined as
Derivatives of Functions w(t)
3
w(t)= u(t)+ iv(t)
w '(t),ord
dtw(t)
w '(t)= u'(t)+ iv '(t)
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Properties
For any complex constant z0=x0+iy0,
Derivatives of Functions w(t)
4
d
dt
[z0w(t)]= [(x0 + iy0 )(u+ iv)]' = [(x0u - y0v)+ i(y0u+x0v)]'
= (x0u - y0v)'+ i(y0u+x0v)'
= (x0u'- y0v ')+ i(y0u'+x0v ')
= (x0 + iy0 )(u'+ iv')=z0w '(t)
u(t) v(t)
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Properties
Derivatives of Functions w(t)
5
d
dtez0t= z
0ez0t
where z0=x0+iy0. We write
0 0 0 0 0( )
0 0cos sinz t x iy t x t x te e e y t ie y t
u(t) v(t)
0 0 0
0 0( cos ) ' ( sin ) 'z t x t x td e e y t i e y t
dt
Similar rules from calculus and some simple algebra then lead us to the expression
0 0 0 0( )
0 0 0( )z t x iy t z td e x iy e z e
dt
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Example
Suppose that the real functionf(t) is continuous on an
interval a t b, if f(t) exists when a
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Example (Cont)
The mean value theorem no longer applies for some
complex functions. For instance, the function
on the interval 0 t 2 .
Please note that
And this means that the derivative w(t) is never zero, while
Derivatives of Functions w(t)
7
w(t)= eit
|w '(t) |=| ieit |=1
(2 ) (0) 0w w (2 ) (0)
'( ) 0, 0 22 0
w w
w
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When w(t) is a complex-valued function of a real variable
t and is written
where u and v are real-valued, the definite integral of
w(t) over an interval a t b is defined as
Definite Integrals of Functions w(t)
8
w(t)= u(t)+ iv(t)
( ) ( ) ( )b b b
a a a
w t dt u t dt i v t dt Provided the individual integrals on the right exist.
Re ( ) Re ( ) & Im ( ) Im ( )b b b b
a a a a
w t dt w t dt w t dt w t dt
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Example 1
Definite Integrals of Functions w(t)
9
1 1 1
2 2
0 0 0
2(1 ) (1 ) 2
3it dt t dt i tdt i
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Integral vs. Anti-derivative
Suppose that
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
Definite Integrals of Functions w(t)
11
( ) ( ) ( ), ( ) ( ) ( )w t u t iv t W t U t iV t
( ) ( ) ( )b b b
a a a
w t dt u t dt i v t dt ( ) ( ) [ ( ) ( )] [ ( ) ( )]b ba aU t iV t U b iV b U a iV a
( ) ( ) ( ) baW b W a W t
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Example 2
Since
one can see that
Definite Integrals of Functions w(t)
12
1 1( )
it
it it it d e de ie e
dt i i dt i
4
4
0
0
it
it ee dt
i
4 1
i
e
i i
1 1(1 )2 2
i
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Example 3Let w(t) be a continuous complex-valued function oftdefined on an
interval a t b. In order to show that it is not necessarily true that
there is a numberc in the interval a
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Arc
A set of points z=(x, y) in the complex plane is said to be an
arc if
where x(t) and y(t) are continuous functions of the realparametert. 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 oft. It is convenient to describe the points of C bymeans of the equation
Contours
14
( ) & ( ),x x t y y t a t b
( ) ( ) ( )z t x t iy t
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Example 1
The polygonal line defined by means of the equations
and consisting of a line segment from 0 to 1+i followed
by one from 1+i to 2+i is a simple arc
Contours
16
, 0 1
, 1 2
x ix if xz
x i if x
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Example 2~4
The unit circle
about the origin is a simple closed curve, oriented in the
counterclockwise direction.
So is the circlecentered at the point z0 and with radius R.
The set of points
This unit circle is traveled in the clockwise direction. The set of point
This unit circle is traversed twice in the counterclockwise
direction.
Contours
17
,(0 2 )iz e
0 , (0 2 )i
z z e
, (0 2 )iz e
2 ,(0 2 )iz e
Note: the same set of points can make up different arcs.
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The parametric representation used for any given
arc C is not unique
To be specific, suppose that
where is a real-valued function mapping an interval
onto a t b.
Contours
18
( ),t
Here we assume is a continuous functions with a continuous
derivative, and ()>0 for each (why?)
: ( ) ( ( )),C Z z
: ( ),C z t a t b
The same arc C
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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
is integrable over the interval a t b.
In fact, according to the definition of a length in calculus, thelength of C is the number
Contours
19
2 2| '( ) | [ '( )] [ '( )]z t x t y t
| '( ) |b
a C
L z t dt ds Note: The value L is invariant under certain changes in the representation for C.
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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 contourWhen 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)
Contours
20
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Jordan Curve Theorem
Contours
21
Refer to: http://en.wikipedia.org/wiki/Jordan_curve_theorem
Jordan Curve Theorem asserts that
every Jordan curve divides the plane intoan "interior" region bounded by the curve
and an "exterior" region containing all of
the nearby and far away exterior points.
Interior of C (bounded)
Jordan curve
Exterior of C (unbounded)
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Consider the integrals of complex-valued functionfof
the complex variable z on a given contour C, extendingfrom a point z=z1 to a point z=z2 in the complex plane.
Contour Integrals
22
( )C
f z dz
or 2
1
( )
z
z
f z dz
When the value of the integral is independent of the choice
of the contour taken between two fixed end points.
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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 parametertas follows
Contour Integrals
23
( ) ( ( )) '( )b
C a
f z dz f z t z t dt
Contour integral
Note the value of a contour integral is invariant under a change
in the representation of its contour C.
On the integral [a b] as defined previously
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Properties
Contour Integrals
24
0 0( ) ( )C C
z f z dz z f z dz
[ ( ) ( )] ( ) g( )C C C
f z g z dz f z dz z dz
-
( ) ( )C C
f z dz f z dz
Note that the value of the contour integrals
depends on the directions of the contour
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S E l
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Example 2
C1 denotes the polygnal line OAB, calculate the integral
Some Examples
27
1
1 ( )C
I f z dz ( ) ( )OA AB
f z dz f z dz
Where2( ) 3 ,( )f z y x i x z x iy
The leg OA may be represented parametrically as z=0+iy, 0y 1
1
0
( )2
OA
if z dz yidy
In this case, f(z)=yi, then we have
S E l
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Example 2 (Cont)
Some Examples
28
1
1 ( )C
I f z dz ( ) ( )OA AB
f z dz f z dz
Similarly, the leg AB may be represented parametrically as z=x+i, 0x 1
In this case, f(z)=1-x-i3x2, then we have
1 2
0
1( ) (1 3 ) 12
AB
f z dz x i x dx i
Therefore, we get
1 1( )
2 2 2
i ii
S E l
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Example 2 (Cont)
C2 denotes the polygonal line OB of the line y=x, with
parametric representation z=x+ix (0 x 1)
Some Examples
29
2
1
22
0
( ) 3 (1 ) 1C
I f z dz i x i dx i
1 2
1 2
OABO -C
-1+i( ) ( ) =I -I =
2C
f z dz f z dz
A nonzero value
S E l
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Example 3
We begin here by letting C denote an arbitrary smooth arc
from a fixed point z1 to a fixed point z2. In order to calculate
the integral
Some Examples
30
( ), ( )z z t a t b
( ) '( )b
C a
zdz z t z t dt
Please note that
2[ ( )]( ) '( )
2
d z tz t z t
dt
S E l
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Example 3 (Cont)
Some Examples
31
2 22 2 2
2 1[ ( )] ( ) ( )( ) '( )2 2 2
b
b
a
C a
z zz t z b z azdz z t z t dt
The value of the integral is only dependent on the two end points z1 and z2
2
1
2 2
2 1
2
z
C z
z zzdz zdz
S E l
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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 Ckextend from wkto wk+1, then
Some Examples
32
1k
n
kC C
zdz zdz
1 2 2
1
1 1 2
k
k
wn n
k k
k kw
w w
zdz
2 2 2 2
1 1 2 1
2 2nw w z z
E l ith B h C t
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Example 1
Let C denote the semicircular path
from the point z=3 to the point z = -3.
Although the branch
of the multiple-valued function z1/2 is not defined at the
initial point z=3 of the contour C, the integral
Examples with Branch Cuts
33
3 (0 )iz e
1/2 1( ) exp( log ), (| | 0, 0 arg 2 )2
f z z z z z
1/2
C
I z dz nevertheless exists.Why?
E l ith B h C t
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Example 1 (Cont)
Note that
Examples with Branch Cuts
34
( ) 3 iz e
/ 2 3 3[ ( )] '( ) ( 3 )(3 ) 3 3 sin 3 3 cos2 2
i if z z e ie i
0
At =0, the real and imaginary component are 0 and3 3
Thus f[z()]z() is continuous on the closed interval 0 when its value at =0
is defined as 3 3i
1/2 3 /2
0
3 3 i
C
I z dz i e d
3 /2
0
22 3(1 )
3
ie ii
Examples with Branch Cuts
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Example 2
Suppose that C is the positively oriented circle
Examples with Branch Cuts
35
Re ( )iz
-R
Let a denote any nonzero real number. Using the principal branch
1( ) exp[( 1) ]af z z a Logz (| | 0, )z Argz
of the power function za-1, let us evaluate the integral
1a
C
I z dz
Examples with Branch Cuts
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Example 2 (Cont)
when z()=Rei, it is easy to see that
where the positive value of Ra is to be taken.
Thus, this function is piecewise continuous on - , theintegral exists.
Ifa is a nonzero integer n, the integral becomes 0
Ifa is zero, the integral becomes 2i.
Examples with Branch Cuts
36
[ ( ) ] ' ( ) a iaf z z iR e
1 2[ ] sinia a
a a ia a
C
e RI z dz iR e d iR i a
ia a
Antiderivatives
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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) throughoutD;
b) the integrals of f (z) along contours lying entirelyin D and
extending from any fixed point z1 to any fixed point z2 all
have the same value, namely
where F(z) is the antiderivative in statement (a);
c) the integrals of f (z) around closed contours lying entirelyin D
all have value zero.
Antiderivatives
37
2
2
1
1
2 1( ) ( ) ( ) ( )
z
z
z
z
f z dz F z F z F z
Antiderivatives
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Example 1
The continuous function f (z) = z2 has an antiderivative
F(z) = z3/3 throughout the plane. Hence
For every contour from z=0 to z=1+i
Antiderivatives
38
1 3
2 10
0
2 ( 1 )3 3
i
iz
z dz i
Antiderivatives
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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,
Where C is the positively oriented circle
Antiderivatives
39
20
C
dz
z
Antiderivatives
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Example 3
Let C denote the circle as previously, calculate the integral
It is known that
Antiderivatives
40
1
C
I dzz
1(log ) ' , ( 0)z z
z
10
C
I dzz
?
For any given branch
Antiderivatives
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Example 3 (Cont)
Let C1 denote
The principal branch
Antiderivatives
41
2 , ( )2 2
iz e
ln , ( 0, )Logz r i r
2
2
2
1 2
1 1(2 ) ( 2 )
i
i
i
C i
dz dz Logz Log i Log i iz z
Antiderivatives
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Example 3 (Cont)
Let C2 denote
Consider the function
Antiderivatives
42
32 , ( )2 2
iz e
ln , ( 0, 0 2 )logz r i r
2
2
2
2 2
1 1(2 ) ( 2 )
i
i
i
C i
dz dz logz log i log i i
z z
Why not Logz?
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Antiderivatives
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Example 4
Let us use an antiderivative to evaluate the integral
where the integrand is the branch
Antiderivatives
44
1/2
1C
z dz
1/2 /21( ) exp( log ) , ( 0, 0 2 )2
if z z z re r
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.
Antiderivatives
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Example 4 (Cont)
Antiderivatives
45
/2
13, ( 0, )
2 2
if re r
1/2 /21( ) exp( log ) , ( 0, 0 2 )2
if z z z re r
f1 is defined and continuous everywhere on C1
3 /21 2 3( ) , ( 0, )
3 2 2iF z r re r
1/2 3
1 3
1
( ) 2 3(1 )C
z dz F z i
Antiderivatives
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Example 4 (Cont)
Antiderivatives
46
/2
25, ( 0, )
2 2
if re r
1/2 /21( ) exp( log ) , ( 0, 0 2 )2
if z z z re r
f2 is defined and continuous everywhere on C2
Proof of the Theorem
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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 tb), since
Proof of the Theorem
47
( ( )) '[ ( )] '( ) ( ( )) '( ), ( )d
F z t F z t z t f z t z t a t bdt
2 1( ) ( ( )) '( ) ( ( )) ( ( )) ( ( )) ( ) ( )b
b
a
C a
f z dz f z t z t dt F z t F z b F z a F z F z
Note: C is not necessarily a smooth one, e.g. it may contain finite number of smooth arcs.
Proof of the Theorem
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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.
Proof of the Theorem
48
1 2
( ) ( )C C
f z dz f z dz
1 2
( ) ( ) ( )C C C
f z dz f z dz f z dz
C=C1-C2 denote any integral around a closed contour C in D
1 2
( ) ( ) 0C C
f z dz f z dz
Proof of the Theorem
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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?)
Proof of the Theorem
49
0
( ) ( )z
z
F z f s ds
We create the following function
and try to show that F(z)=f(z) in D
i.e. (a) holds
Proof of the Theorem
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Proof of the Theorem
50
0
( ) ( )z
z
F z f s ds
0 0
( ) ( ) ( ) ( ) ( )z z z z z
z z z
F z z F z f s ds f s ds f s ds
z z
z
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
1( ) ( )
z z
z
f z f z dsz
( ) ( ) 1( ) [ ( ) ( )]
z z
z
F z z F zf z f s f z ds
z z
Proof of the Theorem
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Proof of the Theorem
51
Please note thatfis continuous at the point z, thus, for each positive number,
a positive number exists such that
| ( ) ( ) |f s f z When | |s z
( ) ( ) 1 1| ( ) | | [ ( ) ( )] | | |
| |
z z
z
F z z F zf z f s f z ds z
z z z
Consequently, if the point z+z is close to z so that | z|
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Cauchy-Goursat Theorem
Give other conditions on a functionfwhich 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 acomplex variable, some modification of it, involving
certain special types of domains, will be given in
Sections 48 and 49.
Cauchy Goursat Theorem
52
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Cauchy-Goursat Theorem
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Cauchy Goursat Theorem
54
Based on Greens 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
( )x yC R
Pdx Qdy Q P dA
( )C C C
f z dz udx vdy i vdx udy
( ) , ( , )x y
R
v u dA P u Q v ( ) , ( , )x yR
u v dA P v Q u
If f(z) is analytic in R and C, then the Cauchy-Riemann equations shows that
,y x x y
u v u v
Both become zeros
( ) 0C
f z dz
Cauchy-Goursat Theorem
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Example
If C is any simple closed contour, in either direction,
then
This is because the composite function f(z)=exp(z3) is
analytic everywhere and its derivate f(z)=3z2exp(z3) is
continuous everywhere.
Cauchy Goursat Theorem
55
3exp( ) 0
C
z dz
Cauchy-Goursat Theorem
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Two Requirements described previously
The functionfis analytic at all points interior to and on
a simple closed contour C, then
The derivativefis continuous there
Goursat was the first to prove that the condition of
continuity onfcan be omitted.
Cauchy Goursat Theorem
56
Cauchy-Goursat Theorem
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Cauchy-Goursat Theorem
If a functionfis analytic at all points interiorto and on a
simple closed contour C, then
Cauchy Goursat Theorem
57
Interior of C (bounded) ( ) 0C
f z dz
Cauchy Integral Formula
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Theorem
Letfbe analytic everywhere inside and on a simple closed contourC, taken in the positive sense. If z0 is any point interior to C, then
which tells us that if a function f is to be analytic within and on a
simple closed contour C, then the values offinterior to C are
completely determined by the values of f on C.
Cauc y teg a o u a
58
0
0
1 ( )( )
2C
f zf z dz
i z z
Cauchy Integral Formula
Cauchy Integral Formula
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y g
59
0
0
1 ( )( )
2C
f zf z dz
i z z
Proof:
Let C denote a positively oriented circle |z-z0|=, where is small enough that Cis interior to C , since the quotient f(z)/(z-z0) is analytic between and on the contours
C and C, it follows from the principle of deformation of paths
0 0
( ) ( )
C C
f z dz f z dz
z z z z
This enables us to write
0 0
0 0 0 0
( ) ( )( ) ( )
C C C C
f z dz dz f z dz dzf z f zz z z z z z z z
2i
00
0 0
[ ( ) ( )]( )2 ( )
C C
f z f z dzf z dzif z
z z z z
0
0
( )2 ( )
C
f zdz if z
z z
pp. 136 Ex. 10
Cauchy Integral Formula
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y g
60
00
0 0
( ) ( )( )2 ( )
C C
f z f z dzf z dzif z
z z z z
Now the fact that f is analytic, and therefore continuous, at z0 ensures
that corresponding to each positive number, however small, there is a
positive number such that when |z-z0|<
0| ( ) ( ) |f z f z
00
0 0
( ) ( )( )| 2 ( ) | | | ( )(2 ) 2
C C
f z f z dzf z dzif z
z z z z
0
0
( )2 ( )
C
f z dzif z
z z
The theorem is proved.
Cauchy Integral Formula
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y g
61
0
0
( )2 ( )
C
f zdz if z
z z
This formula can be used to evaluate certain integrals along simple closed contours.
0
0
1 ( )( )
2 C
f zf z dz
i z z
Example
Let C be the positively oriented circle |z|=2, since the function
2( )
9
zf z
z
is analytic within and on C and since the point z0=-i is interior to C,
the above formula tells us that2
2
/(9 )2 ( )
(9 )( ) -(-i) 10 5C C
z z z idz dz i
z z i z
An Extension of the Cauchy Integral Formula
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y g
62
( )
010
( ) 2( )
( ) !
n
nC
f z idz f z
z z n
0,1,2,...n
The Cauchy Integral formula can be extended so as to provide an
integral representation for derivatives of f at z0
An Extension of the Cauchy Integral Formula
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y g
63
Example 1
If C is the positively oriented unit circle |z|=1 and
then
( ) exp(2 )f z z
4 3 1
exp(2 ) ( ) 2 8'''(0)
( 0) 3! 3C C
z f z i idz dz f
z z
An Extension of the Cauchy Integral Formula
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Example 2
Let z0be any point interior to a positively oriented
simple closed contour C. When f(z)=1, then
And
y g
64
0
2C
dzi
z z
1
0
0, 1, 2,...
( )n
C
dzn
z z
Some Consequences of the Extension
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Theorem 1
If a function f is analytic at a given point, then its derivatives of allorders are analytic there too.
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.
q
65
( )
0 10
! ( )
( ) 2 ( )
n
nC
n f z
f z dzi z z 0,1,2,...n
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