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  • 7/27/2019 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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