chapter 3 integral of complex function §3.1 definition and properties §3.2 cauchy integral theorem...
TRANSCRIPT
Chapter 3 Integral of Complex Function
§3.1 Definition and Properties
§3.2 Cauchy Integral Theorem
§3.3 Cauchy’s Integral Formula
§3.4 Analytic and Harmonic Functions
Review
1
( ) lim ( )nb
i ia ni
f x dx f x
1
( , ) lim ( , )n
i i in
iD
f x y d f
1
( , , ) lim ( , , )n
i i i ini
f x y z dV f V
1 1
( , ) ( , ) lim[ ( , ) ( , ) ]n n
i i i i i iL ni i
P x y dx Q x y dy P x Q y
n
iiiii
nSfdSzyxf
1
),,(lim),,(
1
( , ) lim ( , )n
i i iL ni
f x y ds f s
§3.1 Definition and Properties
0 1 2 1, , , , , , ,k k nz z z z z z
1 ( 1, 2, , )k k kz z k n
11 1
( )( ) ( ) , n n
n k k k k kk k
S f z z f z
1where k k kz z z
11
Let be the length of arc , max{ }k k k kk n
S z z S
( 0)
when & 0, lim exist.nn
n S
1. Def. C smooth (or piecewise smooth)
f : C→C.
( ) integrable on , the limit is called the integral
of ( ) along .
f z C
f z C
C1( 0)
Denote ( )d lim ( ) (3.1.1)n
k kn
k
f z z f z
C —— path of integral f —— integrandZ —— integration variable
the limit is called the integral of f along C,
denoted by .
If C is closed, we can write .
C( )df z z
C( )df z z
2. EvaluationLet i , ( ) ( , ) i ( , ), =1,2,k k k k k k k ky f u y v y k n
1 1 1i ( i )k k k k k k kz z z x y x y
1 1( ) i( ) ik k k k k kx x y y x y
1 1
( ) [ ( , ) i ( , )]( i )n n
k k k k k k k kk k
f z u v x y
1 1
[ ( , ) ( , ) ] i [ ( , ) ( , ) ]n n
k k k k k k k k k k k kk k
u x v y v x u y
1
d d lim [ ( , ) ( , ) ]n
k k k k k kC nk
u x v y u x v y
1
d d lim [ ( , ) ( , ) ]n
k k k k k kC nk
u x v y u x v y
TH 3.1.1 ( ) ( , ) i ( , ) integral on Cf z u x y v x y ( )d d d i d d
C C Cf z z u x v y v x u y (3.1.2
)Corollary: ( )is integral on C if ( ) is continuous on smooth or piecewise smooth arc C.
f z f z
( ) ( ) i ( ) [3.1.3], :z z t x t y t t
( )d ( ), ( ) '( ) ( ), ( ) '( ) dC
f z z u x t y t x t v x t y t y t t
i ( ), ( ) '( ) ( ), ( ) '( ) dv x t y t x t u x t y t y t t
( ), ( ) i ( ), ( ) ( '( ) i '( ))du x t y t v x t y t x t y t t
( ( ) '( )df z t z t t
(3.1.4)
Ex.1 010
d, C: ( 0), ,
( )nC
zz z r r n N
z z
C is counterwise clock.
010
2 , if 0d is independent of , .
0, if 0( )nC
i nzz r
nz z
center of a circle
radius
0
2
1 1100
2 2
0 0
parameter equation of is: ,0 2 ,i
i
n i nnc
inn in n
C z z re
dz ired
r ez z
i id e d
r e r
3. Properties
① ( )d ( )dC C
f z z f z z
② ( )d ( )d
C Ckf z z k f z z
③ ( ) ( ) d ( )d ( )dC C C
f z g z z f z z g z z ④
1 2 1 2
( )d ( )d ( )dC C C C
f z z f z z f z z
⑤ ( )d ( ) ds ML, ( ) M on C
C Cf z z f z f z
the length of C(Pf. P38)
⑥ ( )d independent of parametric representation of CC
f z z
①2
C : , [0,1]x t
ty t
1 20
2 11iI [2 i ] (2 i)d
3t t t
② C : , [0,2]2
x ttty
2 20
1 2 11iI [ i ] (1 i)d
2 2 3
tt t
2I dC
z zEx.
③ 1 2 1C=C C C : , [0, 2], 0 i0
x xx z x
y
2
2C : , [0,1], 2 i
xy z y
y y
1 20
2 11iI [2 i ] (2 i)d
3t t t
2 2 1 20 0
8 11i 2 11id (2 i) id 2
3 3 3x x y y
Note: I independent of integration path.
Ex.3.1.21 2
1Evaluate
1z
zI dz
z
On circle 1 2,
1 1 2 1( 1 2) 2.
1 2 2
the length of circle 1 2 is 4 ,
8 .
z
z zz
z
z L
I LM
Ex.3.1.32
2
Evaluate ( ) , :
(1). (0,1) (1,2) along 1;
(2). (0,1) (1,1) (1,2), .
Cz dz C
A B y x
A N B ANB
2
2 2
1 22 2
0
1 14 2 5 3
0 0
(1).The parametric equation of 1is:
( 1), : 0 1, ( 1), (1 2 ) ,
( ) ( 1) (1 2 )
(3 3 1) ( 2 4 4 )
3 10.
5 3
C
y x
z x i x x z x i x dz xi dx
z dz x i x xi dx
x x dx i x x x dx
i
1 2
12 2 2 2
0
2 2
1
2 .The parametric equation of is: , : 0 1,
the parametric equation of is: 1 , :1 2,
( ) ( ) ( ) ( 1 2 )
7 7[2 (1 ) ] .
3 3
C C C
AN z x i x
NB z iy y
z dz z dz z dz x xi dx
y y i dy i
Note: integration of f(z) dependent on integration path.
§3.2 Cauchy Integral Theorem
d d indepentent of integration pathC
P x Q yP Q
y x
—— continuous
( )d d d i d dC C C
f z z u x v y v x u y v u
y x
v u
x y
C-R equation
TH.3.2.1(Cauchy TH)
( )analytic on (simply connected domain)
( ) 0, .C
f z D
f z dz whereC D
TH 3.2.2 P42
1
1 2 0
( )d ( )d ( )dz
C C zf z z f z z f z z
upper limit
lower limit
0
( ) ( )d -primitive function antiderivativez
zF z f z z
( )analytic on (simply connected domain)
( ) along on is independent of integral path,
and it is only determined by starting point and end point.
then:
C
f z D
f z dz C D
TH 3.2.3 P45
0
( ) analytic on (simply connected domain)
( ) ( )d is analytic function in ,and '( ) ( ).z
z
f z u iv D
F z f z z D F z f z
0 0 0 0 0
0 0 0 0
( , ) ( , )
( , ) ( , )
( , ) ( , )
( , ) ( , )
Pf: ( ) ( )d
( , ) ( , ),
where ( , ) , ( , ) .
integration of ( )is independent of integration path,
so ar
z x y x y
z x y x y
x y x y
x y x y
F z f z z udx vdy i vdx udy
P x y iQ x y
P x y udx vdy Q x y vdx udy
f z
e ( , ) ( , ),
, , , , , .
( ) ( , ) ( , )is analytic function,
and '( ) ( ).
x y x y x y y x
x x
P x y and Q x y
P u P v Q v Q u P Q P Q
F z P x y iQ x y
F z P iQ u iv f z
Def.3.2.1
Properties
① G anti derivative of f on D G analytic on D
② G1and G2 anti derivative of f on D
G1=G2 +constant on D.
TH 3.2.5 (Fundamental Theorem of Contour Integral)
( ) is anti derivative of ( ) on , if '( ) ( ) on .G z f z D G z f z D
0 1
( )analytic on (simply connected domain),
'( ) ( ) on , , ,
f z D
G z f z D z z 0
1 0( )d ( ) ( )z
zf z z G z G z
Ex.3.2.15sin( 2 1)
23 1
.1
z z
z
eI dz
z
5sin( 2 1)
2
The singular points are outside 3 1,
so is analytic on and in 3 1,1
.3.2.1 0.
z z
z i z
ez
zTH I
Ex.3.2.22 2 2
1
1(1). ; (2). along the right curve.
i i
i iz dz dz
z
3
2 2 2 2 2 3 311
1 14(1). | [(2 2 ) (1 ) ] ( 1 ).
3 3 3
i iii
zz dz i i i
(2).Log function is multivalued,analytic on except , 0.
1( )
1| (ln arg 2 ) |
[arg( ) arg( )] .
k
i i ii ii
z x x
dLnz
dz z
dz Lnz z i z k iz
i i i i
C
Generalized Cauchy theorem in multi-connected domains
1 2 nT C C C C
TH 3.2.5 D multi connected with multi closed contours
Γ,f(z) analytic in D and on Γ.
1 2
( )d ( )d 0nC C C C
f z z f z z
1
or ( )d ( )dk
n
C Ck
f z z f z z
1
n=1, ( )d ( )dC C
f z z f z z -Deformation Theorem
2
1(1)
1C C C
dz dzdz
z z z z
2 11C C
dz dz
z z
2 2
0
i i
Closed Deformation Theorem
Ex.3.2.3
2
1 1 1( )
1z z z z
2
1, is positive oriented simple closed curve,
and 1 is in curve .
Cdz C
z zz C
Solutions:
C
1C2C
0 1
(2) .3.2.5TH
1 22 2 2
1C C C
dz dzdz
z z z z z z
1 1 2 21 1C C C C
dz dz dz dz
z z z z
0 2 2 0i i
0
Homework:
P59-60: A1-A7
00 0
( ) ( )Closed Deformation Theorem
C z z
f z f zdz dz
z z z z
0 0f z f z ��������������
0
0 00
1( ) d 2 π ( ).
z z
f z z if zz z
DC
0z
§3.3 Cauchy’s Integral Formula & High Order Derivative
0 , and ( ) is analytic on D,then:z D f z
Analysis:
1. Cauchy’s Integral Formula: (TH 3.3.1)
0( ) analytic on C and on C, inside f z z C
00
1 ( )( ) d
2 i C
f zf z z
z z
Pf. ∵f(z) continuous at z0,
00, 0, z z
0
0
( ) ( ) ,
Let , :
f z f z
R k z z R C
0 0
( ) ( )d d
C K
f z f zz z
z z z z
0 0
0 0
( ) ( ) ( )d d ( 0)
K K
f z f z f zz z
z z z z
00
0 0
( ) ( )( ) ( )d d d 2
K K K
f z f zf z f zz s s
z z z z R
00
( )d 2 i ( )
C
f zz f z
z z
1 ( )
( ) d , : inside points of C2 i C
ff z z z
z
Note 1. f(z) on D depend on f(z) on C
D: domain
2. f =g analytic on C f =g on D
3. f: → C analytic.
00
1 ( )( ) d
2 i C
f zf z z
z z
2 i
00
1( e )dt
2tf z r
average of over [0,2 ]f
dzz
dzz zz
44 3
12
1
1
.62.21.2 iii
4
1 2
1 3zdz
z z
4
1 2,along the positive oriented
1 3
circle 4.
zdz
z z
z
Ex:
Ex: : ( 1, 2)
( 1)( 2)
z
C
edz C z r r
z z z
0 1,when r
C
z
dzz
zz
e
)2)(1(
izz
ei
z
z
0
)2)(1(2
Solution:
0
21 CC
2 1
)2(C
z
dzz
zz
e
i
3C
2
2C1
1C
1 2,when r
ie
i3
2
1)2(
2
z
z
zz
eii
3 2)1(
3
2C
z
dzzzz
e
ie
i
2)1(
23
2
z
z
zz
eii
ei
ie
ie
i33
2 2
2,when r
321 CCC
2. Existence of higher derivative
TH 3.3.2. f analytic on C & on D,
0 z D
0 10
! ( )( ) d , 1, 2,
2 i ( )n
nC
n f zf z z n
z z
Pf. n=1
0 00 0
1 ( ) 1 ( )( ) d , ( ) d
2 i 2 iC C
f z f zf z z f z z z
z z z z z
0 0
0 0
( ) ( ) 1 ( ) ( )d d
2 i C C
f z z f z f z f zz z
z z z z z z z
0 0
1 ( )d
2 i ( )( )C
f zz
z z z z z
2 20 0 0
1 ( ) 1 ( )d d
2 i ( ) 2 i ( ) ( )C C
f z zf zz z
z z z z z z z
I
3
ML0, ( 0)z z
d
20 0
1 ( )I d
2 ( ) ( )C
zf zz
z z z z z
2
0 0
( )1ds
2 C
z f z
z z z z z
0 20
1 ( )'( ) d
2 i ( )C
f zf z z
z z
Note. f(z) analytic on D
f (n)(z) exist on D & analytic on D. n=1,2,
…
-the difference with real function
5
cos(1). not analytic on 1in .
( 1)
but cos analytic in .
zz C
z
z C
.12
)(cos)!15(
2d
)1(
cos 5
1
)4(5 | i
zi
zz
zz
C
5 2 2
cos e(1) d ; ( 2) d ; | | 1,
( 1) ( 1)
C:positive oriented circle.
z
C C
zz z z r
z z
Ex:
2 2(2)
( 1)
z
C
edz
z
1 22 2 2 2( 1) ( 1)
z z
C C
e edz dz
z z
21
2
2
2
2
)(
)(
)(
)(C
z
C
z
dziz
iz
e
dziz
iz
e
2 22
( ) ( )
z z
z i z i
e ei
z i z i
)4
1sin(2 i
1 2C C C
2C
1C
§3.4 Analytic and Harmonic Function
continuous on D), and
Def. real
harmonic
on D, if
D is domain, : D , C R ( , )x y2( , ) C (D) ( , , , exist and x y xx yyx y
=0, xx yy Laplace Equation
is called harmonic function on D.
Def. u, v harmonic on D, v is harmonic conjugate of u
if ' ' , ' ' on D. (C-R equation)x y y xu v u v
( , )x y
Note. v harmonic conjugate of u u harmonic conjugate of v
i.e. u+iv analytic on D v+iu analytic on D
2 2 2
2 2Ex. ( ) i 2 analytic on 2 +i ( ) is not analytic on
f z z x y xyxy x y
CC
Properties:
(1). , , .?
u iv analytic on D u vharmonic on D
(2). i , analytic on D
, harmonic on D & harmonic conjugate of .
f u v
u v v u
TH.3.4.2
(3). v harmonic conjugate of u -u harmonic conjugate of v
i.e. u+iv analytic on D v-iu analytic on D
(4). v harmonic conjugate of u on D u harmonic conjugate of v on D
u, v constants on D.
(5). v1,v2 harmonic conjugate of u on D v1=v2 + constant on D.
Pf. u+iv1 analytic on D, u+iv2 analytic on D i(v1-v2) analytic on D v1-v2 analytic on D (real)
v1-v2 =constant.
Question:
Does u have a harmonic conjugate (ux=vy , uy=-vx) on
D?
Does there exist an analytic f :D →C, u=Re f ?
(v=Im f )Ans. No in general .yes if D is simply connected.
D simply connected domain, u harmonic on D,
find , ( ) i analytic.v f z u v
d d d d dv v u u
v x y x yx y y x
0 0
( , )
( , )( , ) d d independent of integral path
x y
x y
u uv x y x y
y x
0 00( , ) '( , )d '( , )dy C
x y
y xx yv x y u x y x u x y
'( ) ' - ' '( ) ' - 'x y y xf z u iu or f z v iv integration of '( ).f z
Similarly,
0 00( , ) '( , )d '( , )dy C
x y
y xx yu x y v x y x v x y
Ex.3.4.2Prove 2 aharmonic function
and ( ) satisfying ( ) -1.
v xy
f z u iv f i
2 2
2 2
2 2
2 2
2
2
2 2
2 2 2 2
2 , 0; 2 , 0.
0 2 is a harmonic function.
2 2 ( ).
2 ' ( ) .
2 .
( ) 2 ( ) ,
( ) -1
v v v vy x
x x y y
v vv xy
x y
u vx u xdx x g y
x y
v uy g y g y y C
x y
u xdx x y C
f z x y C i xy x iy C z C
f i C
20. ( ) .f z z
Solution:
Ex.3.4.3 2 2is a harmonic function on Re( ) 0,
xu z
x y
1( ) satisfying (1 ) .
2
if z u iv f i
' '
1 0
2 2
2 2 2 2 2 2 20 0
2 2
(1). ( , ) ( ,0) ( , )
( ) .( )
1( ) .
1 1(1 ) 0 ( ) .
2
x y
y x
y y
v x y u x dx u x y dy C
y x y ydy C dy C C
x y y x y x y
x iyf z Ci iC
x y z
if i C f z
z
2 2 2 2
' '2 2 2 2 2 2 2 2 2 2
21
2 2 1(2). '( ) ,
( ) ( ) ( )
1 1( ) 1 .
1 1 1(1 ) ( ) .
2
x y
z
y x xyi y x xyif z u iu
x y x y x y z
f z dz iC iCz z
if i C f z
i z
method2
method1
Homework:
P60-61: A8-A17