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Chapter 5 Residue Theory
—Residue ＆ Application
§5.1 Isolated Singularities
§5.2 Residue
§5.3 Application of Residue Theory to Integrals
§5.1 Isolated Singularities1.Definition ＆ Classification
Def. not analytic at ,but analytic on for some ,then is called the Isolated Singular point of .
( )f z
0z00 z z
0 0z
1( ) , 1
1f z z
z
( ) ln , 0 and 0f z z z z x
0
1 1( ) , 0, ( 1, 2,...)
1sinkf z z z k
kz
0 0z kz
( )f z
Ex. isolated singular point
not isolated singular points
singular points
isolated singular pointsWhere:
not isolated singular point
0
0
0
is isolated singular point of ( ),
( ) analytic on 0
( ) has Laurent expansion: ( ) ( )nn
n
z f z
f z z z
f z f z c z z
(1).Removable Singular points.
If 0, for all n 0.
(2).Pole of order of .
If 0, for some positive integer 0,
but 0, for all - .
(3).Essential Singular points.
If 0, for an infinite num
n
m
n
n
c
m f
c m
c n m
c
ber of negative
value of .n
Def: Classification (according to )-- 0( ) nnc z z
0
0 0
For case (1):
Laurent Series takes the form:
( ) ( ) ... ( ) ...,0 | | .
( ),0 | | ,The sum function ( )
, .
nnf z c c z z c z z z z
f z z zs z
c z z
1 0 0 0
0
0 0 0
00
Redefine: ( ) ( ) ,
( ) ( ) ( ) ( ) ,| | .nn
n
f z s z c
f z s z f z c z z z z
0
00 0Note: lim ( ) removable singular point of ( ).
z zf z c z f z
3 5 2 4
0
sinEx: ( ) , 0 is removable singular point.
sin 1 1 1 1 1( ) 1 no negative parts.
3! 5! 3! 5!sin sin
If define | =1, then analytic on 0.z
zf z z
zz
z z z z zz z
z zz
z z
0
For case (2):
( ) ( ) ... ( ) ( ) ( ) ...
1( ), ( , ).
( - )
mm
mm
f z c z z c z z c z z c c z z
g z m cz z
2 10 2 0 1 0 0 1 0
1 0
0
where ( ) ( ) ( ) ...analytic
on | | and ( ) 0.
Note: has a pole of order at
m m m
m
g z c c z z c z z
z z g z c
f m z
21 0 2 0
0 0
0
00
0 0
( )In some punctured neighborhood of , ( ) ,
( - )
where ( )is analytic at and ( ) 0.
lim ( ) .
m
z z
g zz f z
z z
g z z g z
f z
2
5 1Ex: ( ) ,
(2 1) ( 1)
1is a pole of order 2, 1 is a pole of order 1.
2
zf z
z z
z z
0
0
For case (3):
is an essential singularity of ( )
lim ( ) not exists and .z z
z f z
f z
0
1
Ex: ( ) , 0 is isolated singularity,
( ) , 0
( ) 0, 0 .
lim ( ) not exists and .
0 is essential singularity.
z
z z
f z e z
f z z x
f z z x
f z
z
11 2
Laurent expansion:
1 11 .
2! !nze z z z
n
In conclusion:
0
0
0
0
0
0
is the removable singular point ( ) lim ( ) exists and finite;
is the pole of ( ) lim ( ) ;
is the essential singular point of ( ) lim ( ) not exists and .
z z
z z
z z
z of f z f z
z f z f z
z f z f z
We can use the above different situations to judge the types of isolated singular points.
2.Zeros V.S. Poles
0 0 0
0 0 1 1
00
Def: (1). is zero of : If ( ) analyse at and ( ) 0.
(2). is zero of order m of : ... 0, 0
where ( ) ( ) ,| | , Taylor expansion.
m m
nn
n
z z f f z z f z
z z f c c c c
f z c z z z z
0
0 0
( 1) ( )0 0 0 0
0
0 0
( ) analy. at , is zero of order of
( ) '( ) ... ( ) 0 ( )
where ( ) analy. at and ( ) 0.
m m
m
f z z z m f
f z f z f z f z
f z z z g z
g z z g z
（ ）（ ）（ ）
Note: Zero of order 1 is sometimes called a simple zero.
TH.5.1.1
3( ) ( 1) , 0 is the pole of order 1
and 1is the pole of order 3 of ( ).
f z z z z
z f z
3 211is zero of ( ) 1, ' 1 3 | 3 0,
1is the pole of order 1 of ( ).zz f z z f z
z f z
We can use the above theorem to judge the followings.
0
0
is pole of order of ( )
1is zero of order of .
( )
z z m f z
z z mf z
0 0 0
0
( )analy at and ( ) 0, is zero/pole of order m of ( )
is zero/pole of order m of ( ) ( ).
g z z g z z f z
z f z g z
0
0 1 2 1 2
10 2 1 1 2
2
is zero of order of ( ), ( 1,2),
is zero of order of ( ) ( ),
( )is pole of order - of , ( ).
( )
k kz m f z k
z m m f z f z
f zz m m m m
f z
TH.5.1.2
Corollary 1.
Corollary 2.
Ex.5.1.1
1(1). ( ) ;
sin
: ( 0, 1, 2, ). sin ' | cos | ( 1) 0,
is zero of order 1 of sin ,
1( 0, 1, 2, )is pole of order 1 of .
sin
kz k z k
f zz
Poles z k k z z
z k z
z k kz
1 1
(2). ( ) ( 0);1
1( ) 0 is pole of order 1
1 2! !
2 is zero of order 1 of 1( 0).
2 is pole of order 1 of ( )( 0).
n
z
n m
z
zf z n
e
f z z nz z z m
z k i e k
z k i f z k
.
0
20
1
( )0 0
1
( )0 0
essential singularity.
cos essential singularity.
m z z
m z z
z z e z
z z z
1Judge ( ) on 0( ).
z
m
ef z z m Z
z
0 : ( ) is analytic on 0
1: 0is removable singular point
m f z z
m z
；；
Note:
Ex:
2 1
1 1
11: ( )
2! ! ( 1)!
1 1 1 1
2! ! ( 1)!
0 1
m m
m
m m
z z zm f z z
z m m
z
z z m m
z m
为 阶极点。
§5.2 Residue1.Definition ＆ Evaluation
0
0 0
is isolated singular point of ( ),
( ) has Laurent expansion:
( ) ( ) ,0 .nn
n
z f z
f z
f z c z z z z
0
0
1 1
( ) ( )
2 , 1,1In ( )
0, 1.( )
1( ) 2 , . . ( ) .
2
nnC C
n
nC
C C
f z dz c z z dz
i ndz n N
nz z
f z dz c i i e f z dz ci
0 0
0
0
0
If ( ) has an isolated singularity at ( ), is the
positive oriented simple closed curve in0 | | ,
1then ( ) is called the residue of ( ) at
2and is denoted by Res[ ( ), ],
. .Res[ ( ),
C
f z z z C
z z
f z dz f z zi
f z z
i e f z
0 1
1] ( ) .
2 Cz f z dz c
i
Def:
Note:
0 0
0 0 1
0
(1). is removable singularity Res[ ( ), ] 0.
(2). is essential singularity Res[ ( ), ] .
(3). is a pole,then we can get the following rules.
z f z z
z f z z c
z
0
0
0
0 0
0
1
0 01
(1). is a pole of order 1.[first order pole]
Res[ ( ), ] lim( ) ( ).
(2). is a pole of order ,( 1),
1Res[ ( ), ] lim [( ) ( )].
( 1)!
z z
nn
nz z
z
f z z z z f z
z m m n m
df z z z z f z
n dz
0
10 1 0 0 1 0
10 0 1 0 0 0
On 0 ,
,
( ) ,
mm
n n m n nm
z z
f z c z z c z z c c z z
z z f z c z z c z z c z z
（ ） （ ） （ ） （ ）
（ ） （ ） （ ） （ ）
Rule:
0
1
0 1 0 01
1
1 0 01
{( ) ( )} ( 1)! !( )
1Res[ ( ), ] lim [( ) ( )].
( 1)!
nn
n
nn
nz z
dz z f z n c c n z z
dz
dc f z z z z f z
n dz
When n=m=1,we can get the rule (1).
0 0 0 0
00
(3). ( ) , , analy at , ( ) 0, ( ) 0, ' 0
Res[ ( ), ] .'
P zf z P Q z P z Q z Q z
Q z
P zf z z
Q z
（ ）
0 0 0
0
( ) 0, '( ) 0, is zero of order 1 of ( )
1is pole of order 1 of .
( )
Q z Q z z Q z
zQ z
0
0 00
0
0 0 0
1 1( ), ( ) analy at , ( ) 0
( )
is pole of order 1 of ( ),
and rule(1) Res[ ( ), ] lim( ) ( ),and ( ) 0z z
z z z zQ z z z
z f z
f z z z z f z Q z
0 0
00
0 0
0
( ) ( )lim( ) ( ) lim ( ) .
( ) ( ) '( )z z z z
P z P zz z f z f z
Q z Q z Q zz z
Ex.5.2.1
0
ln(1 )(1). ( ) , .
zf z z
z
cos(2). ( ) , ( integer).
sinis the pole of order 1 of ( ), rule(2)
cosRes[ ( ), ] | 1, 0, 1, 2,...
sin '
k
k
z k
zf z z k k
zz k f z
zf z k k
z
00
0
ln(1 )0 is the isolated singularity of ( ), lim 1
0is the removable singularity Res[ ( ),0] 0.
z
zz f z
zz f z
0 12(3). ( ) , 0, 1.
( 1)
zef z z z
z z
0
20
1
21 1
0 is the pole of order 1 of ( ), rule(1)
Res[ ( ),0] lim 1;( 1)
1is the pole of order 2 of ( ), rule(2)
( 1)Res[ ( ),1] lim ( ) lim 0
z
z
z z
z z
z f z
ef z
z
z f z
d e e zf z
dz z z
Dz1
z2z3
zn C1
C2
C3
Cn
C
TH.5.2.1 Residue Theorem
1 2
1
If is a simple closed positively oriented contour
and ( )is analytic inside and on except at isolated
singularities, , ,..., inside , then
( )d 2 π Res[ ( ), ].
n
n
kkC
C
f z C
z z z C
f z z i f z z
2
1ln 1 can never use the residue theorem.
zdz
z
Satisfy the conditions of the residue theorem, then use it.
Note:
Ex:
1
the residue of ( ) on singular points
in Laurent Series
f z
c
And if we know the type of the singular points, then we can get the residue conveniently.
0 0
0
0
(1). is removable singular point of ( ), then Res[ ( ), ] 0 .
(2). is essential singular point of ( ) then expand ( )
to Laurent Series to get the residue.
(3). is pole of ( ), then use rule(1)(2
z f z f z z
z f z f z
z f z
,
)(3).
22
5 2(1). d ;
( 1)z
zI z
z z
20
Inside 2, 0is pole of order 1 and 1is the
pole of order 2.
5 2Res[ ( ),0] lim( 0) 2,
( 1)z
z z z
zf z z
z z
220
5 2Res[ ( ),1] lim [( 1) ] 2,
( 1)
2 {Res[ ( ),0]+Res[ ( ),1]} 2 ( 2 2) 0.
z
d zf z z
dz z z
I i f z f z i
Ex.5.2.2
|z|=n
(2). tan , .I zdz z N 1
( 0, 1, 2, )are the singularity points of2sin 1
( )= ,sin and cos analy on ( 0, 1, 2, ),cos 2
andsin 0,cos =0, (cos )' 0.
z k k
zf z z z z k k
zz z z
1'2
1( 0, 1, 2, )are poles of order 1 of ( ).
21 sin 1
Rule(3) Res[tan , ] | -2 (cos ) z k
z k k f z
zz k
z
1| | | |2
1tan 2 Re [tan , ]
2
22 (- ) -4 .
z n k n
zdz i s z k
ni ni
2
1
|z|=1
(3). zI e dz
2
1
2 4
0 is essential singular point of ( ) in 1 .
1 1=1+ + +
2!Res[ ( ),0] 0
0
z
z f z z
ez z
f z
I
Homework:
P117-118:
A1(1)(3)(5)(7)(9)(11),A2.(1)(3)(5),A3-A6, A7(1)(3)(5),A8(1)(3)(5)
§5.3 Application of Residue Theorem to Integrals
1
1 2
( ) , [ , ](line ),
[ , ] must be part of ,
b
af x dx x a b l
a b l l l
0a b1l
2l
Residue Theorem is the theorem of complex function and is
related to closed loop integral.
So if we want to use this theorem in definite integral of real variable
function, the real variable function must be transformed to complex
function and definite integral must be part of closed loop integral.For example,
2
then real integral is part of closed loop integral:
( ) ( ) ( ) .b
l la
f z dz f x dx f z dz
1.Trigonometric Integrals over [0,2 ]
2 π
0
i
(cos ,sin ) where (cos ,sin )is a rational function of
sin and cos . Let ,0 2 . i
R d R
z e dz ie d izd
，
2 21 1 1 1sin (e e ) ,cos (e e ) .
2 2 2 2i i i iz z
i iz z
1| | 1
( )d 2π Res[ ( ), ],n
kkz
f z z i f z z
where ( 1,2,..., )are isolated singularities inside of 1.kz k n z
0 2 01 1
i
i
2 22
0| | 1
1 1 d(cos ,sin )d ,
2 2z
z z zR R
z iz iz
Ex.5.4.12
0 2 cos
dI
2
2| | 1
2
1 2Let = , then ,cos , .
2 4 1
1-2 3are poles of order 1 of ( ) ,
4 1
=-2 3 inside 1.
i
z
dz z dzz e d I
iz z i z z
z f zz z
z z
2-2 3
1Res[ ( ), -2 3] lim - (-2 3)
4 1
1,
2 3
2 1 2 32 .
32 3
xf z z
z z
I ii
2 2 4
1 22| | 1 | | 1 | | 1
1 1( )
2 2 (1 )( )1 2
2z z z
z z dz zI dz f z dz
z z iz iz pz z pp p
2 2
12
4 4
2 2 2 2
4
2
1 1{
21 2
2
1 1 1 1
2 2 (1 ) (1 )
1 1}
2 ( )(1 )
z z
z z izp p
z z
iz z pz p p z iz z pz p pz
z
iz z p pz
2
2 22 2
0< <1 1 2 cos 0 on 0 2 .
1 zcos 2 (e e )
2 2i i
p p p p
z
2
20
cos 2(0 1)
1 2 cosI d p
p p
，Ex:
There are three poles, , inside the circle , is
pole of order 2 and pole of order 1.
42
20
2 2 3 4 2 2
2 2 2 20
d 1Res[ ( ),0] lim
d 2 (1 )( )
( )4 (1 )(1 2 ) 1lim ,
2 ( ) 2
z
z
zf z z
z iz pz z p
z pz p p z z z pz p p
i z pz p p z ip
4 4
2 2 2
1 1Res[ ( ), ] lim ( ) ,
2 (1 )( ) 2 (1 )z p
z pf z p z p
iz pz z p ip p
2 4 2
2 2 2 2
1 1 2π2π
2 2 (1 ) 1
p p pI i
ip ip p p
0, ,1/z p p 0,z p| | 1z 0z z p
Ex:20
1
25 3sin
d
2
20
2 1Let
2 5 3sind
2 21
2Let
(3 ) ( 3 )i
z
zz e dz
i z i z i
is pole of order 2 on 1.3
iz z
64
5
256
522]
3),([Res2
ii
izfi
( )d , Improper integralR x x
( )where ( ) , , are real polynomials,
( )
no common factor, ( ) has no pole in ,deg - deg 2.
P xR x P Q
Q x
R x R Q P
1
( )d 2 π Res[ ( ), ],
where ( 1,2,..., )are isolated singularities
lie in upper half-plane.
kk
k
R x x i R z z
z k n
2.
z1z2
z3
yCR
R RO x
Integration path is as the figure. CR is the upper-half circle
in the upper-half plane, with O as the center and R as the radius. Let R be big properly to make sure that all the singular points zk of R(z) in the upper-half are in the integration
path.
z1z2
z3
yCR
R RO x
11
11
Suppose ( ) ,
2,no common factor.
n nn
m mm
z a z aR z
z b z b
m n
then ( )d ( ) 2π Res[ ( ), ]R
R
kR CR x x R z dz i R z z
2
π( )d | ( ) | d π 0
R RRC C
M MR z z R z s R
R R
11
11
11
11
2
|1 |1| ( ) |
| | |1 |
1 | |1
| | 1 | |
1(when | | is enough big)
| | | |
nn
m n mm
nn
m n mm
m n
a z a zR z
z b z b z
a z a z
z b z b z
MM z
z z
( )d 2 π Res[ ( ), ].kR x x i R z z
0
For even function ( ),
1( )d ( )d π Res[ ( ), ].
2 k
R x
R x x R x x i R z z
Ex.5.4.2
4
2 4
2is pole of order 4 of ( )
3 (2 3 )
in the upper half-plane.
zz i R z
z
4
2 4(2 3 )
xI dx
x
3 44
3 2 42
3
2Res[ ( ), ]
3
1 2lim [( - ) ]
(4 -1)! 3 (2 3 )
-
576 6
x i
R z i
d zz i
dz z
i
4
2 4-
-2
2 3 576 6 288 6
x idx i
x
（ ）
Ex: 2 10
1
( 1)ndx
x
2 1
1is the only pole of order +1 of ( )
( 1)
in the upper-half plane
nz i n f z
z
.
2 1
1 1
2 ( 1)nI dx
x
Solution:
1
2 1
2 1
1 1 ( 1) ( 1)( 2) 2Res[ ( ), ]
! ! (2 )
( 1)( 2) 2 (2 1)!!
! 2 2 (2 )!!
nn n
n n
z i
n
d n n ni f z i i i
n dz z i n i
n n n n
n n
Ex.5.4.32 20 (1 )
dxI
x
22 2
2 20
( ) is even, is pole of order 2 of ( ),
1 1Res[ ( ), ] lim [( ) ] .
(1 ) 4
1Res[ ( ), ] .
(1 ) 4 4
z i
R x z i R z
dR z i z i
dz z i
dxi R z i i
x i
3. ( ) , 0.Improper Integrali xR x e dx
( )where ( ) rational function, ( ) no common factor,
( )
( ) has no pole in ,deg - deg 1.
P xR x R x
Q x
R x R Q P
( ) e d 2 π Res[ ( )e , ] .
where ( 1,2,..., )are isolated singularities of ( ) e
lie in upper half-plane.
ix izk
izk
R x x i R z z
z k n R z
( ) cos d ( )sin d
2 π Res[ ( ) , ].aizk
R x ax x i R x ax x
i R z e z
Or:
Ex.5.4.4 2
cos
2 10
x xI dx
x x
2
2 2
( ) , 2, 1, 1.2 10
( ) has no singular points in real axis,
exists,and is the real part of .2 10 2 10
ix ix
xR x s t s t
x xR z
xe xedx I dx
x x x x
3+
21 3
1 3 is pole of order 1 of ( ) in the upper half-plane,
(1 3 )Res[ ( ) ,1 3 ] lim {[ (1 3 )] } .
2 10 6
iz iiz
z i
z i R z
ze i eR z e i z i
z z i
3+
32
3 3
(1 3 )2 (1 3 )(cos1 sin1)
2 10 6 3
(cos1 3sin1) (3cos1 sin1).3 3
ix ixe i edx i e i i
x x i
e i e
32
32
cos(cos1 3sin1).
2 10 3sin
and (3cos1 sin1).2 10 3
x xdx e
x xx x
dx ex x
Ex.5.4.5 20
cos, ( 0).
1
mxI dx m
x
2 2 20
2 2
cos cos 1 cos is even,so .
1 1 2 1
cosis the real part of .
1 1
imx
mx mx mxdx dx
x x x
mx edx dx
x x
2
2 2 2
1is pole of order 1 of ( ) ,
1
2 Res[ , ] 2 lim[( - ) ] ,1 1 1
imx imz imzm
x
z i R zz
e e edx i i i z i e
x z z
2 2
20
cos sin. . ,
1 1cos
.1 2
m
m
mx mxi e dx dx e
x xmx
dx ex
4.Integrals where Integral has poles in C
0
sin xI dx
x
0
sin sin 1 sineven, .
2
sinis imaginary part of , we consider the contour
integral over a closed contour.
ix
iz
c
x x xdx dx
x x x
x e
x x
edz
z
Ex.
Solutions ：
0.
Let ,
0.
sin2 0.
R r
R r
R r
iz ix iz ixr R
C R C r
ix it itr r R
R R r
ix ix iz izR
r C C
iz izR
r C C
e e e edz dx dz dx
z x z x
e e ex t dx dt dt
x t t
e e e edx dz dz
x z z
x e ei dx dz dz
x z z
sin
0
sin (2 / )2 2
0 0
1
0
1
lim 0
| | 1{
| |
2 2 (1 ).}
1 1lim { ( )
2! !
where ( ) .anal2! !
R
R R R
r
iz
CR
iz izy R
C C C
R R R
iz iz n n
Cr
n n
edz
z
e edz ds e ds e d
z z R
e d e d eR
e e z i zdz i i z
z z z n z
z i zz i
n
y at 0,and (0) ,
when sufficent small, ( ) 2, ( )r r r
iz
C C C
z i
e dzz z dz z dz
z z
0
0 0
0 0
( ) | ( ) | 2 2 0( 0)
lim ( ) 0 lim }
sin sin2 .
2
r
r r r
r r
i
iC
C C C
iz
C Cr r
dz ired i
z re
z dz z ds ds r r
ez dz dz i
zx x
i dx i dxx x
Homework:
P118:A9-A11, A12(1)(3)(5)(7)
Summary:
1.Classification of isolated singularities.
2.Residue definition and evaluation.
3.Residue Theorem.
4.Application of Residue Theorem to integrals.