chapter 6. residues and poles
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Chapter 6. Residues and Poles. Weiqi Luo ( 骆伟祺 ) School of Software Sun Yat-Sen University Email ： [email protected] Office ： # A313. Chapter 6: Residues and Poles. Isolated Singular Points Residues Cauchy’s Residue Theorem Residue at Infinity - PowerPoint PPT PresentationTRANSCRIPT
Chapter 6. Residues and Poles
Weiqi Luo (骆伟祺 )School of Software
Sun Yat-Sen UniversityEmail ： [email protected] Office ： # A313
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Isolated Singular Points Residues Cauchy’s Residue Theorem Residue at Infinity The Three Types of Isolated Singular Points Residues at Poles; Examples Zeros of Analytic Functions; Zeros and Poles Behavior of Functions Near Isolated singular Points
2
Chapter 6: Residues and Poles
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Singular Point A point z0 is called a singular point of a function f if f
fails to be analytic at z0 but is analytic at some point in every neighborhood of z0.
Isolated Singular Point A singular point z0 is said to be isolated if, in addition,
there is a deleted neighborhood 0<|z-z0|<ε of z0 throughout which f is analytic.
68. Isolated Singular Points
3
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Example 1 The function
has the three isolated singular point z=0 and z=±i.
Example 2 The origin is a singular point of the principal branch
68. Isolated Singular Points
4
3 2
1
( 1)
z
z z
ln , ( 0, )Logz r i r
Not Isolated.
x
y
xx
y
x
y
εεεε
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Example 3 The function
has the singular points z=0 and z=1/n (n=±1,±2,…), all lying on the segment of the real axis from z=-1 to z=1.
Each singular point except z=0 is isolated.
68. Isolated Singular Points
5
1
sin( / )z
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If a function is analytic everywhere inside a simple closed contour C except a finite number of singular points : z1, z2, …, zn
then those points must all be isolated and the deleted neighborhoods about them can be made small enough to lie entirely inside C.
Isolated Singular Point at ∞
If there is a positive number R1 such that f is analytic for R1<|z|<∞, then f is said to have an isolated singular point at z0=∞.
68. Isolated Singular Points
6
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Residues When z0 is an isolated singular point of a function f, there is a
positive number R2 such that f is analytic at each point z for which 0<|z-z0|<R2. then f(z) has a Laurent series representation
where the coefficients an and bn have certain integral representations.
where C is any positively oriented simple closed contour around z0 hat lies in the punctured disk 0<|z-z0|<R2.
69. Residues
7
1 20 2
0 0 0 0
( ) ( ) ... ...( ) ( ) ( )
n nn n
n
bb bf z a z z
z z z z z z
10
1 ( ), ( 1,2,...)
2 ( )n nC
f z dzb n
i z z 1
0
1 ( ), ( 0,1,2,...)
2 ( )n nC
f z dza n
i z z
Refer to pp.198
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Residues (Cont’)
69. Residues
8
1 20 2
0 0 0 0
( ) ( ) ... ...( ) ( ) ( )
n nn n
n
bb bf z a z z
z z z z z z
1 1 10
1 ( ) 1( )
2 ( ) 2C C
f zb dz f z dz
i z z i
1
1( )
2 C
b f z dzi
1( ) 2C
f z dz ib
0
( ) 2 Re ( )z z
C
f z dz i s f z
Then the complex number b1 is called the residues of f at the isolated singular point z0, denoted as
01 Re ( )
z zb s f z
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Example 1 Consider the integral
where C is the positively oriented unit circle |z|=1. Since the integrand is analytic everywhere in the finite plane except z=0, it has a Laurent series representation that is valid when 0<|z|<∞.
69. Residues
9
2 1sin( )
C
z dzz
2 2
0
1 1sin( ) 2 Re ( sin( ))
zC
z dz i s zz z
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Example 1 (Cont’)
69. Residues
10
3 5 7
sin ..., (| | )3! 5! 7!
z z zz z z
23 5
1 1 1 1 1 1 1sin( ) ..., (0 | | )
3! 5! 7!z z z
z z z z
2 2
0
1 1 1sin( ) 2 Re ( sin( )) 2
3! 3zC
iz dz i s z i
z z
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Example 2 Let us show that
when C is the same oriented circle |z|=1. Since the 1/z2 is analytic everywhere except at the origin, the same is true of the integrand.
One can write the Laurent series expansion
69. Residues
11
2
1exp( ) 0
C
dzz
2
1 ..., (| | )1! 2!
z z ze z
2 2 4
1 1 1 1 1exp( ) 1 ..., (0 | | )
1! 2!z
z z z 2
1exp( ) 0
C
dzz
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Example 3 A residues can also be used to evaluate the integral
where C is the positively oriented circle |z-2|=1.
Since the integrand is analytic everywhere in the finite plane except at the point z=0 and z=2. It has a Laurent series representation that is valid in the punctured disk
69. Residues
12
4( 2)C
dz
z z
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Example 3 (Cont’)
69. Residues
13
4 4 4
1 1 1 1 1, (0 | 2 | 2)
2( 2) ( 2) 2 ( 2) 2( 2) 1 [ ( )]2
zzz z z z z
41
0
( 1)( 2)
2
nn
nn
z
0
1, (| | 1)
1n
n
z zz
4 42
1 12 Re ( ) 2 ( )
( 2) ( 2) 16 8zC
dz ii s i
z z z z
4
54
( 1)( 2) , (0 | 2 | 2)
2
nn
nn
z z
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Theorem Let C be a simple closed contour, described in the
positive sense. If a function f is analytic inside and on C except for a finite number of singular points zk (k = 1, 2, . . . , n) inside C, then
70. Cauchy’s Residue Theorem
14
1
( ) 2 Re ( )k
n
z zkC
f z dz i s f z
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Theorem (Cont’) Proof: Let the points zk (k=1,2,…n) be centers of positively
oriented circles Ck which are interior to C and are so small that no two of them have points in common (possible?).
Then 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
70. Cauchy’s Residue Theorem
15
1
( ) ( ) 0k
n
kC C
f z dz f z dz
( ) 2 Re ( ), 1,2,...,
kk
z zC
f z dz i s f z k n
1
( ) 2 Re ( )k
n
z zkC
f z dz i s f z
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Example Let us use the theorem to evaluate the integral
70. Cauchy’s Residue Theorem
16
5 2
( 1)C
zdz
z z
0 1
( ) 2 [Re ( ) Re ( )]z z
C
f z dz i s f z s f z
5 2 5 2 1 2 1( ) ( ) (5 )( )
( 1) 1 1
z z
z z z z z z
22(5 )( 1 ...)z z
z
0Re ( ) 2
zs f z
5 2 5( 1) 3 1 3( ) (5 )
( 1) ( 1) 1 ( 1) 1
z z
z z z z z z
23(5 )(1 ( 1) ( 1) ...)
1z z
z
1
Re ( ) 3z
s f z
10 i
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Example (Cont’) In this example, we can write
70. Cauchy’s Residue Theorem
17
5 2 2 3 2 3( ) +
( 1) 1 1C C C C
zdz dz dz dz
z z z z z z
2=2 i(2)=4 i
C
dzz
3
2 (3) 61C
dz i iz
5 24 6 10
( 1)C
zdz i i i
z z
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Definition Suppose a function f is analytic throughout the finite plane except for a
finite number of singular points interior to a positively oriented simple close contour C. Let R1 is a positive number which is large enough that C lies inside the circle |z|=R1
The function f is evidently analytic throughout the domain R1<|z|<∞. Let C0 denote a circle |z|=R0, oriented in the clockwise direction, where R0>R1. The residue of f at infinity is defined by means of the equation
71. Residue at Infinity
18
0
( ) 2 Re ( )z
C
f z dz i s f z
0
1Re ( ) ( )
2 z
C
s f z f z dzi
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71. Residue at Infinity
19
0 0
( ) ( ) ( )C C C
f z dz f z dz f z dz
Based on the definition of the residue of f at infinity
0
( ) ( ) 2 Re ( )
zC C
f z dz f z dz i s f z
1( ) , ( | | )nn
n
f z c z R z
0
1
1 ( ), ( 0, 1, 2,...)
2n nC
f z dzc n
i z
22 2
1
1 1 1( ) ( ) , (0 | | )n n
n nn n
c cF z f z
z z z z R
Refer to the Corollary in pp.159
20 0
1 1Re ( ) Re [ ( )]
z zs F z s f
z z
1c
0 0
1
1 1Re ( ) ( ) ( )
2 2
z
C C
s f z f z dz f z dz ci i
20
1 1Re ( ) Re [ ( )]
z zs f z s f
z z
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Theorem If a function f is analytic everywhere in the finite plane
except for a finite number of singular points interior to a positively oriented simple closed contour C, then
71. Residue at Infinity
20
20
1 1( ) 2 Re [ ( )]
zC
f z dz i s fz z
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Example In the example in Sec. 70, we evaluated the integral of
around the circle |z|=2, described counterclockwise, by finding the residues of f(z) at z=0 and z=1, since
71. Residue at Infinity
21
5 2( )
( 1)
zf z
z z
2
1 1 5 2 5( ) 3 3 ...(0 | | 1)
(1 )
zf z z
z z z z z
5 22 (5) 10
( 1)C
zdz i i
z z
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pp. 239-240
Ex. 2, Ex. 3, Ex. 5, Ex. 6
71. Homework
22
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Laurent Series If f has an isolated singular point z0, then it has a
Laurent series representation
In a punctured disk 0<|z-z0|<R2.
is called the principal part of f at z0 In the following, we use the principal part to identify the isolated singular point z0
as one of three special types.
72. The Three Types of Isolated Singular Points
23
1 20 2
0 0 0 0
( ) ( ) ... ...( ) ( ) ( )
n nn n
n
bb bf z a z z
z z z z z z
1 22
0 0 0
... ...( ) ( ) ( )
nn
bb b
z z z z z z
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Type #1:
If the principal part of f at z0 at least one nonzero term but the number of such terms is only finite, the there exists a positive integer m (m≥1) such that
72. The Three Types of Isolated Singular Points
24
0mb and 0,kb k m
1 20 2
0 0 0 0
( ) ( ) ...( ) ( ) ( )
n mn m
n
bb bf z a z z
z z z z z z
Where bm ≠ 0, In this case, the isolated singular point z0 is called a pole of orderm. A pole of order m=1 is usually referred to as a simple pole.
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Example 1 Observe that the function
has a simple pole (m=1) at z0=2. It residue b1 there is 3.
72. The Three Types of Isolated Singular Points
25
2 2 3 ( 2) 3 3 32 ( 2) , (0 | 2 | )
2 2 2 2
z z z zz z z
z z z z
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Example 2 From the representation
72. The Three Types of Isolated Singular Points
26
2 32 2 2
1 1 1 1( ) (1 ...)
(1 ) 1 ( )f z z z z
z z z z z
22
1 11 ..., (0 | | 1)z z z
z z
One can see that f has a pole of order m=2 at the origin and that
0Re ( ) 1
zs f z
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Type #2 If the principal part of f at z0 has no nonzero term
z0 is known as a removable singular point, and the residues at a removable singular point is always zero.
72. The Three Types of Isolated Singular Points
27
20 0 1 0 2 0 0 2
0
( ) ( ) ( ) ( ) ..., (0 | | )nn
n
f z a z z a a z z a z z z z R
Note: f is analytic at z0 when it is assigned the value a0 there. The singularity z0 is, therefore, removed.
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Example 4 The point z0=0 is a removable singular point of the
function
Since
when the value f(0)=1/2 is assigned, f becomes entire.
72. The Three Types of Isolated Singular Points
28
2
1 cos( )
zf z
z
2 4 6 2 4
2
1 1( ) [1 (1 ...)] ..., (0 | | )
2! 4! 6! 2! 4! 6!
z z z z zf z z
z
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Type #3
If the principal part of f at z0 has infinite number of nonzero terms, and z0 is said to be an essential singular point of f.
Example 3 Consider the function
has an essential singular point at z0=0. where the residue b1 is 1.
72. The Three Types of Isolated Singular Points
29
1/2
0
1 1 1 1 1 11 ..., (0 | | )
! 1! 2!z
nn
e zn z z z
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pp. 243
Ex. 1, Ex. 2, Ex. 3
72. Homework
30
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Theorem An isolated singular point z0 of a function f is a pole of
order m if and only if f (z) can be written in the form
where φ(z) is analytic and nonzero at z0 . Moreover,
73. Residues at Poles
31
0
( )( )
( )m
zf z
z z
0
0
( 1)0
( ), 1
Re ( ) ( ), 1
( 1)!
m
z z
z m
s f z zm
m
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Proof the Theorem Assume f(z) has the following form
where φ(z) is analytic and nonzero at z0, then it has Taylor series representation
73. Residues at Poles
32
0
( )( )
( )m
zf z
z z
( 1) ( )2 ( 1)0 0 0 0
0 0 0 0 0
'( ) ''( ) ( ) ( )( ) ( ) ( ) ( ) ... ( ) ( )
1! 2! ( 1)! !
m nm n
n m
z z z zz z z z z z z z z z
m n
( 1) ( )0 0 0 0 0
1 20 0 0 0 0
( ) '( ) /1! ''( ) / 2! ( ) / ( 1)! ( ) / !( ) ...
( ) ( ) ( ) ( ) ( )
m n
m m m m nn m
z z z z m z nf z
z z z z z z z z z z
b1
a pole of order m, φ(z0)≠0
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On the other hand, suppose that
The function φ(z) defined by means of the equations
Evidently has the power series representation
Throughout the entire disk |z-z0|<R2. Consequently, φ(z) is analytic in that disk, and, in particular, at z0. Here φ(z0) = bm≠0.
73. Residues at Poles
33
1 20 0 22
0 0 0 0
( ) ( ) ... , (0 | | )( ) ( ) ( )
n mn m
n
bb bf z a z z z z R
z z z z z z
0 0
0
( ) ( ),( )
,
m
m
z z f z z zz
b z z
2 11 0 2 0 1 0 0
0
( ) ( ) ... ( ) ( ) ( )m m n mm m n
n
z b b z z b z z b z z a z z
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Example 1 The function
has an isolated singular point at z=3i and can be written
Since φ(z) is analytic at z=3i and φ(3i)≠0, that point is a simple pole of the function f, and the residue there is
The point z=-3i is also a simple pole of f, with residue B2= 3+i/6
74. Examples
34
2
1( )
9
zf z
z
( ) 1( ) , ( )
3 3
z zf z z
z i z i
1
3 1 3(3 )
6 6
i i iB i
i i
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Example 2 If
then
The function φ(z) is entire, and φ(i)=i ≠0. Hence f has a pole of order 3 at z=i, with residue
74. Examples
35
3
3
2( )
( )
z zf z
z i
33
( )( ) , ( ) 2
( )
zf z z z z
z i
''(3 ) 63
2! 2!
i iB i
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Example 3 Suppose that
where the branch
find the residue of f at the singularity z=i.
The function φ(z) is analytic at z=i, and φ(i)≠0, thus f has a simple pole there, the residue is B= φ(i)=-π3/16.
74. Examples
36
3
2
(log )( )
1
zf z
z
log ln , ( 0,0 2 )z r i r
3( ) (log )( ) , ( )
z zf z z
z i z i
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Example 5 Since z(ez-1) is entire and its zeros are
z=2nπi, (n=0, ±1, ±2,… )
the point z=0 is clearly an isolated singular point of the function
From the Maclaurin series
We see that
Thus
74. Examples
37
1( )
( 1)zf z
z e
2 3
1 ..., (| | )1! 2! 3!
z z z ze z
22( 1) (1 ...), (| | )
2! 3!z z z
z e z z
2 2
( ) 1( ) , ( )
1 / 2! / 3! ...
zf z z
z z z
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Example 5 (Cont’)
since φ(z) is analytic at z=0 and φ(0) =1≠0, the point z=0 is a pole of the second order. Thus, the residue is B= φ’(0)
Then B=-1/2.
74. Examples
38
2 2
1(1/ 2! 2 / 3! ...)'( )
(1 / 2! / 3 ...)
zz
z z
2 2
( ) 1( ) , ( )
1 / 2! / 3! ...
zf z z
z z z
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pp. 248
Ex. 1, Ex. 3, Ex. 6
74. Examples
39
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Definition Suppose that a function f is analytic at a point z0. We known
that all of the derivatives f(n)(z0) (n=1,2,…) exist at z0. If f(z0)=0 and if there is a positive integer m such that f(m)
(z0)≠0 and each derivative of lower order vanishes at z0, then f is said to have a zero of order m at z0.
75. Zeros of Analytic Functions
40
00
( ) ( )
nn
n
f z a z z ( )0( )
0, ( 0,1,2,..., 1)!
n
n
f za n m
n
0( ) ( )
nn
n m
f z a z z
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Theorem 1 Let a function f be analytic at a point z0. It has a zero of
order m at z0 if and only if there is a function g, which is analytic and nonzero at z0 , such that
Proof: 1) Assume that f(z)=(z-z0)mg(z) holds,
Note that g(z) is analytics at z0, it has a Taylor series representation
75. Zeros of Analytic Functions
41
0( ) ( ) ( )mf z z z g z
20 00 0 0 0
'( ) ''( )( ) ( ) ( ) ( ) ..., (| | )
1! 2!
g z g zg z g z z z z z z z
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75. Zeros of Analytic Functions
42
0( ) ( ) ( )mf z z z g z
1 20 00 0 0 0 0
'( ) ''( )( )( ) ( ) ( ) ..., (| | )
1! 2!m m mg z g z
g z z z z z z z z z
( 1)0 0 0 0( ) '( ) ''( ) ... ( ) 0mf z f z f z f z
Thus f is analytic at z0, and ( )
0 0( ) ! ( ) 0 mf z m g z
Hence z0 is zero of order m of f.
2) Conversely, if we assume that f has a zero of order m at z0, then( )
00
( )( ) ( )
!
nn
n m
f zf z z z
n
( ) ( 1) ( 2)
20 0 00 0 0
( ) ( ) ( )( ) [ ( ) ( ) ...]
! ( 1)! ( 2)!
m m mm f z f z f z
z z z z z zm m m
g(z)
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75. Zeros of Analytic Functions
43
( ) ( 1) ( 2)20 0 0
0 0 0
( ) ( ) ( )( ) ( ) ( ) ..., (| | )
! ( 1)! ( 2)!
m m mf z f z f zg z z z z z z z
m m m
The convergence of this series when |z-z0|<ε ensures that g is analyticin that neighborhood and, in particular, at z0, Moreover,
( )0
0
( )( ) 0
!
mf zg z
m
This completes the proof of the theorem.
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Example 1 The polynomial
has a zero of order m=1 at z0=2 since
where
and because f and g are entire and g(2)=12≠0. Note how the fact that z0=2 is a zero of order m=1 of f also follows from the observations that f is entire and that f(2)=0 and f’(2)=12≠0.
75. Zeros of Analytic Functions
44
3 2( ) 8 ( 2)( 2 4)f z z z z z
( ) ( 2) ( )f z z g z
2( ) 2 4g z z z
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Example 2 The entire function
has a zero of order m=2 at the point z0=0 since
In this case,
75. Zeros of Analytic Functions
45
( ) ( 1)zf z z e
(0) '(0) 0f f ''(0) 2 0f
2( ) ( 0) ( )f z z g z
( 1) / , 0( )
1, 0
ze z zg z
z
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Theorem 2 Given a function f and a point z0, suppose that
a) f is analytic at z0 ;
b) f (z0) = 0 but f (z) is not identically equal to zero in any neighborhood of z0.
Then f (z) ≠ 0 throughout some deleted neighborhood 0 < |z − z0| < ε of z0.
75. Zeros of Analytic Functions
46
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75. Zeros of Analytic Functions
47
Proof:
Since (a) f is analytic at z0, (b) f (z0) = 0 but f (z) is not identically equal to zero in any neighborhood of z0 , f must have a zero of some finite orderm at z0 (why?). According to Theorem 1, then
0( ) ( ) ( )mf z z z g z
where g(z) is analytic and nonzero at z0.
Since g(z0)≠0 and g is continuous at z0, there is some neighborhood|z-z0|<ε, g(z) ≠0.
Consequently, f(z) ≠0 in the deleted neighborhood 0<|z-z0|<ε (why?)
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Theorem 3 Given a function f and a point z0, suppose that
a) f is analytic throughout a neighborhood N0 of z0
b) f (z) = 0 at each point z of a domain D or line segment L containing z0.
75. Zeros of Analytic Functions
48
Then in N0 0f
That is, f(z) is identically equal to zero throughout N0
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75. Zeros of Analytic Functions
49
Proof:
We begin the proof with the observation that under the stated conditions,f (z) ≡ 0 in some neighborhood N of z0.
For, otherwise, there would be a deleted neighborhood of z0 throughout which f(z)≠0, according to Theorem 2; and that would be inconsistent with the condition that f(z)=0 everywhere in a domain D or on a line segment L containing z0.
Since f (z) ≡ 0 in the neighborhood N, then, it follows that all of the coefficients in the Taylor series for f (z) about z0 must be zero.
( )0( )
, ( 0,1,2,...)!
n
n
f za n
n
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Lemma (pp.83) Suppose that
a) a function f is analytic throughout a domain D;
b) f (z) = 0 at each point z of a domain or line segment contained in D.
Then f (z) ≡ 0 in D; that is, f (z) is identically equal to zero throughout D.
75. Zeros of Analytic Functions
50
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Theorem 1 Suppose that
a) two functions p and q are analytic at a point z0;
b) p(z0)≠0 and q has a zero of order m at z0.
Then the quotient p(z)/q(z) has a pole of order m at z0.
Proof:
76. Zeros and Poles
51
0 0
( ) ( ) ( )
( ) ( ) ( ) ( )m m
p z p z z
q z z z g z z z
0( ) ( ) ( )mq z z z g z Since q has a zero of order m at z0
where g is analytic at z0 and g(z0) ≠0
where φ(z)=p/g is analytic and φ(z0)≠0 Why?
Therefore, p(z)/q(z) has a pole of order m at z0
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Example 1 Two functions
are entire, and we know that q has a zero of order m=2 at the point z0=0.
Hence it follows from Theorem 1 that the quotient
Has a pole of order 2 at that point.
76. Zeros and Poles
52
( ) 1, ( ) ( 1)zp z q z z e
( ) 1
( ) ( 1)z
p z
q z z e
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Theorem 2
Let two functions p and q be analytic at a point z0 . If
then z0 is a simple pole of the quotient p(z)/q(z) and
76. Zeros and Poles
53
0 0 0( ) 0, ( ) 0, '( ) 0p z q z q z
0
0
0
( )( )Re
( ) '( )z z
p zp zs
q z q z
a zero of order m=1 at the point z0
0( ) ( ) ( ) q z z z g z
0 0
( ) ( ) ( )
( ) ( ) ( ) ( )
p z p z z
q z z z g z z z
pp. 252Theorem 1
0
00
0
( )( )Re ( )
( ) ( )
z z
p zp zs z
q z g zpp. 244Theorem (m=1)
0
0
( )
'( )
p z
q z
0 0( ) '( )g z q z
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Example 2 Consider the function
which is a quotient of the entire functions p(z) = cos z and q(z) = sin z. Its singularities occur at the zeros of q, or at the points z=nπ (n=0, ±1,±2,…)
Since p(nπ) =(-1)n ≠ 0, q(nπ)=0, and q’(nπ)=(-1)n ≠ 0,
Each singular point z=nπ of f is a simple pole, with residue Bn= p(nπ)/ q’(nπ)= (-1)n/(-1)n=1
76. Zeros and Poles
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cos( ) cot
sin
zf z z
z
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Example 4 Since the point
is a zero of polynomial z4+4. it is also an isolated singularity of the function
writing p(z)=z and q(z)=z4+4, we find that
p(z0)=z0 ≠ 0, q(z0)=0, and q’(z0)=4z03 ≠ 0
And hence that z0 is a simple pole of f, and the residue is
B0=p(z0)/ q’(z0)= -i/8
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/40 2 1iz e i
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zf z
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pp. 255-257
Ex. 6, Ex. 7, Ex. 8
76. Homework
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