SECTION 8 Residue Theory (1) The Residue
34
1 (1) The Residue (2) Evaluating Integrals using the Residue (3) Formula for the Residue (4) The Residue Theorem Section 8 SECTION 8 Residue Theory
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What is a Residue? Section 8 The residue of a function is the coefficient of the term in the Laurent series expansion (the coefficient b1). Examples:
Transcript of SECTION 8 Residue Theory (1) The Residue
1
(1) The Residue
(2) Evaluating Integrals using the Residue
(3) Formula for the Residue
(4) The Residue Theorem
Section 8
SECTION 8Residue Theory
2
Section 8What is a Residue?The residue of a function is the coefficient of the term
in the Laurent series expansion (the coefficient b1).0
1zz
221
21 842
11 zz
z01 b
842111
2332 2
22
zzzzzz
z 11 b
40
43
0
32
0
2
0
1
303
202010
)()()(
)()()()(
zzb
zzb
zzb
zzb
zzazzazzaazf
Examples:
3
Section 8What is a Residue?The residue of a function is the coefficient of the term
in the Laurent series expansion (the coefficient b1).0
1zz
221
21 842
11 zz
z01 b
842111
2332 2
22
zzzzzz
z 11 b
40
43
0
32
0
2
0
1
303
202010
)()()(
)()()()(
zzb
zzb
zzb
zzb
zzazzazzaazf
Examples:
4
Section 8What’s so great about the Residue?The formula for the coefficients of the Laurent series saysthat (for f (z) analytic inside the annulus)
40
43
0
32
0
2
0
1
303
202010
)()()(
)()()()(
zzb
zzb
zzb
zzb
zzazzazzaazf
C
nn
Cnn dzzzzf
jbdz
zzzf
ja 1
010
))((21,
)()(
21
So
12)( jbdzzfC
C0z
We can use it to evaluate integrals
5
Section 8What’s so great about the Residue?The formula for the coefficients of the Laurent series saysthat (for f (z) analytic inside the annulus)
40
43
0
32
0
2
0
1
303
202010
)()()(
)()()()(
zzb
zzb
zzb
zzb
zzazzazzaazf
C
nn
Cnn dzzzzf
jbdz
zzzf
ja 1
010
))((21,
)()(
21
So
12)( jbdzzfC
C0z
We can use it to evaluate integrals
6
Section 8Example (1)
jjbdzzC
221
11
Integrate the function counterclockwise about z 2z11
2z
zzzz
zzzz
z 111111
11
32
32
By Cauchy’s Integral Formula:
jfjdzz
zfjdzzzzf
CC
2)1(21
1)(2)(0
0
singularpointcentre
7
Section 8
2z
zzzz
zzzz
z 111111
11
32
32
singularpointcentre
8
Section 8Example (1)
jjbdzzC
221
11
Integrate the function counterclockwise about z 2z11
2z
zzzz
zzzz
z 111111
11
32
32
By Cauchy’s Integral Formula:
jfjdzz
zfjdzzzzf
CC
2)1(21
1)(2)(0
0
singularpointcentre
9
Section 8Example (1) cont.
jjbdzzC
221
11
We could just as well let the centre be at z1
2z
10
,1
11
1)(
zzz
zf
centre /singular
point
- a one-term Laurent series
- as before
10
Section 8Example (2)
jjbdzzz
z
C
2223
3212
Integrate the function counterclockwise about z 3/2
By Cauchy’s Integral Formula:
jfjdzz
dzz
zfjdzzzzf
CCC
2)1(21
12
1)(2)(0
0
2332
2 zz
z
2/3z
zzzzz
zzzzz
zzz
zzz
29532
21842
111
189
45
23
2332
432
2
2
2
2
0
11
Section 8Example (2)
jjbdzzz
z
C
2223
3212
Integrate the function counterclockwise about z 3/2
By Cauchy’s Integral Formula:
jfjdzz
dzz
zfjdzzzzf
CCC
2)1(21
12
1)(2)(0
0
2332
2 zz
z
2/3z
zzzzz
zzzzz
zzz
zzz
29532
21842
111
189
45
23
2332
432
2
2
2
2
0
12
Section 8
So the Residue allows us to evaluate integrals of analyticfunctions f (z) over closed curves C when f (z) has one singularpoint inside C.
12)( jbdzzfC
C0z
b1 is the residue of f (z) at z0
13
Section 8
That’s great - but every time we want to evaluate an integraldo we have to work out the whole series ?
No - in the case of poles - there’s a quick and easy wayto find the residue
We’ll do 3 things:
1. Formula for finding the residue for a simple pole
2. Formula for finding the residue for a pole of order 2
3. Formula for finding the residue for a pole of any order
1sin4z
z
7)3(2
jze z
e.g.
e.g.
2)1(33
zz
e.g.
14
Section 8Formula for finding the residue for a simple pole
If f (z) has a simple pole at z0, then the Laurent series is
Rzzzz
bzzaazf
00
1010 0)()(
12
01000 )()()()( bzzazzazfzz
10 )()(lim0
bzfzzzz
)()(lim)(Res 000
zfzzzfzzzz
we’re putting the centre atthe singular point here
15
Section 8Formula for finding the residue for a simple pole
If f (z) has a simple pole at z0, then the Laurent series is
Rzzzz
bzzaazf
00
1010 0)()(
12
01000 )()()()( bzzazzazfzz
10 )()(lim0
bzfzzzz
)()(lim)(Res 000
zfzzzfzzzz
we’re putting the centre atthe singular point here
16
Section 8Formula for finding the residue for a simple pole
If f (z) has a simple pole at z0, then the Laurent series is
Rzzzz
bzzaazf
00
1010 0)()(
12
01000 )()()()( bzzazzazfzz
10 )()(lim0
bzfzzzz
)()(lim)(Res 000
zfzzzfzzzz
we’re putting the centre atthe singular point here
17
Section 8Example (1)
Find the residue of at zj
4)()2(lim
)1)(()2)((lim
)()(lim)(Res
22
000
jjzjz
zjzjzjz
zfzzzf
iziz
zzzz
)1)((2)( 2
zjzjzzf
Check: the Laurent series is
2
3
3
2
2
222
222
)(21)(
165
411
4
)2()(4
)2()(3
2)(21
)2)(()(2
)2/()(11
)2)(()(2
)(21)(2
)(1)(2
)1)((2)(
jzjzjz
i
jjz
jjz
jjz
jjzjjz
jjzjjzjjz
jzjjzjjz
jzjzjjz
zjzjzzf
20 jz
18
Section 8Example (2)
Find the residue at the poles of
21
21lim
)2(1lim)(Res
000
z
zzzzzzf
zzz
zzzzf
21)( 2
Check: the Laurent series are
163
83
43
21
221
21
2/11
21
)2(1)(
2
2
2 zzz
zzz
zzz
zzzzzf
20 z
231lim
)2(1)2(lim)(Res
222
z
zzzzzzf
zzz
8)2(
4)2(
21
)2(23
2)2(
221
)2(23)2(
2/)2(11
)2(23)2(
)2(21
)2(3)2(
)2(1)(
2
2
2 zzz
zzz
z
zzz
zzz
zzzzf
220 z
19
Section 8Example (2)
Find the residue at the poles of
21
21lim
)2(1lim)(Res
000
z
zzzzzzf
zzz
zzzzf
21)( 2
Check: the Laurent series are
163
83
43
21
221
21
2/11
21
)2(1)(
2
2
2 zzz
zzz
zzz
zzzzzf
20 z
231lim
)2(1)2(lim)(Res
222
z
zzzzzzf
zzz
8)2(
4)2(
21
)2(23
2)2(
221
)2(23)2(
2/)2(11
)2(23)2(
)2(21
)2(3)2(
)2(1)(
2
2
2 zzz
zzz
z
zzz
zzz
zzzzf
220 z
20
Section 8Example (2)
Find the residue at the poles of
21
21lim
)2(1lim)(Res
000
z
zzzzzzf
zzz
zzzzf
21)( 2
Check: the Laurent series are
163
83
43
21
221
21
2/11
21
)2(1)(
2
2
2 zzz
zzz
zzz
zzzzzf
20 z
231lim
)2(1)2(lim)(Res
222
z
zzzzzzf
zzz
8)2(
4)2(
21
)2(23
2)2(
221
)2(23)2(
2/)2(11
)2(23)2(
)2(21
)2(3)2(
)2(1)(
2
2
2 zzz
zzz
z
zzz
zzz
zzzzf
220 z
21
Section 8Example (2)
Find the residue at the poles of
21
21lim
)2(1lim)(Res
000
z
zzzzzzf
zzz
zzzzf
21)( 2
Check: the Laurent series are
163
83
43
21
221
21
2/11
21
)2(1)(
2
2
2 zzz
zzz
zzz
zzzzzf
20 z
231lim
)2(1)2(lim)(Res
222
z
zzzzzzf
zzz
8)2(
4)2(
21
)2(23
2)2(
221
)2(23)2(
2/)2(11
)2(23)2(
)2(21
)2(3)2(
)2(1)(
2
2
2 zzz
zzz
z
zzz
zzz
zzzzf
220 z
22
Section 8Example (2)
Find the residue at the poles of
21
21lim
)2(1lim)(Res
000
z
zzzzzzf
zzz
zzzzf
21)( 2
Check: the Laurent series are
163
83
43
21
221
21
2/11
21
)2(1)(
2
2
2 zzz
zzz
zzz
zzzzzf
20 z
231lim
)2(1)2(lim)(Res
222
z
zzzzzzf
zzz
8)2(
4)2(
21
)2(23
2)2(
221
)2(23)2(
2/)2(11
)2(23)2(
)2(21
)2(3)2(
)2(1)(
2
2
2 zzz
zzz
z
zzz
zzz
zzzzf
220 z
23
Section 8
Find the residue at the pole z01 of )1(3)(
2
zz
zzf
Question:
24
Section 8Formula for finding the residue for a pole of order 2
If f (z) has a pole of order 2 at z0, then the Laurent series is
20
2
0
1010 )()()(
zzb
zzbzzaazf
)()(lim)(Res 20
00
zfzzdzdzf
zzzz
2013
012
002
0 )()()()()( bzzbzzazzazfzz
now differentiate:
12
01002
0 )(3)(2)()( bzzazzazfzzdzd
12
0 )()(lim0
bzfzzdzd
zz
25
Section 8Example
Find the residue of at z1
92
)2(2lim
2lim
)()(lim)(Res
211
20
00
zzz
dzd
zfzzdzdzf
zz
zzzz
2)1)(2()(
zzzzf
Check: the Laurent series is
81)1(2
272
)1(92
)1(31
3)1(
31
)1(31
)1(1
31)1(
3)1(
311
)1(31)1(
)3/)1((11
)1(31)1(
)1(31
)1(1)1(
)1)(2()(
2
3222
2
2
222
zzz
zzz
zzzz
z
zzz
zzz
zzzzf
310 z
26
Section 8Formula for finding the residue for a pole of any order
If f (z) has a pole of order m at z0, then the Laurent series is
mm
zzb
zzb
zzbzzaazf
)()()()(
02
0
2
0
1010
)()(lim)!1(
1)(Res 0)1(
)1(
00
zfzzdzd
mzf m
m
m
zzzz
mm
mmmm
bzzb
zzbzzazzazfzz
2
02
101
101000
)(
)()()()()(
now differentiate m1 times and let zz0 to get:
10)1(
)1(
)!1()()(lim0
bmzfzzdzd m
m
m
zz
27
Section 8
We saw that the integral of an analytic function f (z) over a closed curve C when f (z) has one singular point inside C is
12)( jbdzzfC
C0z
b1 is the residue of f (z) at z0
The Residue Theorem
C
Residue Theorem: Let f (z) be an analyticfunction inside and on a closed path Cexcept for at k singular points inside C.Then
k
i zzC
zfjdzzfi1
)(Res2)(
28
Section 8
Example
Integrate the function around
C
zzz
2
2
zzz
zzzjdz
zzz
zzC
21202
2Res2Res22
2z
32lim2Res
21
2lim2Res
121
020
zz
zzz
zz
zzz
zz
zz
jdzzzz
C
222
29
Section 8
Example
Integrate the function around
C
zzz
2
2
zzz
zzzjdz
zzz
zzC
21202
2Res2Res22
2z
32lim2Res
21
2lim2Res
121
020
zz
zzz
zz
zzz
zz
zz
jdzzzz
C
222
30
Section 8
Example
Integrate the function around
C
zzz
2
2
zzz
zzzjdz
zzz
zzC
21202
2Res2Res22
2z
32lim2Res
21
2lim2Res
121
020
zz
zzz
zz
zzz
zz
zz
jdzzzz
C
222
31
Section 8
Example
Integrate the function around
C
zzz
2
2
zzz
zzzjdz
zzz
zzC
21202
2Res2Res22
2z
32lim2Res
21
2lim2Res
121
020
zz
zzz
zz
zzz
zz
zz
jdzzzz
C
222
32
Section 8
Example
Integrate the function around
C
zzz
2
2
zzz
zzzjdz
zzz
zzC
21202
2Res2Res22
2z
32lim2Res
21
2lim2Res
121
020
zz
zzz
zz
zzz
zz
zz
jdzzzz
C
222
33
Section 8Proof of Residue TheoremEnclose all the singular pointswith little circles C1, C1, Ck.
f (z) is analytic in here
By Cauchy’s Integral Theorm for multiply connected regions:
kCCCC
dzzfdzzfdzzfdzzf )()()()(21
C
But the integrals around each of the small circles is just theresidue at each singular point inside that circle, and so
k
i zzC
zfjdzzfi1
)(Res2)(
34
Section 8
Topics not Covered
(1) Another formula for the residue at a simple pole (when f (z) is a rational function p(z)q(z),
(2) Evaluation of real integrals using the Residue theorem
(3) Evaluation of improper integrals using the Residue theorem
)()()(Res
0
0
0 zqzpzf
zz
2
0 sin2de.g. using jez
dxxx
x45
124
2
e.g.
