(1) the residue (2) evaluating integrals using the residue (3) formula for the residue
DESCRIPTION
Section 8. SECTION 8 Residue Theory. (1) The Residue (2) Evaluating Integrals using the Residue (3) Formula for the Residue (4) The Residue Theorem. What is a Residue?. Section 8. The residue of a function is the coefficient of the term - PowerPoint PPT PresentationTRANSCRIPT
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
1
zz
2
21
21 842
1
1zz
z01 b
842
111
23
32 2
22
zz
zzzz
z 11 b
40
43
0
32
0
2
0
1
303
202010
)()()(
)()()()(
zz
b
zz
b
zz
b
zz
b
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
1
zz
2
21
21 842
1
1zz
z01 b
842
111
23
32 2
22
zz
zzzz
z 11 b
40
43
0
32
0
2
0
1
303
202010
)()()(
)()()()(
zz
b
zz
b
zz
b
zz
b
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
)()()(
)()()()(
zz
b
zz
b
zz
b
zz
b
zzazzazzaazf
C
nn
Cnn dzzzzf
jbdz
zz
zf
ja 1
010
))((2
1,
)(
)(
2
1
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
)()()(
)()()()(
zz
b
zz
b
zz
b
zz
b
zzazzazzaazf
C
nn
Cnn dzzzzf
jbdz
zz
zf
ja 1
010
))((2
1,
)(
)(
2
1
So
12)( jbdzzfC
C0z
We can use it to evaluate integrals
6
Section 8Example (1)
jjbdzzC
221
11
Integrate the function counterclockwise about z 2z1
1
2z
zzzz
zzzz
z 1111
11
1
1
32
32
By Cauchy’s Integral Formula:
jfjdzz
zfjdzzz
zf
CC
2)1(21
1)(2
)(0
0
singularpointcentre
7
Section 8
2z
zzzz
zzzz
z 1111
11
1
1
32
32
singularpointcentre
8
Section 8Example (1)
jjbdzzC
221
11
Integrate the function counterclockwise about z 2z1
1
2z
zzzz
zzzz
z 1111
11
1
1
32
32
By Cauchy’s Integral Formula:
jfjdzz
zfjdzzz
zf
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
1
1
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
zfjdzzz
zf
CCC
2)1(21
1
2
1)(2
)(0
0
23
322
zz
z
2/3z
zzzzz
zzz
zz
zzz
zz
z
29532
21842
111
18
9
4
5
2
3
23
32
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
zfjdzzz
zf
CCC
2)1(21
1
2
1)(2
)(0
0
23
322
zz
z
2/3z
zzzzz
zzz
zz
zzz
zz
z
29532
21842
111
18
9
4
5
2
3
23
32
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
1
sin4
z
z
7)3(
2
jz
e z
e.g.
e.g.
2)1(
33
z
ze.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
0
0
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
0
0
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
0
0
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
j
jz
jz
zjz
jzjz
zfzzzf
iziz
zzzz
)1)((
2)(
2
zjz
jzzf
Check: the Laurent series is
2
3
3
2
2
222
222
)(2
1)(
16
5
4
11
4
)2(
)(4
)2(
)(3
2
)(21
)2)((
)(2
)2/()(1
1
)2)((
)(2
)(2
1)(2
)(
1)(2
)1)((
2)(
jzjzjz
i
j
jz
j
jz
j
jz
jjz
jjz
jjzjjz
jjz
jzjjz
jjz
jzjz
jjz
zjz
jzzf
20 jz
18
Section 8Example (2)
Find the residue at the poles of
2
1
2
1lim
)2(
1lim)(Res
000
z
z
zz
zzzf
zzz
zz
zzf
2
1)(
2
Check: the Laurent series are
16
3
8
3
4
3
2
1
221
2
1
2/1
1
2
1
)2(
1)(
2
2
2 zz
z
zz
z
z
zz
z
zz
zzf
20 z
2
31lim
)2(
1)2(lim)(Res
222
z
z
zz
zzzf
zzz
8
)2(
4
)2(
2
1
)2(2
3
2
)2(
2
21
)2(2
3)2(
2/)2(1
1
)2(2
3)2(
)2(2
1
)2(
3)2(
)2(
1)(
2
2
2 zz
z
zz
z
z
zz
z
zz
z
zz
zzf
220 z
19
Section 8Example (2)
Find the residue at the poles of
2
1
2
1lim
)2(
1lim)(Res
000
z
z
zz
zzzf
zzz
zz
zzf
2
1)(
2
Check: the Laurent series are
16
3
8
3
4
3
2
1
221
2
1
2/1
1
2
1
)2(
1)(
2
2
2 zz
z
zz
z
z
zz
z
zz
zzf
20 z
2
31lim
)2(
1)2(lim)(Res
222
z
z
zz
zzzf
zzz
8
)2(
4
)2(
2
1
)2(2
3
2
)2(
2
21
)2(2
3)2(
2/)2(1
1
)2(2
3)2(
)2(2
1
)2(
3)2(
)2(
1)(
2
2
2 zz
z
zz
z
z
zz
z
zz
z
zz
zzf
220 z
20
Section 8Example (2)
Find the residue at the poles of
2
1
2
1lim
)2(
1lim)(Res
000
z
z
zz
zzzf
zzz
zz
zzf
2
1)(
2
Check: the Laurent series are
16
3
8
3
4
3
2
1
221
2
1
2/1
1
2
1
)2(
1)(
2
2
2 zz
z
zz
z
z
zz
z
zz
zzf
20 z
2
31lim
)2(
1)2(lim)(Res
222
z
z
zz
zzzf
zzz
8
)2(
4
)2(
2
1
)2(2
3
2
)2(
2
21
)2(2
3)2(
2/)2(1
1
)2(2
3)2(
)2(2
1
)2(
3)2(
)2(
1)(
2
2
2 zz
z
zz
z
z
zz
z
zz
z
zz
zzf
220 z
21
Section 8Example (2)
Find the residue at the poles of
2
1
2
1lim
)2(
1lim)(Res
000
z
z
zz
zzzf
zzz
zz
zzf
2
1)(
2
Check: the Laurent series are
16
3
8
3
4
3
2
1
221
2
1
2/1
1
2
1
)2(
1)(
2
2
2 zz
z
zz
z
z
zz
z
zz
zzf
20 z
2
31lim
)2(
1)2(lim)(Res
222
z
z
zz
zzzf
zzz
8
)2(
4
)2(
2
1
)2(2
3
2
)2(
2
21
)2(2
3)2(
2/)2(1
1
)2(2
3)2(
)2(2
1
)2(
3)2(
)2(
1)(
2
2
2 zz
z
zz
z
z
zz
z
zz
z
zz
zzf
220 z
22
Section 8Example (2)
Find the residue at the poles of
2
1
2
1lim
)2(
1lim)(Res
000
z
z
zz
zzzf
zzz
zz
zzf
2
1)(
2
Check: the Laurent series are
16
3
8
3
4
3
2
1
221
2
1
2/1
1
2
1
)2(
1)(
2
2
2 zz
z
zz
z
z
zz
z
zz
zzf
20 z
2
31lim
)2(
1)2(lim)(Res
222
z
z
zz
zzzf
zzz
8
)2(
4
)2(
2
1
)2(2
3
2
)2(
2
21
)2(2
3)2(
2/)2(1
1
)2(2
3)2(
)2(2
1
)2(
3)2(
)2(
1)(
2
2
2 zz
z
zz
z
z
zz
z
zz
z
zz
zzf
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 )()()(
zz
b
zz
bzzaazf
)()(lim)(Res 20
00
zfzzdz
dzf
zzzz
2013
012
002
0 )()()()()( bzzbzzazzazfzz
now differentiate:
12
01002
0 )(3)(2)()( bzzazzazfzzdz
d
12
0 )()(lim0
bzfzzdz
dzz
25
Section 8Example
Find the residue of at z1
9
2
)2(
2lim
2lim
)()(lim)(Res
211
20
00
zz
z
dz
d
zfzzdz
dzf
zz
zzzz
2)1)(2()(
zz
zzf
Check: the Laurent series is
81
)1(2
27
2
)1(9
2
)1(3
1
3
)1(
3
1
)1(3
1
)1(
1
3
1)1(
3
)1(
3
11
)1(3
1)1(
)3/)1((1
1
)1(3
1)1(
)1(3
1
)1(
1)1(
)1)(2()(
2
3222
2
2
222
z
zz
z
zz
zzz
z
z
zz
z
zz
z
zz
zzf
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
zz
b
zz
b
zz
bzzaazf
)()()()(
02
0
2
0
1010
)()(lim)!1(
1)(Res 0)1(
)1(
00
zfzzdz
d
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
bmzfzzdz
d mm
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
C
0z
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
izz
C
zfjdzzfi1
)(Res2)(
28
Section 8
Example
Integrate the function around
C
zz
z
2
2
zz
z
zz
zjdz
zz
zzz
C21202
2Res
2Res2
2
2z
32
lim2
Res
21
2lim
2Res
121
020
z
z
zz
zz
z
zz
z
zz
zz
jdzzz
z
C
222
29
Section 8
Example
Integrate the function around
C
zz
z
2
2
zz
z
zz
zjdz
zz
zzz
C21202
2Res
2Res2
2
2z
32
lim2
Res
21
2lim
2Res
121
020
z
z
zz
zz
z
zz
z
zz
zz
jdzzz
z
C
222
30
Section 8
Example
Integrate the function around
C
zz
z
2
2
zz
z
zz
zjdz
zz
zzz
C21202
2Res
2Res2
2
2z
32
lim2
Res
21
2lim
2Res
121
020
z
z
zz
zz
z
zz
z
zz
zz
jdzzz
z
C
222
31
Section 8
Example
Integrate the function around
C
zz
z
2
2
zz
z
zz
zjdz
zz
zzz
C21202
2Res
2Res2
2
2z
32
lim2
Res
21
2lim
2Res
121
020
z
z
zz
zz
z
zz
z
zz
zz
jdzzz
z
C
222
32
Section 8
Example
Integrate the function around
C
zz
z
2
2
zz
z
zz
zjdz
zz
zzz
C21202
2Res
2Res2
2
2z
32
lim2
Res
21
2lim
2Res
121
020
z
z
zz
zz
z
zz
z
zz
zz
jdzzz
z
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
izz
C
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 zq
zpzf
zz
2
0 sin2
de.g. using jez
dxxx
x
45
124
2
e.g.