october 21 residue theorem 7.1 calculus of residues
DESCRIPTION
Chapter 7 Functions of a Complex Variable II. October 21 Residue theorem 7.1 Calculus of residues. Suppose an analytic function f ( z ) has an isolated singularity at z 0 . Consider a contour integral enclosing z 0. z 0. - PowerPoint PPT PresentationTRANSCRIPT
October 21 Residue theorem
7.1 Calculus of residues
Chapter 7 Functions of a Complex Variable II
1
Suppose an analytic function f (z) has an isolated singularity at z0. Consider a contour integral enclosing z0 .
z0
)( sRe22)(
1 ,2)ln(
1 ,01
)(
)(
)()()(
01
1
'
'01
'
'
10
0
00
zfiiadzzf
niazza
nn
zza
dzzza
dzzzadzzzadzzf
C
z
z
z
z
n
n
C
nn
nC
nnC
n
nnC
The coefficient a-1=Res f (z0) in the Laurent expansion is called the residue of f (z) at z = z0.
If the contour encloses multiple isolated singularities, we have the residue theorem:
n
nCzfidzzf )( sRe2)(
z0 z1
Contour integral =2i ×Sum of the residuesat the enclosed singular points
2
Residue formula:
To find a residue, we need to do the Laurent expansion and pick up the coefficient a-1. However, in many cases we have a useful residue formula(Problem 6.6.1):
)()(lim)!1(
1)(sRe
)!1())(2()1)((lim)(lim
)(lim)!1(
1)()(lim
)!1(
1
:Proof
.)()(lim)(sRe
,pole simple afor ly,Particular
)()(lim)!1(
1)(sRe
,order of pole aFor
01
1
01
11
1001
1
01
1
01
1
00
01
1
0
0
00
00
0
0
zfzzdz
d
mzfa
mazznmnmnazzadz
d
zzadz
d
mzfzz
dz
d
m
zfzzzf
zfzzdz
d
mzf
m
mm
m
zz
n
nn
zzmn
mnnm
m
zz
mn
mnnm
m
zz
mm
m
zz
zz
mm
m
zz
3
.0 ,)()(lim!
1
:tscoefficien theall find way toa is e that therprovedactually We
.)()(lim)!1(
1 us gives 1 upPick . Also
.)()(lim!
1 ,)()()(
expansionTaylor by analytic, is )()( Because
)()()(
)()(
:#2 Method Proof
0
01
1
1
000
0
0
00
0
0
0
0
kzfzzdz
d
ka
a
zfzzdz
d
mamkab
zfzzdz
d
kbzzbzfzz
zfzz
zzazfzz
zzazf
mk
k
zzmk
mm
m
zzmkk
mk
k
zzk
k
kk
m
m
mn
mnn
m
n
mnn
4
.1 of residue a with 0at polse simple a has 1
1 therefore,1
1
1lim
.2
1 of residue a with at polse simple a has
1
1 therefore,
2
1
1
1)(lim
:Examples
.)()(lim)(sRe
.at pole simple a has )(
0)()(lim
.2for 0
0 and exists )(lim)()(lim
then,)()(Let
:Proof
.)()(lim)(sRe)2
.at pole simple a has )( 1)
then0, toequalnot is and exists )()(lim If
:Fact
0
22
00
0
10
100
0
00
0
0
00
00
0
0
zee
z
iiz
ziziz
zfzzzf
zzzf
azfzz
na
zzazfzz
zzazf
zfzzzf
zzzf
zfzz
zzz
iz
zzzz
n
n
nn
zzzz
nn
nn
zz
zz
5
Residue at infinity:Stereographic projection:
Residue at infinity:Suppose f (z) has only isolated singularities, then its residue at infinity is defined as
nn
CC
zf
dzzfi
dzzfi
f
)( sRe
)(2
1)(
2
1)( sRe ~
Another way to prove it is to use Cauchy’s integral theorem. The contour integral for a small loop in an analytic region is
)( sRe)( sRe20)(
0
fzfidzzfn
nCz
C~
.1)( sRe ,1
)( :Example fz
zf
One other equivalent way to calculate the residue at infinity is
.0at 11
Res11
2
1)(
2
1)( sRe
22
/1let
~0
ZZZ
fdZZZ
fi
dzzfi
fC
Zz
C
By this definition a function may be analytic at infinity but still has a residue there.
6
Read: Chapter 7: 1Homework: 7.1.1Due: October 28
.1 ,
even ,1
2
.1odd ,0
)1(
1
)1()1(
)(
)(let )()(
0. )( )(
0
0000
0
0
0
0
0
0
0
)1(
)1()1()1())(1()1(
)1(
)1(1
00
)(
n
nen
i
n
ni
ee
ni
ee
ni
ede
deiadeiea
ezzdzzzadzzf
dzzfdzzf
ni
ninininini
ni
nin
nn
ini
nn
i
S
n
nnS
SC
7
Cauchy principle value:
Suppose f (z) has an isolated singularity z0 lying on a closed contour C. The contour integral is then not well defined. To solve this problem, we can remove a small segment of the contour for a distance of on each side of the singularity and create a new contour C(). We define the Cauchy principle value of as
C dzzf )(
C dzzf )(
exists.)limit the(If )(lim)()(0
CC
dzzfdzzfP
October 24 Cauchy principle value
z0
C()
S C
Let us first see on what condition this limit exists. We draw a semicircle path S with radius around z0. Suppose contourC=C()+S does not enclose any singularity, then
8
).(Res22
1)(Res )(lim)(
condition, Under this
.even negativefor 0 then isexist to)(limfor condition The
1
2 )( )(
001)(0
)(0
)1(1
even1)(
0
zfizfiaidzzfdzzfP
nadzzf
en
aaidzzfdzzf
CC
nC
nin
nnSC
2) We assumed that the contour is smooth at z =z0. If not the value should be3) It is easy to remember: when the singularity is on the contour, it contributes half as if it were inside the contour.4) Similar definitions exist for open contour integrals. Example:
Also for infinity integration limits. Example:
).(Res 0zfi
.)(sRe2)(sRe)(1
0
n
iiC
zfizfidzzfP
0
0
)()(lim)(0
x
a
b
x
b
adxxfdxxfdxxfP
a
aadxxfdxxfP )(lim)(
More about Cauchy principle values:1) If C originally encloses isolated singularities, then
9
Example:
dx
xxP
)1)(1(
12
.24
12
2
1
.4
1)()(lim)( Res
,2
1)()1(lim)1( Res
)( Res2)1( Res
.)even negativefor 0(
)1)(1(
1
)1)(1(
1
)1)(1(
1
)1)(1(
1lim
)1)(1(
1lim
1
2
222
2
iiiI
izfizif
zfzf
ififi
na
dzzz
P
dzzz
dzzz
dxxx
P
dxxx
PI
iz
z
n
C
R
RR
R
RR
1
Ci
-iR
10
Read: Chapter 7: 1Homework: 7.1.4Due: November 4
11
Cauchy’s integral theorem and Cauchy’s integral formula revisited:
(in the view of the residue theorem):
.!
)()(
)!11(
1lim
is at residue its formula, residue the toAccording
1.order of pole a isIt .)(')()(
)'3
!
)(22
)()(
)( )3
)(2)(
)(')()(
)2
0)(Res2)( )1
))((')()()( :function Analytic
0)(
10
101)1(
1)1(
0
0
01
0
01
0
0)(
100
101
0
00
00
0
0
0
0000
0
0 n
zf
zz
zfzz
dz
d
n
zz
nzz
zf
zz
zf
zz
zf
n
zfiiadz
zz
zfzza
zz
zf
zifdzzz
zfzf
zz
zf
zz
zf
zfidzzf
zzzfzfzzazf
n
n
n
n
n
zz
nnn
n
nC nm
nmmn
C
C
m
mm
October 26 Evaluation of definite integrals -1
7.1 Calculus of residues
12
L’Hospital’s rule:
exsists.limit theuntil )(
)(lim
)(''
)(''lim
)('
)('lim
)(
)(lim ,Repeatedly
.)('
)('lim
)(
)(lim then exists,
)('
)('lim and ,or 0)(lim)(lim If
)(
)(
)or ()or ()or ()or ()or (
0000
00000
zg
zf
zg
zf
zg
zf
zg
zf
zg
zf
zg
zf
zg
zfzgzf
n
n
zzzzzzzz
zzzzzzzzzz
.2
1
2lim
2
1lim
1lim
:Example
0020
z
z
z
z
z
z
e
z
e
z
ze
I. Integrals of trigonometric functions :
2
0)cos,(sin df
.1
2
/1,
2
/1 Res2
2
/1,
2
/1)cos,(sin
.2
/1cos ,
2
/1sin , , ,
circle.unit thearound integralcontour a usingby done becan This
circleunit
2
0
iz
zz
i
zzfi
iz
dzzz
i
zzfdf
zz
i
zz
iz
dzddiedzez
C
ii
C
r=1
13
.3
2
32
122
32
111lim)(Res
circle). theof(out 32
,circle) the(inside 32014
142
2/1
2
1
cos2
1
:1 Example
2
2
2
0
iiI
zzzzzzzzzf
z
zzz
zz
dzi
iz
dzzz
dI
zz
CC
C
r=1z+z-
.224
121
4
1
12
4
1
2
/1course) (of cos
:2 Example
3
3
2422
0
2
ii
dzzzzi
dzz
zz
iiz
dzzzdI
C
CC
C
r=1
14
.1
2
1
1
22
2
.1
1
2
1
))((
1lim)( Res
circle. theofout is circle, in the is |||| ,1||
.111
01/2 poles, simple 2 have We
))((
2
12
2
2/1
1
1
ok.)our textbo (From 1.|| and real is ,cos1
:3 Example
22
2
22
2
2
0
ii
I
zzzzzzzzzf
zzzzzz
zzz
zzzz
dz
izz
dz
iiz
dzzz
I
dI
zz
CCC
C
r=1z+z-
15
2222
2
22
2
22
2
22
2
0
220
2
2
2
2
2
0
111
2
1112
1 ,
111
12
11
)1(1
)(
1111)( Res )0( Res
.)(
1
))((
1lim)( Res
.1
))((
1lim)0( Res
circle. theofout is circle, in the is |||| ,1||
.1101/2 ,0 poles, simple 3 have We
)1/2(
11
2
111
2//11
2//1
1.|| and real is ,sin1
sin
:4 Example
aa
a
aa
ai
iaI
aa
a
aa
ia
aizzz
zz
zz
z
zzzff
zzz
z
zzzzz
zzzzf
zzzzzzz
zzf
zzzzzz
aa
izaizzz
dzaizzz
z
iadz
aizazz
z
iiz
dz
izza
izzI
aaa
dI
zz
z
CCC
C
r=1
z+
z-
z0
16
Read: Chapter 7: 1Homework: 7.1.7,7.1.8,7.1.10Due: November 4
17
October 31 Evaluation of definite integrals -2
7.1 Calculus of residues
II. Integrals along the whole real axis:
dxxf )(
Assumption 1: f (z) is analytic in the upper or lower half of the complex plane, except isolated finite number of poles.
∩
R
Condition for closure on a semicircular path:
dzzfdzzfdzzfdzzfdzzfdxxf
RCR
R
RR)(lim)(lim)()()(lim)(
.0 ,1
~)(lim0lim) (lim
) (lim ) (lim)(lim
1max0
00
zzfRfRdeRf
deiReRfdeiReRfdzzf
RR
i
R
ii
R
ii
RR
Assumption 2: when |z|, | f (z)| goes to zero faster than 1/|z|.
Then, plane. halfupper on the )( of esiduesR2)(lim)( zfidzzfdxxfCR
18
.arctan1
Or
.))((
1lim2)( Res2
plane halfupper on the 1
1 of esiduesR2
1
:1 Example
2
2
2
xx
dx
iziziziifi
ziI
x
dxI
iz
.
2'
)()(
1lim2)( Res2
plane halfupper on the 1
of esiduesR2
.0 ,
:2 Example
322
2
222
222
aaizaiziaziiafi
aziI
aax
dxI
aiz
19
III. Fourier integrals: .0 ,)(
kdxexf ikx
Assumption 1: f (z) is analytic in the upper half of the complex plane, except isolated finite number of poles.
dzezfdzezfdxexf ikz
RC
ikz
R
ikx )(lim)(lim)(
Jordan’s lemma:
.0)(lim number,given any becan Since
)1(/2
1222
)(
then,Let .)( have we that whenso exists there0given any For
:Proof
0. with 0)(lim then ,0)(lim If
2/
0
/22/
0
sin
0
sin
0
sincos
dzezf
ke
kkR
eRdeRdeR
deRdeiRedzedzezf
MRzfMzM
kdzezfzf
ikz
R
kRkR
kRkR
kRikRikRikzikz
ikz
Rz
plane. halfupper on the )( of esiduesR2)(lim)( Then,
.0)(lim :2 Assumption
ikz
C
ikz
R
ikx
z
ezfidzezfdxexf
zf
∩
R
20
.))((
lim2)( Res2
plane halfupper on the of esiduesR2
.0,0 ,
:1 Example
22
22
kaikz
iaz
ikz
ikx
eaiaziaz
eiaziiafi
az
eiI
akax
dxeI
Question: How about
Answer: We can go the lower half of the complex plane using a clockwise contour.
?0)( ,)(
kdxexf ikx
plane. halflower on the )( of esiduesR2)(lim)( ikz
C
ikz
R
ikx ezfidzezfdxexf
.))((
lim2)( Res2
plane halflower on the of esiduesR2
.0,0 ,
:2 Example
22
22
kaikz
iaz
ikz
ikx
eaiaziaz
eiaziiafi
az
eiI
akax
dxeI
21
Question: How about
Answer:
?0)( ,sin)( ,cos)(
kkxdxxfkxdxxf
.)(2
1sin)( ,)(
2
1cos)(
.)(Imsin)( ,)(Recos)(
dxeexfi
kxdxxfdxeexfkxdxxf
dxexfkxdxxfdxexfkxdxxf
ikxikxikxikx
ikxikx
.Re
.0,0 ,cos
:3 Example
22
22
kaikx
eaax
dxeI
akax
kxdxI
.2
.lim)0(Res
,Im2
1
.sin
:4 Example
0
0
Iix
exifi
x
dxeP
x
dxePI
dxx
xPI
ix
x
ix
ix
22
Read: Chapter 7: 1Homework: 7.1.11,7.1.12,7.1.13,7.1.14,7.1.16Due: November 11
23
November 2 Evaluation of definite integrals -3
7.1 Calculus of residues
IV. Rectangular contours: Exponential and hyperbolic forms
is a periodic function. We may use rectangular contours and hope that the integral reappears in some way on the upper contour line.
ixx ee 2
ae
eiI
eie
eiziifi
e
eiIe
e
dxee
e
dxe
e
idye
e
dxe
e
idye
e
dxe
e
dze
ae
dxeI
ia
ai
aiz
az
iz
z
aziaR
R x
axiaR
R x
ax
R
iyR
iyRaR
R ix
ixa
iyR
iyRaR
R x
ax
Rz
az
R
x
ax
sin1
2
)(21
lim2)( Res2
rectangle in the 1 of esiduesR2)1(
11lim
1111lim
1lim
.10 ,1
:1 Example
2
22
0
2
)(
)2(
)2(2
0
)(
R
i
24
)2/cosh(1
)(2
)(2)(2
cosh2
3lim
cosh2lim2
)2
3( Res)
2( Res2
rectangle in the cosh
of esiduesR2
)1(coshcosh
lim
)cosh()2cosh()cosh(coshlim
coshlim
.0 ,cosh
:2 Example
2
2/32/
2/32/2/32/
2
3
2
22
0
2
)()2(2
0
)(
ae
eeI
eeieiei
z
eiz
z
eizi
if
ifi
z
ei
Iex
dxee
x
dxe
iyR
idye
ix
dxe
iyR
idye
x
dxe
z
dze
ax
dxeI
a
aa
aaaa
iaz
iz
iaz
iz
iaz
aR
R
iaxaR
R
iax
R
iyRiaR
R
ixiaiyRiaR
R
iax
R
iaz
R
iax
R
ii3
This integral can also be done by using the line y=i and the fact .cosh)cosh( ziz
25
V. Sector contours
For functions involving we may use sector contours and hope that the integral reappears in some way on the upper radius of the sector.
,)( because , /2 nninn xexx
)/sin(1
2
21
lim2 Res2
sector in the 1
of esiduesR2
)1(lim
)() (
limlim
2. andinteger an is 0, ,
:1 Example
1/2
1
/
1
///
/20/2
0
0
/2
/2/2
00
0
/
nnae
nae
i
I
na
ei
azaeziaefi
azi
Iear
dre
ax
dx
are
dre
aeR
deiR
ax
dx
az
dz
nnaax
dxI
nni
n
ni
n
ni
nnni
aez
ni
nn
ni
R nnniR
nnR
R nnni
nin
nni
iR
nnRnnR
nn
ni
R
2/n
26
VI. Contours avoiding branch points
When the integrands have branch points and branch cuts, contours need to be designed to avoid the branch points and the branch cuts.
contour in the of esiduesR22
.limlimlim
.0limlim
. when 1
~ because ,0lim
lim
cut.branch theas axis positive theuse Wepoint.branch a is 0
0. ,
:1 Example
22showncontour 22
22
0
22
0
22
0
0
2 222
2/
0220
2/32222
22222222
0showncontour 22
0 22
az
ziI
az
dzz
Iar
drr
ar
drer
ax
dxx
ae
deie
az
dzz
Rzaz
z
az
dzz
ax
dxx
ax
dxx
az
dzz
az
dzz
az
dzz
xz
aax
dxxI
R
RR
i
RRR
i
ii
C
CR
R
RCCR
C L+
L-
ia
-ia
27
aI
aa
i
a
ii
az
ziaz
az
ziazi
iafiafiaz
ziI
az
dzz
iaziaz
2
2
22
1
22
12
limlim2
Res Res2
contour in the of esiduesR22
2222
22showncontour 22
C L+
L-
ia
-ia
28
Read: Chapter 7: 1Homework: 7.1.17 (b),7.1.18,7.1.19,7.1.21,7.1.22,7.1.25,7.1.26Due: November 11
29
Reading: Dispersion relations
7.2 Dispersion relations
00
00
000
00
00
0000
)(1)(
)(1)(
)()()()(
)()(
0
) (lim
)(lim
)()()(
lim)(
lim
xx
dxxuPxv
xx
dxxvPxu
xivxuixx
dxxivxuPxif
xx
dxxfP
xeR
deiReRf
xz
dzzf
xifxz
dzzf
xx
dxxfP
xz
dzzfP
i
ii
RR
R
RRCR
Suppose f (z) = u(z)+iv(z) is analytic, and in the upper half plane, then
0)(lim||
zfz C
Rx0
These are the dispersion relations. u(z) and v(z) are sometimes called a Hilbert transform pair.
30
Symmetry relations:
0 20
20
00
00
00
0
000
0 20
2
00
00
00
0
000
*
)(2
)()(1)()(1)(1)(
)(2
)()(1)()(1)(1)(
).()( ),()(then
function), real a of ransform(Fourier t )()( Suppose
xx
dxxuxP
xx
dxxu
xx
dxxuP
xx
dxxu
xx
dxxuP
xx
dxxuPxv
xx
dxxxvP
xx
dxxv
xx
dxxvP
xx
dxxv
xx
dxxvP
xx
dxxvPxu
xvxvxuxu
xfxf
31
In optics, the refractive index can be chosen to be a complex number where its real part is the normal refractive index, determined by the electric permittivity (), and its imaginary part is responsible for absorption, determined by the electric conductivity ():
)()(4
)()(2 fin
0 20
20
0
0 20
20
0 20
20
0
0 20
20
)(Re2)(Im
)(Im2)(Re
)(2)(
)(2)(
df
Pf
dfPf
xx
dxxuxPxv
xx
dxxxvPxu
These are the Kramers-Kronig relations. These relations state that knowledge of the absorption coefficient of a material over all frequencies will allow one to obtain the (normal) refractive index of the material for any frequency, and vise versa.
32
Hi, Everyone:
My previous student Mahendra Thapa brought to me this problem, which may be in Arfken 3rd edition:
It is used in the black body radiation:
If you can crack this integral (using contour integral on the complex plane) by yourself, please come to show me.
0
3
1dx
e
xx