7. Calculus of Residues
0 , 1
0
2
circles once
n
n
C
z z id z
C z
Residue Theorem :
0
n
nn
f z a z z
12
C
f z i ad z a1 = residue of f at z0 .
For a meromorphic function f with n isolated singularities,one can deform C to circumvent them all so that
1
0
i
n
iC C
f z f zd z d z
1,1
2n
ii
C
f z i ad z
Residue Theorem
a1, i = residue at singularity i.
Computing Residues 0
n
nn
f z a z z
1
0 1 00
n
nn
z z f z a a z z
0
1 0limz z
a z z f z
If f has a simple pole at z0 , 01
n
nn
f z a z z
simple poles
If f has a pole of order m at z0 , 0
n
nn m
f z a z z
2
1
0 0 1 0 00 0
mm k m n m
nm kk n
z z f z a z z a z z a z z
0
1
1 01
1lim
1 !
mm
mz z
da z z f z
m d z
poles of order m
Example 11.7.1. Computing Residues
Some examples :
1/4
1 1 1 1Res ; lim
4 1 4 4 4 1zz z
z z
1
4
0
1 1Res ; 0 lim
sin sinzz z
z z
1
/2
/22 22
ln lnRes ; 2 lim 2
4 4i
i
z e
z zz e z i
z z
ln 2 / 2
4
i
i
22 2Res ; lim
sin sinz
z d zz z
z dz z
2lim 3 2 cot cscz
z z z z z z
1
/22
lnlim
2iz e
z
z i
Mathematica
2 31
cotRes ; 0
2
1 1 11
2 2 3
zz
z z
z za of O z O z
z z
31cot
3
xx O x
x
1
4
21 11
2 2 2
zO z
z
1/ 21
1Res ; 0 1ze z a of O z
z
1
Alternatively, z = 0 is a pole of 2nd order :
2
0
cot cotRes ; 0 lim
2 2z
z d zz z
z z d z z z
2
20
cot cot csclim
2 22z
z z z z z
z zz
1
4
Mathematica
Cauchy Principal ValueSimple pole on path :
Eg.0
1b
a
I d xx x
Path: a b on real axis.
Simple pole at x = x0.
At x0, integral diverges as 0ln x x
Cauchy principal value of a real integral with a simple pole at x0 is defined as
0
0
0lim
xb b
a a x
d x f x d x f x
P
But the divergence cancels out if the pole is approached from both sides in a limiting process.
Caution: The real part of such integral is often defined by its Cauchy principal value.
However, the integral itself is actually not convergent. For example, the value of the limit changes if we use different values of for different sides of x0 .
Mathematica
Example 11.7.2. A Cauchy Principal Value
0
sin xI d x
x
1 sin
2
xd x
x
0
1 sinlim
2
xI d x
x
P
02
i x i xe ed x
ix
0 0
1
2
i x i xe ed x d x
i x x
1
2
i xed x
i x
0lim
2
i xed x
ix
Contour Integral & Cauchy Principal Value
Contour C = ( ) + C (x0) + ( + ) + C .
CC C
f z d x f x d z f z d z f zd z
P
0
Res
2Res Res
jj
jj
z
iz x
zj = poles inside C (excluding x0 )
0
0
Res
Res 2Res Res
jj
jj
z
d x f x i x iz x
P
0
C
d z f z
0Res
rC
d z f z i x
0Res 2 Res jj
d x f x i x i z
P
Mathematica
Contribution of a pole is halved if it lies on the contour.
Pole Expansion of Meromorphic Functions
Expansion of f about
1.A regular point Taylor series.
2.A pole Laurent series.
3.All simple poles (Mittag-Leffler) pole expansion.
Consider where CN is a circle centered at w=0, of radius RN,
that encloses the 1st N poles of f closest to w=0.
Mittag-Leffler Theorem
Let f be meromorphic with simple poles at zj 0, (with residue bj) ; j = 1,..., M.If for large |z| , then
1
1 10
M
jj j j
f z f bz z z
Proof :
N
N
C
f wI d w
w w z
For RN ,
0NN
N
f RI
R
0
f z
z
1
02
Nj
Nj j j
bf f zI i
z z z z z
1
00
Mj
j j j
bf f z
z z z z z
QED
Example 11.7.3.Pole Expansion of tan z
1
1 10
M
jj j j
f z f bz z z
sin
tancos
i z i z
i z i z
z e ez
z i e e
poles at 2 12jz j 0,1, 2,j
Res tan ; 2 12jb z j
2 1 2 1
2 2
2 1 12lim limcos sinz j z j
z j
z z
1jb
Let
0
1 1 1 1tan
2 1 / 2 2 1 / 2 2 1 / 2 2 1 / 2j
zz j j z j j
0
1 1
2 1 / 2 2 1 / 2j z j z j
2 2 20
2tan
1 / 2j
zz
j z
tan 0 0
tan 1lim lim 0z z
z
z z
Example 11.7.4.Pole Expansion of cot z
1
1 10
M
jj j j
f z f bz z z
coscot
sin
zz
z
has poles at jz j 1, 2,j
Res ;jb f z j
2 4
3 5
11cos 2
1sin6
z j O z jz
z z j z j O z j
1jb
Let
Similarly
1
1 1 1 1
j
f zz j j z j j
0 0f
lim 0z
f z
z
has at z = 0 with residue 0 0
coslim lim cos 1
sinz z
z zz
z
1cotf z z
z
cos sin
sin
z z z
z z
2 2 21
2
j
z
z j
2 2 2
1
1 2cot
j
zz
z z j
2 2 2
0
2 1sec
1 / 2
j
j
jz
j z
2 2 2
1
1 2csc
j
j
zz
z z j
if only zero or pole enclosed by C is z0.
Set where g is regular at z0.
Counting Poles & Zeros
Let z0 be either a pole or zero of f (z).
0f z z z g z
0
f z g z
f z z z g z
1
0 0f z z z g z z z g z
2
C
f zid z
f z
2 f f
C
f zi N Pd z
f z
Nf = Sum of the orders of zeros of f enclosed by C.
Pf = Sum of the orders of poles of f enclosed by C.
zero 0order of the
pole 0
Rouche’s Theorem
If f (z) & g(z) are analytic in the region R bounded by C and | f (z) | > | g(z) | on C,
then f (z) & f (z) + g(z) have the same number of zeros in R.
2 f
C
f zi Nd z
f z
f (z) & g(z) are analytic Proof :
2 f f
C
f zi N Pd z
f z
lnC f argCi f
C x = change of x going around C once.
arg
2C
f
fN
arg
2C
f g
f gN
1g
f g ff
arg arg arg 1
gf g f
f
f g Re 1g g
f f
Re 1 0
g
f
arg 1
2
g
f
arg 1C
g
f
arg arg
2 2C Cf g f
f g fN N
integer
integer
To find NF for , let with F z f z g z
Let . Find NF for
Example 11.7.5. Counting Zeros
3 2 11F z z z 1 3z
1z
11f z 3 2g z z z so that on C : |z| = 1f g
i.e., 0F fN N
To find NF for , let with F z f z g z 3z
3f z z 2 11g z z so that on C : |z| = 3f g
1 3 3F FN z N z
3ln 13
C C
d zd z d z
d z z 3 2 i 3FN
Product Expansion of Entire Functions
f (z) is an entire function if it is analytic for all finite z.
Let f (z) be an entire function. Then
f zF z
f z
is meromorphic (analytic except for poles for all finite z).
Poles of F are zeros of f.
If the M zeros, zj , of f are all simple (of order 1), then zj are simple poles of F with
1
1 10
M
j j j
F z Fz z z
1
1 10
M
jj j j
f z f bz z z
1jb so that
0
ln ln 0
zf z
d z f z ff z
10
0 ln
zM
j
j j j
z z zd z F z zF
z z
1
ln ln 0 0 lnM
j
j j j
z z zf z f zF
z z
f zF z
f z
1
exp 0 exp ln 10
M
j j j
f z z zz F
f z z
/
1
00 exp 1
0j
Mz z
j j
f zf z f z e
f z
/
; 0
sin1 z j
j j
z ze
z j
2
1
1j
z
j
Ex.11.7.5
2
1
cos 11 / 2j
zz
j
LikewiseGamma function
Set
8. Evaluation of Definite Integrals
Trigonometric Integrals, Range (0,2)
2
0
sin , cosI d f
f = rational function & finite everywhere
iz e id z ie d
d zd i
z
1
sin2
z z
i
1
cos2
z z
1 1
,2 2
C
d z z z z zI i f
z i
C = unit circle at z = 0.
12 Res ; j
j
I f zz
zj = poles inside C
Example 11.8.1.Integral of cos in Denominator
12 Res ; j
j
I f zz
2
01 cos
dI
a
1a
1
1
1 / 2C
d zI i
z a z z
2
2 1
2 / 1C
i d za z a z
Poles at 2
1 2 44
2z
a a
211 1 a
a inside
outsideC
1 2 1
Res ;2 2 /
f zz a z a
2
1
1 a
2
2
1I
a
Example 11.8.2.Another Trigonometric Integral
12 Res ; j
j
I f zz
2
0
cos2
5 4cos
dI
2 2
1
/ 2
5 4 / 2C
z zd zi
z z z
4
4 3 2
1 1
4 5 / 2C
zi d z
z z z
4
2
1 1
4 1 / 2 2C
zi d z
z z z
Only poles at 0 & ½ are inside C.
4
0
1 1 1Res ; 0
4 1 / 2 2z
d zf
z d z z z
3 4 4
2 2
0
1 4 1 1
4 1 / 2 2 1 / 2 2 1 / 2 2z
z z z
z z z z z z
1 1 1
4 1 / 4 2 1 / 2 4
1 5
4 2
5
8
2
0
cos2
5 4cos
dI
1 5
Res ; 08
fz
4
2
1 1
4 1 / 2 2C
zi d z
z z z
4
21/2
1 1 1 1Res ; lim
2 4 2z
zf
z z z
1 /16 11
4 1 / 4 3 / 2
1 17
4 6
17
24
12 Res ; j
j
I f zz
5 17
28 24
I 6
Integrals, Range (, )
I d x f x
f is meromorphic in upper half z-plane, where lim 0z
z f z
RC C
Id z f z d z f z C = Cx + CR R
2 Res ; jj
i f z z zj in upper plane
Since (see next page for proof )
lim 0
R
R
C
d z f z
lim 0z
z f z
on upper plane
2 Res ; jj
I i f z z zj in upper planeObviously, one can also close the contour in the lower half plane.
lim 0
RC
RI d z f z
lim 0
zz f z
on arc from 1 2
R
1
2
CRProof :
2
1
lim i i
RI i d Re f R e
iz R e
2
1
lim i
RI d R f R e
2
1
limz
d z f z
0QED
Example 11.8.3. Integral of Meromorphic Function
2
01
d xI
x
2
1
2 1
d x
x
2
1
2 1C
d z
z
1 1Res
2 2z
z i
Poles are at . But only z+ is inside C.z i
2
I
Alternatively 1
0tanI x
2
2 ResiII z
1 1Res
2 2z
z i
i
i
2 Res zII i
Integral with Complex Exponentials
i a xI d x e f x
a > 0 & real.
f analytic in upper plane, where lim 0z
f z
R lim limi a z ia x a y
z ze f z e e f z
In upper plane, y > 0
0
2 Res ;i a zj
j
I i e f z z zj = poles of f in upper plane
lim a y
ze f z
lim 0
R
i a z
C
Rd z e f z
Jordan’s lemma
Example 11.8.4. Oscillatory Integral
2
0
cos
1
xI d x
x
2
1 cos
2 1
xd x
x
2
1Re
2 1
i xed x
x
2
1Re
2 1
i z
C
ed z
z
Poles are at z i
Only z+ is in upper plane with
1Re 2 Res
2I i z
Res2
i zez
z
1
2ei
2
Ie
Example 11.8.5. Singularity on Contour of Integration
0
sin xI d x
x
1 sin
2
xd x
x
P1
Im2
i xed x
x
P
i z i z
C
e ed z d z
z z
P 0
i zed z i
z
1Im
2I i
0i ze
d zz
2
I
i xe
d x ix
P
Another Integration TechniqueExample 11.8.6. Evaluation on a Circular Sector
3
01
d
xI
x
3
0
31 1rCC CC
d z d z
z z
Poles are at exp 23j
iz j
/3 5 /3, ,i i ie e e
for j = 0,1,2
On C , iz r e 3 3 3 iz r e id z e d r
0
3 3 31 1
i
i
C
d z ed r
z r e
Set & 3 1ie 2
3
2 /3i Ie 2 /33 3
0
1
1 1C
id ze d r
z r
Branch cut
2 /3
3 3 3
0
lim1 1
C
i
iR
d z R ei d
z R e
2 /301 2 Resi Ie i z
0 20
1Res
3z
z 2 /31
3ie
2 /3
2 /3
2
3 1
i
iI
i e
e
2 /33 1
C
id z
zIe
0
3 3 30
2 /3
lim1 1
r
i
ir
C
d z r ei d
z r e
On Cr , iz r e id z i r e d
0
On C , iz R e id z iR e d
0
/3 /3
2
3
i
i i
i e
e e
1
3 sin / 3
2
3 3I
3
01
d
xI
x
/30
iz e
Avoidance of Branch PointsExample 11.8.7. Integral Containing Logarithm
3
0
ln
1
x
xI
d x
3
0
3
ln ln
1 1rCC C C
z d z z d z
z z
Poles are at exp 23j
iz j
/3 5 /3, ,i i ie e e
for j = 0,1,2
On C , iz r e 3 3 3 iz r e id z e d r
0
3 3 3
lnln
1 1i
C
ir i ez d zd r
z r e
Set 3 1ie 2
3
22 /3 2
33i I
ie
2 /33 3
0
ln ln 2 / 3
1 1C
iz d z r ie d r
z r
Branch cut
3
0
2
1 3 3
d x
x
2 /3
3 3 3
0
lnlnlim
1 1C
i
iR
R e R iz d zi d
z R e
2
2 /3 2 /30
21 2 Res
33i i i
e e zI i
00 2
0
lnRes
3
zz
z 2 /3
9ii
e 2 2
2 /3 2 /32 /3
1 2 2
1 3 93i i
i
ie
eI e
22 /3
3
2
1 33i
C
d z ie
zI
0
3 3 30
2 /3
lnlnlim
1 1rC
i
ir
r e r iz d zi d
z r e
On Cr , iz r e id z i r e d
0
On C , iz R e id z iR e d
0
Branch cut
3
0
ln
1
x
xI
d x
0lim ln 0r
r r
lnlim 0R
R
R
/30
iz e
22
27I
Ex.11.8.6
Exploiting Branch CutsExample 11.8.8. Using a Branch Cut
2
01
px d xI
x
0 1p
2 21 1r CCC C C
p pz d z z d z
z z
Branch cut
i
i
22 20
lim1 1
rp p
i
C
p
r
z d z x d xe
z x
On C , z x i d z d x 2p p i pz x e
2i pe I
On C+ , z x i d z d x p pz x
2 20lim
1 1
p
C
p
r
r
z d z x d x
z x
I
2
01
px d xI
x
0 1p
2 21 1r CCC C C
p pz d z z d z
z z
Branch cut
i
i
0
2 2 20
2
lim1 1
r
p p i p
ir
C
z d z r e d
z r e
On Cr , iz re id z ire d p p i pz r e
22
11
p
C
i p
C
z d ze I
z
0
2
2 2 2
0
lim1 1
C
p p i p
iR
z d z R e d
z R e
On C , iz Re id z iRe d p p i pz R e
0 22
11
pi p
C
z d ze I
z
2
01
px d xI
x
0 1p
Branch cut
i
i
22
11
pi p
C
z d ze I
z
21 2 Resi pj
j
e I i z
z iPoles are at
Res2
pzz
z
/2
3 /2
1
2
i p
i p
e
i e
1
2pz
i
/2 3 /22
2 1
1 2i p i p
i p
iI e e
e i
/2 3 /2
21
i p i p
i p
e e
e
/2 /2i p i p
i p i p
e e
e e
sin / 2
sin
p
p
2cos / 2I
p
/2
3 /2
i pp
i p
ez
e
Example 11.8.9. Introducing a Branch Point
Ex.11.8.6-7 evaluated 1st ,
then use it to get with a contour that avoids the branch cut.
3
01
d x
x
3
0
ln
1
xd x
x
Here, we pretend to evaluate using a branch cut.
But ends up with .
3
0
ln
1
xd x
x
3
01
d x
x
3 3
ln ln
1 1rC CC CC
z zd z d z
z z
3 30
ln ln 2lim
1 1
r
C
r
z x id z d x
z x
On C , z x i d z d x ln ln 2z x i
3
0
21
d xI i
x
On C+ , z x i d z d x
3 30
ln lnlim
1 1r
rC
z xd z d x I
z x
3
0
ln
1
xI d x
x
ln lnz x
3 3
0
ln2
1 1CC
z d xd z i
z x
3 3
0
ln2
1 1CC
z d xd z i
z x
0
3 3 30
2
lnlnlim
1 1rC
iir
ir r izd z d e
z r e
On Cr , iz re id z ire d
0
2
3 3 3
0
lnlnlim
1 1i
iR
C
iR R izd z d e
z R e
On C , iz Re id z iRe d
0
3 3
0
ln2
1 1C
z d xd z i
z x
ln lnz r i
0lim ln 0r
r r
ln lnz R i
lnlim 0R
R
R
2 Res jj
i z
3
0
Res1 j
j
d xz
x
Poles are at exp 23j
iz j
/3 5 /3, ,i i ie e e for j = 0,1,2
2
lnRes
3j
jj
zz
z 0 2 /3
/ 3Res
3 i
iz
e
3
0
Res1 j
j
d xz
x
2 /3
9ii e
3i
2 10 /3
5 / 3Res
3 i
iz
e
10 /35
9ii e 1 2
Res3 i
iz
e
2 /3 4 /33
0
3 51 9
i id xi e e
x
4 /35
9ii e
1 3 1 33 5
9 2 2 2 2i i i
3
0
2
1 3 3
d x
x
Exploiting PeriodicityExample 11.8.10. Integrand Periodic on Imaginary Axis
0R R
i + R i R
0sinh
xI d x
x
1
2 sinh
xd x
x
sinh sinh cosh cosh sinhx i y x i y x i y sinh cos cosh sinx y i x y
sinh sinhn
x ni x n = integers
sinh i n = 0 but x = 0 is not a pole since
i
0lim 1
sinhx
x
x
since sinh x is odd sinh sinhiC
z x id z d x
z x i
Psinh
x id x
x
P 2I
10
sinhd x
x
P
On C / , & 0,y
limsinh sinhR
z R i y
z R i y
lim 2
RR
R
e 0
0sinh sinh
C C
z zd z d z
z z
2sinh
iC
zd z I
z 0
sinh sinhC C
z zd z d z
z z
sinh sinhix C CC C C
z zd z
z z
Resi i Pole on path
counted as half.
2sinh
xC
zd z I
z
Rescosh
ii
i
i2
4I
Consider
where f is meromorphic with poles at zj .
9. Evaluation of Sums
is meromorphic with simple poles atcos
cotsin
zz
z
0, 1,nz n
cos 1Res cot ;
cos
nz n
n
can be used to evaluate series.
cot
N
N
C
I d z f z z
inside
2 Res cot ;j N
N
N jn N z C
I i f n f z z z
lim 0NNI
Res cot ; j
n j
f n f z z z
Example 11.9.1.Evaluating a Sum
2 21
1
n
Sn a
a integer
2 2
1
n n a
S 2
12S
a
1
2 2
1
n
Sn a
Res cot ; jn j
f n f z z z
coth a
a
S
Poles of at withz i a2 2
1
z a
2 2
cot cotRes ;
2
z zz
z a z
cot
2
i a
i a
coth
2
a
a
2
1 coth 1
2
aS
a a
Other Sum Formulae
Res cot ; jn j
f n f z z z
cosRes cot ; 1
cos
nz n
n
Res csc ;n
jn j
f n f z z z
1Res csc ;
cosn
z nn
1Res tan ;
2 jn j
f n f z z z
sin 1 / 21Res tan ; 1
2 sin 1 / 2
nz n
n
1 1Res sec ;
2 sin 1 / 2n
z nn
1Res sec ;
2n
jn j
f n f z z z
has 2nd order poles at z = 0, 1.
Example 11.9.2. Another Sum
1
1
1n
Sn n
1
1f z
z z
has poles at z = 0, 1.
cotf z z
,0, 1
12
1nn
Sn n
2
2
1 1
1 1n nn n n n
1
1
1m m m
S
0, 1
cotRes ;
1j
jz
zz
z z
0
cot cotRes ; 0
1 1z
z d z z
z z d z z
1
1
1 cotcotRes ; 1
1z
z zz d
z z d z z
1
1S 1
1 1
1n n n
Mathematica
Schwarz reflection principle :
If f (z) is analytic over region R with R x-axis , and f (x) is real,
then
10. Miscellaneous Topics
* *f z f z
Proof :
For any 0,z x R 0
00 !
nn
n
f xf z z x
n
f (x) is real f (n) (x) is real .
0
00
* *!
nn
n
f xf z z x
n
*f z
Mapping
w z u i v can be considered as a mapping : , ,w x y u v
Example 1
wz
Dotted circle = unit circle
iz re 1 iw er
Mathematica
1w
z
2 iz r e
iz i r e Mathematica