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