(1) the residue (2) evaluating integrals using the residue (3) formula for the residue

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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 Presentation

TRANSCRIPT

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.

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