(1) negative powers (2) convergence within an annulus (3) laurent’s theorem (4) singular points

1

(1) Negative Powers

(2) Convergence within an Annulus

(3) Laurent’s Theorem

(4) Singular Points

Section 7

SECTION 7Power Series II - Laurent Series

2

Section 7

Notice that (a) we always have positive powers of (z z0) (b) the series converges inside a disk

3211

1zzz

z

We saw that we can expand a function in a Taylorseries about a centre.

For example,

2

21

21 842

1

1zz

z

Also, we can expand functions about different centres.

For example21oz

0oz

3

Section 7But there is another type of series we can have which(a) includes negative powers of (z z0) (b) converges within an annulus

Such series are called Laurent Series

842

111

23

32 2

22

zz

zzzz

z

singular points at z1, 2 centre

Example

converges for 1z 2

4

Section 7But there is another type of series we can have which(a) includes negative powers of (z z0) (b) converges within an annulus

Such series are called Laurent Series

842

111

23

32 2

22

zz

zzzz

z

converges for 1z 2

singular points at z1, 2 centre

Example

5

Section 7

If we take a function and plot its singular points, we’ll be able toseparate the complex plane into different regions of convergence.

Laurent Series always converge within an annulus

Example

zzf

1

1)(

centre

inside a disk z 1 - ordinaryTaylor series with positive powers

centre

in an annulus 1z - Laurent series

3211

1zzz

z

32

111

1

1

zzzz

6

Section 7



Example

zzf

1

1)(

centre


centre


3211

1zzz

z

32

111

1

1

zzzz

7

Section 7



Example

zzf

1

1)(

centre


centre


3211

1zzz

z

32

111

1

1

zzzz

8

Section 7Of course we could have different centres ...

zzf

1

1)(

centre

inside a disk z1 2- Taylor series

in an annulus 2z1 - Laurent series

centre

8

)1(

4

1

2

1

1

1 2zz

z

32 )1(

4

)1(

2

1

1

1

1

zzzz

9

Section 7We could even have the centre at the singular point ...

zzf

1

1)(

centre

In this case the series is only be validfor 0z1 - a disk with the

singular point / centre punched out of it

In fact the series in this case is simply the single term !z1

1

10

Section 7We could even have the centre at the singular point ...

zzf

1

1)(

centre

In this case the series is only be validfor 0z1 - a disk with the

singular point / centre punched out of it

In fact the series in this case is simply the single term !z1

1

11

Section 7

Example (1)

How many series does the function have about

the centre z14 ?2

sin2 zz

z

The function has two singularities (simple poles), at 1, 2.

z14 54 5/4z14 7/4 7/4z14

12

Section 7The annulus is always between two singularpoints

Example (1)


the centre z14 ?2

sin2 zz

z

The function has two singularities (simple poles), at 1, 2.

z14 54 5/4z14 7/4 7/4z14

13

Section 7Example (2)


the centre z0 ?

2)2( z

e z

The function has one singularity (second order pole).

z2 2z

14


How many series does the function

have about the centre z2 ?)4)(1)((

3

zziz

The function has three singularities (simple poles).

z21 1z22 2z25

5z2

15

Section 7


the centre z1 ?

3)(

2

jz

jz

Question:

16

Section 7

Suppose that the function f(z) is analytic in an annulus withcentre z0. Then the Laurent series is

How do we find these Laurent Series ?Laurent’s Theorem

40

43

0

32

0

2

0

1

303

202010

)()()(

)()()()(

zz

b

zz

b

zz

b

zz

b

zzazzazzaazf

where

C

nn

Cnn

dzzzzfj

b

dzzz

zf

ja

10

10

))((2

1

)(

)(

2

1

C

Pierre Alphonse Laurent (1843)compare with Section 6, slide 15

17

Section 7

As with the Taylor series, there are many ways to find the Laurentseries of a function. We don’t actually use the complicated formulaeon the previous slide. One method is to use the geometric series,as we did with Taylor series.

Finding Laurent Series

Example (1)

Expand the function 1(1z) in negative powers of z

432

321

1111

1111

1

)1(

1

1

1

zzzz

zzzzzz z

Since converges for z 1, the series

converges for 1z 1, or z 1

211

1zz

z

18


Expand the function 1(iz) in powers of z2

3

2

2

2

2

22

)2(

)2(

)2(

2

2

1

)2(

)2(

2

21

2

1

1)2(

1

)2(2

11

i

z

i

z

i

i

z

i

z

iizizi iz

Since converges for

z 1, the series converges for

(z 2)(i2)1, or (z 2)5

211

1zz

z

19

Section 7Example (2) cont.

But there is another possibility - expand the function1(iz) in negative powers of z2

3

2

2

2

2

22

)2(

)2(

)2(

2

2

1

)2(

)2(

2

21

2

1

1)2(

1

)2(2

11

z

i

z

i

z

z

i

z

i

zzzizi zi

Since converges for

z 1, the series converges for(i2) (z2)1, or (z2)(i2) 1,or z 25

211

1zz

z

20


Expand the function about the centre z1

converges for 0z 1

3

2

)1( z

e z

)1(!4

2

!3

2

)1(!2

2

)1(

2

)1(

1

!2

)1(2)1(21

)1()1()1(

432

232

2

3

2

3

)1(22

3

2

zzzz

e

zz

z

e

z

ee

z

e zz

21


Expand the function about the centre z1

converges for 0z 1

3

2

)1( z

e z

)1(!4

2

!3

2

)1(!2

2

)1(

2

)1(

1

!2

)1(2)1(21

)1()1()1(

432

232

2

3

2

3

)1(22

3

2

zzzz

e

zz

z

e

z

ee

z

e zz

Here, the centre is the actualsingular point !

22

Section 7Note - this will help make sense of Laurent Series

Each Laurent series consists of two parts:

303

202010 )()()()( zzazzazzaazf

4

0

43

0

32

0

2

0

1

)()()( zz

b

zz

b

zz

b

zz

b

positive powers (Taylor series)

negative powers (the “Principal Part”)

INSIDE

OUTSIDE

23


Expand the function about the centre z0)3)(1(

1

zz

How many ways can we do this ?

centre (a) z 1

(b) 1z 3

(c) 3z

3

1

2

1

1

1

2

1

)3)(1(

1

zzzz

24



1

zz


centre (a) z 1

(b) 1z 3

(c) 3z

3

1

2

1

1

1

2

1

)3)(1(

1

zzzz

25



1

zz


centre (a) z 1

(b) 1z 3

(c) 3z

3

1

2

1

1

1

2

1

)3)(1(

1

zzzz

26



1

zz


centre (a) z 1

(b) 1z 3

(c) 3z

3

1

2

1

1

1

2

1

)3)(1(

1

zzzz

27

Section 7(a) z 1

2

2

22

27

13

9

4

3

1

331

3

11

2

1

)3/(1

1

3

1

)(1

1

2

1

3

1

1

1

2

1

)3)(1(

1

zz

zzzz

zz

zzzz

inside disk - positive terms

28

Section 7(a) z 1

2

2

22

27

13

9

4

3

1

331

3

11

2

1

)3/(1

1

3

1

)(1

1

2

1

3

1

1

1

2

1

)3)(1(

1

zz

zzzz

zz

zzzz

inside disk - positive terms

29

Section 7(b) 1z 3

54186

1

2

1

2

1

2

1

331

3

1111

1

2

1

)3/(1

1

3

1

)/1(1

11

2

1

3

1

1

1

2

1

)3)(1(

1

2

23

2

2

2

zz

zzz

zz

zzz

zzz

zzzz

negative powers1z

positive powersz 3

30

Section 7On the previous slide, how did we know which term toexpand in negative powers and which, if any, to expandin positive powers ?

The term is “outside”

- negative terms

The term is “inside”

- positive terms

3

1

z

1

1

z

The final annulus iswhere they overlap

31

Section 7(c) 3z

432

2

2

2

1341

331

1111

1

2

1

)/3(1

11

)/1(1

11

2

1

3

1

1

1

2

1

)3)(1(

1

zzz

zzzzzz

zzzz

zzzz

negative powers3z

positive powersz

32

Section 7

Singular Points

A Singular Point z0 of a function f (z) is where f (z) is not analytic.There are two main different types of singular point.

Isolated Singularity Non-isolated Singularity

2)2(

1)(

zzzf

2,00 z

)/sin(/1)( zzf

0

,,,1

0

41

31

21

0

z

zisolated

not isolated

33

Section 7

Isolated SingularitiesThere are two types of isolated singularity

Pole of order m

Essential Singularity

mm

n zz

b

zz

b

zz

bazf

)()()(

02

0

2

0

1

2

0

2

0

1

)()(

zz

b

zz

bazf n

The Laurent series “stops” (at the mth negative power)

The Laurent series is infinite (in negative powers)

We can form the Laurent series with centre z0, valid or 0z z0 R

here, centre issingular point

34

Section 7


Pole of order m


mm

n zz

b

zz

b

zz

bazf

)()()(

02

0

2

0

1

2

0

2

0

1

)()(

zz

b

zz

bazf n





35

Section 7


Pole of order m


mm

n zz

b

zz

b

zz

bazf

)()()(

02

0

2

0

1

2

0

2

0

1

)()(

zz

b

zz

bazf n





36


Classify the singularity of the functionz

zf1

)(

The Laurent series with z00 as centre is simply

the one term , valid for 0z . This is asimple pole

z

1

Example (2)

Classify the singularity of the function1

1

)1(

1)(

3

zzzf


the two terms valid for 0z1 .

This is a pole of order 3

3)1(

1

1

1

zz

0z

0z1

37



zf1

)(



z

1

Example (2)


1

)1(

1)(

3

zzzf




3)1(

1

1

1

zz

0z

0z1

38



zf1

)(



z

1

Example (2)


1

)1(

1)(

3

zzzf




3)1(

1

1

1

zz

0z

0z1

39


Classify the singularity of the function 4

sinh)(

z

zzf

This is a pole of order 30z

!7!5!3

11sinh)(

3

34

zz

zzz

zzf

Example (4)

Classify the singularity of the function )/(1)( izezf

32 )(

1

!3

1

)(

1

!2

111)(

izizizzf

!7!5!3

sinh753 zzz

zz

This is an essential singularity

!3!2

132 zz

ze z

0zi

40



sinh)(

z

zzf


!7!5!3

11sinh)(

3

34

zz

zzz

zzf

Example (4)


32 )(

1

!3

1

)(

1

!2

111)(

izizizzf

!7!5!3

sinh753 zzz

zz


!3!2

132 zz

ze z

0zi

41



sinh)(

z

zzf


!7!5!3

11sinh)(

3

34

zz

zzz

zzf

Example (4)


32 )(

1

!3

1

)(

1

!2

111)(

izizizzf

!7!5!3

sinh753 zzz

zz


!3!2

132 zz

ze z

0zi

42

Section 7

Note: there are a couple of good reasons for classifying singularities into poles and essential singularities.

(1) When we have poles we have lots of formulae for evaluating integrals (see next section)

(2) Functions with poles as z0 is approached (from any direction) - those with essential singularities take on many different values depending on the direction of approach

43

Section 7

Topics not Covered

(1) Proof of Laurent’s Theorem (formulae for Laurent’s series)- slide 8

(3) Singularities “at infinity”

(4) Zeros

(2) Removable Singularities

z

zzf

sin)( removable singularity at z00

2

1)(

zzf has a 2nd order pole at 0, so

2)( zzf

has a 2nd order pole at

2)2()( zzf has 2nd order zeros at 2

(1) negative powers (2) convergence within an annulus (3) laurent’s theorem (4) singular points

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