(1) negative powers (2) convergence within an annulus (3) laurent’s theorem (4) singular points
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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
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
7
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
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)
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
13
Section 7Example (2)
How many series does the function have about
the centre z0 ?
2)2( z
e z
The function has one singularity (second order pole).
z2 2z
14
Section 7Example (3)
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
How many series does the function have about
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
Section 7Example (2)
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
Section 7Example (3)
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
Section 7Example (3)
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
Section 7Example (4)
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
Section 7Example (4)
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
25
Section 7Example (4)
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
26
Section 7Example (4)
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
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
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
35
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
36
Section 7Example (1)
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 Laurent series with z01 as centre is simply
the two terms valid for 0z1 .
This is a pole of order 3
3)1(
1
1
1
zz
0z
0z1
37
Section 7Example (1)
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 Laurent series with z01 as centre is simply
the two terms valid for 0z1 .
This is a pole of order 3
3)1(
1
1
1
zz
0z
0z1
38
Section 7Example (1)
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 Laurent series with z01 as centre is simply
the two terms valid for 0z1 .
This is a pole of order 3
3)1(
1
1
1
zz
0z
0z1
39
Section 7Example (3)
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
Section 7Example (3)
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
41
Section 7Example (3)
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
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
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