part 3-2

I. Hm bin phc

II. Chui phc

III. Tch phn ng

IV. im bt thng, zeros v thng d

V. ng dng ca l thuyt thng ds

Part 3:

Hm bin phc

Created and edited by: Nguyen Phuoc Bao Duy

Chui phc

1. nh ngha

2. Chui ly tha

3. Chui Taylor

4. Chui Laurent


Tiu chun DAlembert

V d 2.02: Tm min hi t (region of converge - ROC):

p n: ROC: Re{z} > 0.


Xt chui phc:

Tm gii hn:

L < 1: Chui hi t L > 1: Chui phn k L = 1: Cha th kt lun v tnh hi t ca chui.

1 20

1

( ) ( ) ( ) ... ( ) ...

( )lim

( )

n nn

n

nn

f z f z f z f z

f zL

f z

2

0

1 ...nz z z

n

e e e

2. Chui ly tha

ROC ca chui ly tha:


Mt chui phc c dng

vi cc h s ai l hng s v z0 l mt im cnh trong mt phng phc, c gi l mt chui lytha quanh im z0.

2

0 0 1 0 2 00

( ) ( ) ( ) ...nn

n

a z z a a z z a z z

0

1

lim nn

n

az z R with R

a

V d 2.03: Tm chui ly tha biu din hm 1/(z 3)trong 3 min sau:

a. |z| < 3

b. |z 2| < 1

c. |z| > 3

p n:


2

0

2

0

2 30

1 1 1. ... 3

3 3 3 3 9 27

1. 2 1 ( 2) ( 2) ... 2 1

3

1 1 3 1 3 9. ... 3

3

n

n

n

n

n

n

z z za zz

b z z z zz

c zz z z z z z

2. Chui ly tha

V d 2.03 (tt):

Nhn xt: Chng ta c th chn bt k mt im z0trong mt phng phc v xc nh 1 chui ly tha hit v 1/(z 3) vi min hi t l mt hnh trn tm z0,bn knh R.


2. Chui ly tha

Khai trin chui nh thc:

vi

Hai trng hp c bit thng s dng:


2( 1) ( 1)( 2)...( 1)(1 ) 1 ... ...2! !

k rk k k k k k rz kz z zr

1k and z

2

0

2

0

11 ... 1

1

11 ... ( ) 1

1

n

n

n

n

z z z zz

z z z zz

2. Chui ly tha

3. Chui Taylor

ROC ca chui Taylor:


Mt chui ly tha c dng

trong f(z) l mt hm phc gii tch bn trongv trn bin mt ng cong kn C, c gi l khaitrin chui Taylor ca f(z) quanh im z0.

( )

0 00

1( ) ( )( )

!n n

n

f z f z z zn

0z z R

V d 2.04: Xc nh khai trin chui Taylor ca hm

quanh im z = j, n s hng (z - j)4.

p n:


1( )

( 2 )f z

z z j

(1)

(2) (3) 2 4

(4)

1 1 1 1( )

( 2 ) 2 2

( ) 1; ( ) 0

( ) 2; ( ) 0 ( ) 1 ( ) ( ) ...

( ) 24

f zz z j j z j z

f j f j

f j f j f z z j z j

f j

3. Chui Taylor

V d 2.05: Xc nh khai trin chui Maclaurin cacc hm s sau:

a. ez b. sinz c. cosz

p n:


Mt chui Taylor vi z0 = 0 c gi l chuiMaclaurin.

2 3

0

2 1 3 5

0

2 2 4

0

. 1 ...! 2 6

.sin ( 1)(2 1)! 6 120

.cos ( 1) 1 ...(2 )! 2 24

nz

n

nn

n

nn

n

z z za e z

n

z z zb z z

n

z z zc z

n

3. Chui Taylor

Khai trin chui Taylor c th c xc nh thngqua php khai trin chui nh thc

Example 2.06: Tm khai trin Maclaurin ca cc hm sauv xc nh min hi t (ROC) tng ng:

p n:


1 2 2

3 4 2

1. ( ) . ( )

3 41 1

. ( ) . ( )1 3 2

za f z b f z

z zz

c f z d f zz z z

2

10

1 1 1 1 1. ( ) ...

3 3 3 3 3 9 271

3

ROC : 3

n

n

z z za f z

zz

z

3. Chui Taylor

p n v d 2.06 (tt):


2 1

2 2 10

30

4 2 10

. ( ) 1 ; : 24 4

1. ( ) 1 2 ; : 1

1

1 1. ( ) 1 ; : 1

3 2 2

nn

nn

n

n

n

nn

z zb f z ROC z

z

zc f z z ROC z

z

d f z z ROC zz z

3. Chui Taylor

V d 2.07: Tm khai trin Taylor ca cc hm sau quanhim z0 cho trc v xc nh min hi t (ROC) tngng:


02

0

0

1. ( ) ; 2

3 21

. ( ) ;( 2 )

. ( ) ; 4( 1)(2 )

a f z zz z

b f z z jz z j

zc f z z

z z

3. Chui Taylor

4. Chui Laurent

ROC ca chui Laurent:

r1 c th bng 0

r2 c th bng

Xc nh chui Laurent bng

cch khai trin chui nh thc.


Nu f(z) gii tch trong hnh vnh khn gii hn bi2 ng trn C1 v C2 c bn knh ln lt l r1 v r2 (r1 3

c. 0 < |z + 1| < 2

d. |z| < 1


1( )

( 1)( 3)f z

z z

4. Chui Laurent

part 3-2

Documents