part 3-1

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

Hm bin phc

1. nh ngha

2. Gii hn, lin tc v kh vi

3. Phng trnh Cauchy-Riemann

4. Hm lin hp v hm iu ha

5. Mt s hm bin phc c bn


1. nh ngha

V d 1.01:


Hm bin phc l hm c bin l s phc.

w = f(z) = u(x,y) + j.v(x,y) with z = x + jy

2 2 2

2 2 2 2

( ) .2

1.

cos .sinaz ax

w z x y j xy

yxw j

z x y x y

w e e ay j ay


Cn lu rng, trongmt phng phc, z c thtin n z0 theo nhiu quo khc nhau, do giihn L l khng duy nhttrong nhiu trng hp.


Nu f(z) l mt hm bin phc, f(z) c gii hn l Lkhi z tin ti z0 nu vi mi > 0, lun tn ti mt > 0sao cho:

|f(z) L| < vi mi z trong min S:0 < |z z0| <

iu kin y l f(z0) phi c xc nh v f(z)tin n f(z0) khi z tin n z0 theo qu o bt k.

Vi z l s phc v tin v 0 theo qu o bt k.


Nu

ta ni rng f(z) lin tc ti z0.0

0lim ( ) ( )z z

f z f z

f(z) kh vi ti z0 nu tn ti gi tr L:

0 000

( ) ( )lim '( )

z

f z z f zL f z

z


Khng phi hm phc no cng kh vi!

V d 1.02:

Xt hm phc sau

Gi s f(z) kh vi ti z0, khi :

Vy hm phc ny khng kh vi ti bt k gi tr no.


( )f z z

0 0 0 00 0 0 0

( ) ( )'( ) lim lim lim

1 0

1 0

z z z

f z z f z z z z zf z

z z z

if z along the real axis

if z along the imaginary axis



Nu f(z) = u(x,y) + j.v(x,y) tha mn phng trnhCauchy-Riemann ti z0 v ti mi im thuc lncn ca z0, ta ni rng f(z) gii tch ti z0.

Nu f(z) gii tch ti mi gi tr ca z, f(z) l mthm gii tch.


Cho w = f(z) = u(x,y) + j.v(x,y) kh vi ti z = x + jy, khi ti gi tr (x,y):

v

(*) c gi l phng trnh Cauchy-Riemann

(*)

'( )

u v v uand

x y x y

u v v uf z j j

x x y y

V d 1.03:

Dng phng trnh Cauchy-Riemann kim tra tnhkh vi ca cc hm s sau:

p n:

a. Khng kh vi ti bt k im no

b. Ch kh vi ti z = 0, f(0) = 0.

c. Kh vi ti mi im (hm gii tch)f(z) = 2z.

d. Hm gii tch; f(z) = aeaz.


2

. ( ) . ( ) .Re{z}

. ( ) . ( ) aza f z z b f z z

c f z z d f z e


Hai hm thc u(x,y) v v(x,y) tha mn phngtrnh Cauchy-Riemann vi mi gi tr x, y R cgi l mt cp hm lin hp.

Mt hm tha mn phng trnh Laplace cgi l hm iu ha.

u(x,y) l mt hm iu ha nu:

Nu f(z) = u(x,y) + j.v(x,y) l hm gii tch, khi c u(x,y) v v(x,y) u l hm iu hoi, khi chngc gi l hai hm iu ha lin hp.


4. Hm gii tch v hm iu ha

2 2

2 20

u uLaplace equation

x y

V d 1.04: Cho u(x,y) = x2 y2 + 2x, tm hm lin hpv(x,y) sao cho f(z) = u + j.v l hm gii tch theo bin z.

p n:

vi F(x) l hm theo mt bin x.

trong C l hng s. V bi khng cho thm iukin no xc nh C, nn ta c th chn C bng 0.

F(z) = x2 y2 + 2x + j(2xy + 2y) = z2 + 2z


2 2 2 2 ( )u v

x v xy y F xx y

2 2 0v u dF dF

y y F Cx y dx dx


V d 1.05: Given u(x,y) = ex(xcosy ysiny)

a. Chng minh u(x,y) l hm iu ha

b. Tm hm lin hp v(x,y) sao cho w = u + j.v l hmgii tch.

c. Ch ra rng v(x,y) cng l mt hm iu ha.

p n:

a. Kim tra phng trnh Laplace

b. Dng phng trnh Cauchy-Riemann tm v(x,y):


2 2

2 20

u u

x y

( , ) ( sin cos )

( ) ( cos sin ) . ( sin cos )

x

x x z

v x y e x y y y

w f z e x y y y j e x y y y ze


V d 1.06: Tm hm gii tch f(z) = u + j.v, vi 2u + v =ex(cosy siny) v f(1) = 1.

p n:

f(z) l hm gii tch ux = vy and uy = -vx:


1

2

2 (cos sin )

2 ' ' (cos sin )

2 ' ' ( sin cos ) ' 2 '

(cos 3sin ) ( )5

(3cos sin ) ( )5

x

x

x x

x

y y x x

x

x

u v e y y

u v e y y

u v e y y u v

eu y y F y

ev y y F y


V d 1.06 (tt):

Kim tra ux = vy v uy = -vx F1(y) = c1, F2(y) = c2 (c1,c2: hng s)

Kim tra vi f(1) = 1


1 2( ) (cos 3sin ) . (3cos sin )

5 5

x xe ef z y y c j y y c

1 2

31

5 5

e ec and c

3( ) (cos 3sin ) 1 . (3cos sin )

5 5 5 5

1 3( ) ( ) 1

5

x x

z

e e e ef z y y j y y

jf z e e


5.1. Hmm phc

Tnh cht ca ez:

e0 = 1

ez + w = ezew

ez 0, z

e-z = 1/ez

ez = 1 z = 2nj , n Z

Nu ez + p = ez , z p = 2nj

ez l hm tun hon vi chu k 2nj.



ez = excosy + jexsiny

V d 1.07: gii phng trnh ez = 1 + 2j

p n:

V d 1.08: Chng minh cc hm sau y l cc hmgii tch:


2

1

cos sin 1 2

cos 1 5

sin 2 tan 2

1ln 5 tan 2

2

z x x

x x

x

e e y je y j

e y e

e y y

z j

21

. ( ) . ( ) . ( )z zza f z e b f z e c f z ze


5.2. Hm cosin v sin phc

Vit di dng u + jv:

cosz = cosx.coshy jsinx.sinhy

sinz = sinx.coshy + jcosx.sinhy

trong

Tnh cht ca sinz v cosz:

sinz = 0 z = n; n Z

cosz = 0 z=(2n + 1)/2; n Z

sinz v cosz l hm tun hon vi chu k 2n; n Z*


cos ; sin2 2

jz jz jz jze e e ez z

j

cosh ; sinh2 2

y y y ye e e ey z


V d 1.09: Tnh:

a. sin(1 4j)

b. cos(3 + 2j)

V d 1.10: Chng minh rng:

a. sin(z + w) = sinz.cosw + cosz.sinw

b. cos(z + w) = cosz.cosw sinz.sinw

V d 1.11: Chng min f(z)=cos2z l hm gii tch.



5.3. Logarithm phc

Calculate the complex logarithm: let z = rei and w = u + jv

is any argument of z and n can be any integer, soeach nonzero z has infinitely many different complexlogarithms.


ln ww z z e

ln

2

ln ln ( 2 )

uj jvw u

j jv

r e u rz e re e e

v ne e

w z r j n


Example 1.12: Calculate lnz with

a. z = 1 + j

b. z = -3

Solution:

In complex plane, we can take the logarithm of anegative number!


4. 1 2 ln(1 ) ln 2 24

. 3 3 ln( 3) ln 3 (2 1)

j

j

a z j e j j n

b z e j n


5.4. nth Roots of a complex number

Calculate the nth roots: let z = rej = rej( + 2k)

nth roots of z has n distinct value!


1

; : integernnw z w z n

( 2 )1 1

1

0,1,..., 1

2 2cos sin

j k

n n n

n

z r e with k n

k kr j

n n


Example 1.13:

a. Calculate 4th roots of 1 + j

b. Calculate 5th roots of 1 (5th roots of unity)

Solution:


1 244 2

1 2 /444 8

1 1

8 8

1 1

8 8

. 1 2 2

1 2 0,1,2,3

9 92 cos sin ; 2 cos sin ;

16 16 16 16

17 17 25 252 cos sin ; 2 cos sin

16 16 16 16

j kj

j k

a j e e

w j w e with k

w j j

j j


Example 1.13 (cont):


0 2

25 5

. 1

1 0,1,2,3,4

2 21; cos sin ;

5 54 4

cos sin ;5 5

6 6cos sin ;

5 58 8

cos sin ;5 5

j j k

j k

b e e

w w e with k

w j

j

j

j


5.5. Complex Power zw

Example 1.14: Compute all values of (1 j)1 + j

Solution


ln ,w w zz e with z w are complex numbers

1 (1 )ln(1 )

24

(1 ) ln 2 2 ln 2 224(1 )ln(1 ) 4ln 2 4

21 4

(1 )

1 2 ln(1 ) ln 2 24

(1 ) 2 cos ln 2 sin ln 24 4

j j j

j n

j j n j nnj j

nj

j e

j e j j n

e e e e e

j e j


part 3-1

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