Bài tập hàm biến phức ppt

Baứi taọp haứm bieỏn phửực Trang 1

BI TP CHNG I

1.1. Thc hin cỏc phộp tớnh
1.
(5 6 ) (2 4 )
i i
+ +
2.
(2 3 )(4 )
i i
+
3.
2 3
(1 ) (1 )
i i
+
4.
2 3
5 4
i
i
+
+


5.
(3 2 )(1 2 )
4 3
i i

i
+

6.
(1 )(1 2 )
(2 )(4 3 )
i i
i i
+
+
7.
3
2
4 5
(2 )
i i
i
+ +

8.
3
(1 )
(2 )(1 2 )
i
i i
+
+ +


9.

5
(1 2 )
i
+
10.
25
2
1 2
i
i

+










11.
9
8
(1 )
( 2)
i
i
+

12.
3 2
( 7 3) ( 7 3)
i i
+

13.
3 4
i
+
14.
5 12
i

15.
2
i
16.
1 2 2
i



1.2. Vit s phc di dng lng giỏc v dng m
1.
4
2.
3
i
3.

2
i

4.
2 2
i
+


5.
6 2
i

6.
2 3 2
i
+
7.
2 5
i
+
8.
12 5
i



1.3. Vit di dng m
1.
(2 2 )(3 3 3 )

A i i
= +
2.
(4 4 )( 1 )
B i i
= + +
3.
1
i
C
i

=
+
4.
2 6
1 3
i
D
i
+
=
+


1.4. Tỡm modul ca cỏc s phc
1.
2
3 4
i

i

2.
(1 3)( 3 )
i i
+
3.
2 4
(2 3 ) (3 )
i i
+

4.
3
2
(3 4 )
(1 3)
i
i
+
+
5.
1 2 2
1 1
i i
i i

+
+
6.

1 1 5
1 1 1 2
i i i
+ +
+ +


1.5. Gii cỏc phng trỡnh
1.
2 5 6 0
z z i
+ =
2.
2
4
z z
=
3.
2
2
1 3
i
z z
i

+ =
+

4.
| | 3

z z i
= +
5.
2
| | 1 6 2
z i z
+ + =


1.6. Chng minh vi mi s phc
1 2
, ,
z z z

1.
| | | |
z z
=
2.
| | | |
z z
= 3.
2
| | .
z z z
=
4.
1 2 1 2
| | | | | |
z z z z

+ +
5.
1 2 1 2
| | | | | |
z z z z



1.7. Chng minh rng
1. N
u
| | 1
z
=
thỡ
3
2 | 3 | 4
z

. 2. Nu
| | 2
z
=
thỡ
| 6 8 | 12
z i
+ +
.

1.8. Cho

1
z i
w
iz
+
=
+
. Chng minh rng nu
Im( ) 0
z

thỡ
| | 1
w

.
Baứi taọp haứm bieỏn phửực Trang 2

1.9. Chng minh rng nu
( )
n
u iv x iy
+ = +
thỡ
2 2 2 2
( )
n
u v x y
+ = +
, vi

n
l s nguyờn.

1.10. Chng minh rng
2
0 1 2
( ) 0
n
n
f z a a z a z a z
= + + + + =
, nu
( ) 0
f z
=
, vi
( 0, 1, , )
k
a k n
=

.

1.11. Bng cỏch xột tớch ca
1
1
2
i
+
v

1
1
3
i
+
, chng minh
1 1
arctan arctan
2 3 4

+ =
.

1.12. Biu din qua ly tha ca
cos , sin
x x

1.
cos2 , sin2
x x
2.
cos 3 , sin 3
x x
3.
cos 4 , sin 4
x x


1.13. Tớnh cỏc s phc
1.

3
(1 3)
i

2.
4
1

3.
5
( 3 )
i


4.
4
( 2 6)
i
+


5.
5
(2 2 )
i

6.
7
(1 3)
i

+
7.
1
2
(1 )
i
+
8.
6
( 3 )
i

+

9.
4
1
( 3 )
i

10.
1
3
( )
i

11.
3
i

12.
5
1
i


13.
10
1 1
2 2
i




+






14.
5
1
1
i
i

+

15.
4
1 3
1
i
i

+











16.

3 2
( 1 ) ( 3 )
i i

+ +

17.
4
3
(1 3)
(2 2 )
i
i

+
18.
10
1 3
1 3
i
i

+

1.14. Tớnh v vit di dng m
1.
3
2 2
i
+
2.
4
3
i
+
3.
5
4 3
i
+


1.15. Gii phng trỡnh trong


1.
8
16 0

x
=
2.
3
1 0
x
+ =
3.
4 2
1 0
x x
+ =


1.16. Cho biu thc
1 3
1
i
A
i
+
=

.
1. Vit biu thc trờn di dng
A x iy
= +
.
2. Vit dng m ca
1 3

i
+
v
1
i

. Suy ra dng lng giỏc ca
A
, t ú tớnh
7
cos
12

,
7
sin
12

.

1.17. Tớnh ln lt cn bc 2, 3, 4, 5, 6 ca s phc 1 v biu din cỏc giỏ tr ú trờn ng trũn lng giỏc.

1.18. Cỏc giỏ tr ca
1
n
l
2 2
cos sin , 0, 1, , 1
k
k k

w i k n
n n

= + =
.
1. Tớnh t
ng
0 1 1

n
w w w

+ + +
.
2. Chng minh rng
k
w
v
n k
w

(
1, , 1
k n
=
) l cp s phc liờn hp v nghch o ca nhau.
Baứi taọp haứm bieỏn phửực Trang 3

3. Tớnh
1 1

k n k
w w

+ ,
1, , 1
k n
=
.
4. Tớnh
1 1
1 1
k n k
w w

+

,
1, , 1
k n
=
.

1.19. Xỏc nh cỏc im trong mt phng
Oxy
biu din s phc
z

1.
Re( ) Im( )
z z

=
2.
| | 3
z
<
3.
| 1 | 1
z i
+

4.
Re( ) 2
z i
=
5.
Re( ) Im( ) 1
z z
+ <
6.
| 2 | 4
z i
=

7.
0 Re( ) 1
iz
< <
8.
Im( ) 3
z i

9.
| | | 1 |
z i z
=

10.
| |
z z
=
11.
2
Im( ) 2
z
=
12.
| | Re( )
z z
=

13.
arg
z z
=
14.
| 1 | | 1 | 4
z z
+ + =
15.
arg

3
z

=

16.
arg( 1)
4
z

=
17.
| | | | 6
z i z i
+ + <
18.
arg
6 4
z

< <


1.20. Xỏc nh ng cong
C
cho bi phng trỡnh
1.
2
2 , 0z t it t
= + < < +

2.
3 2 , 1 2
z t it t
= <

3.
, 0
i
z t t
t
= < <
4.
2 4
,z t it t
= + < <

5.
1
,
1
z t
it
= < <
+
6.
3
3(cos sin ),
2 2
z t i t t

= + < <

7.
cos 2 sin , 0
z t i t t

= + < <
8.
2
2 sin i sin ,
2 2 2
t
z t t


= + < <

9.
2
1 , 1 0
z t i t t
= +
10.
2( ),
it
z t i ie t

= + < <

BI TP CHNG II

2.1. Tớnh giỏ tr ca hm ti im
0
z

1.
( ) 3 Im( )
f z z z
=

a)
0
1
z
=
b)
0
2
z i
=
c)
0
1 2
z i
=

2.
2

( ) 2
f z z z i
=
.
a)
0
2
z i
=
b)
0
1
z i
= +
c)
0
3 2
z i
=

3.
2
( ) | | 2Re( )
f z z iz z
= +

a)
0
3 4
z i

=
b)
0
1 2
z i
= +


2.2. Tỡm ph
n thc v phn o ca cỏc hm
1.
( ) 5 3 4
f z z i
= +
2.
( ) 2 3
f z z z i
= +
3.
2
( ) (1 )
f z z i z
=

4.
( ) | | 2 Im( )
f z z iz
= 5. ( )
1
z

f z
z
=
+
6.
1
( )f z z
z
=

Baứi taọp haứm bieỏn phửực Trang 4

2.3. Bit
i
z re

=
, tỡm phn thc v phn o ca cỏc hm theo
,
r


1.
( )
f z z
=
2.
( ) | |
f z z
=

3.
4
( )
f z z
=

4.
2 2
( )
f z x y iy
= +
5.
1
( )f z z
z
= +
6.
2
( ) Re( )
f z z
=


2.4. Tỡm min giỏ tr ca cỏc hm
1.
( )
f z z
=
xỏc nh trong na mt phng trờn
Im( ) 0

z
>
.
2.
( ) Im( )
f z z
=
xỏc nh trong hỡnh trũn úng
| | 2
z

.
3.
( ) | |
f z z
=
xỏc nh trong hỡnh vuụng
0 Re( ) 1
z

,
0 Im( ) 1
z

.

2.5. Tỡm min xỏc nh v min giỏ tr ca cỏc hm
1.
( )
z

f z
z
=
2. ( )
z z
f z
z z
+
=



2.6. Tỡm nh
B
ca tp
A
qua phộp bin hỡnh
( )
w f z
=

1.
( )
f z z
=
;
a)
A
l ng thng
3

y
=
b)
A
l ng thng
y x
=

2.
( ) (1 )
f z i z
= +

a)
A
l ng thng
2
x
=
b)
A
l ng thng
2 1
y x
= +

3.
( ) 2
f z z
=

a)
A
l na mt phng trờn
Im( ) 1
z
>
b)
{ : 0 Re( ) 1}
A z z
= < <


4.
( )
f z iz
=

a)
A
l ng trũn
| 1 | 2
z
=
b)
{ : 1 Im( ) 0}
A z z
= < <

2.7. Tỡm nh qua phộp bin hỡnh
2
w z
=

1. ng thng
1
x
=
2. ng thng
2
y
=
3. tp
: 0 arg
4
A z z




=






2.8. Tỡm nh qua phộp bin hỡnh

1
w
z
=

1. ng thng
y x
=
2. ng thng
1
x
=
3. on thng t im
1
i

n im
2 2
i


4. ng trũn
| | 2
z
=
5. ng trũn
| 1 | 1
z
=
6. tp

{
}
: 1 Re( ) 2
A z z
= < <



2.9. Tỡm gii hn ca cỏc hm s
1.
2
2
lim ( )
z i
z z


2.
2
1
lim (| | )
z i
z iz
+

3.
1
lim
z i
z z

z z
+

+
4.
2
Im( )
lim
Re( )
z i
z
z z

+


2.10. Ch
ng minh
1.
0
lim
z z
c c

=
(
c


) 2.

0
0
lim
z z
z z

=
3.
0
1
lim
z
z

=
4.
1
lim 0
z
z

=



Baứi taọp haứm bieỏn phửực Trang 5

2.11. S dng kt qu bi 2.10, tỡm cỏc gii hn
1.
2

2
lim ( )
z i
z z
+

2.
5
lim( 2 1)
z i
z z

+
3.
4
1
lim
z i
z
z i


+

4.
2
2
1
1
lim

1
z i
z
z

+

5.
2
2
2
lim
(1 2 )
z
z iz
i z

+
+
6.
1
lim
2
z
iz
z i

+

2.12. Xột xem cỏc hm s sau cú gii hn khi
0
z

hay khụng?
1.
Re( )
( )
Im( )
z
f z
z
=
2.
( )
z
f z
z
=


2.13. Chng t rng cỏc hm s sau liờn tc ti
0
z

1.
2
( ) 3 2
f z z iz i

= +
;
0
2
z i
=
2.
( ) 3 Re( )
f z z z i
= +
;
0
3 2
z i
=

3.
3
1
, | | 1
( )
1
3, | | 1
z
z
f z
z
z

=




=



;
0
1
z
=


2.14. Xột tớnh liờn tc ca cỏc hm s
1.
( )
f z z
=
2.
2
( ) Im( )
f z z

=


2.15. Xột tớnh kh vi ca hm
( )
f z
v tớnh o hm (nu cú)
1.
( )
f z z
=
2.
2
( ) ( )
f z z
=
3.
( ) .Im( )
f z z z
=
4.
( ) .
f z z z
=


5.
2
( ) Re( )
f z z

=
6.
2
| 1|
( )
z
f z e

=
7.
5
( )
f z z z
= +
8.
( ) | | .
f z z z
=


2.16. Chng t cỏc hm s sau khụng gii tớch ti bt k im no trong mt phng phc
1.
( ) Re( )
f z z
=
2.
( )
f z y ix
= +
3.

2 2
( ) 2 ( )
f z x y i y x
= + +
4.
( ) 4 6 3
f z z z
= +


2.17. Cỏc hm s sau gii tớch trong min no ca mt phng phc
1.
2
( )
f z z
=
2.
( ) .Re( )
f z z z
=
3.
2
( ) Im( )
f z z
=
4.
2
( )
f z z zz
= +

2.18. Chng minh rng
1. Nu
f
gii tớch trong min
D
v
| ( ) |
f z
l hng s trong
D
thỡ
f
cng l hng s trong
D
.
2. Nu
f
gii tớch trong min
D
v
( ) 0
f z

=
thỡ
f
l hng s trong
D

.

2.19. Cho
i
z x iy re

= + =
v
( ) ( , ) ( , )
f z u x y iv x y
= +
.
Chng minh iu kin C - R tng ng vi
1
u v
r r


=

,
1
v u
r r


=

.

2.20. Tỡm hm gi
i tớch
( ) ,
f z u iv
= +
bit
1.
u x y
= +
2.
2
v x y
=
3.
2 3
u xy x
= +
4.
y
x
v e

=

5.
3 2
3
v x xy
=
6.

arctan
y
u
x
= 7.
2 2
ln( ) 2
u x y x y
= + +
8.
cos 2
x
v e y x y

= +

Baứi taọp haứm bieỏn phửực Trang 6

2.21. Tỡm cỏc hng s
, ,
a b c
v
d
cỏc hm sau gii tớch
1.
( ) ( )
f z x ay i bx cy
= + + +
2.
( ) 3 ( 3)

f z x y i ax by
= + +

3.
2 2 2 2
( ) ( )
f z x axy by i cx dxy y
= + + + + +


2.22. Tỡm phn thc v phn o ca cỏc hm s
1.
3 4
( )
i
f z e
+
=
2.
( ) cos(2 )
f z i
= +
3.
( ) sin 2
f z i
=
4.
( ) sh(1 4 )
f z i
=

2.23. Tớnh
sin
z
, nu
ln(3 2 2)
2
z i

= +
.

2.24. Chng minh
1.
sin( ) sin ch cos sh
x iy x y i x y
+ = +
2.
cos( ) cos ch sin sh
x iy x y i x y
+ = +


2.25. Gii phng trỡnh
1.
sin 2
z
=
2.

cos 3
z
=
3.
1
z
e
=
4.
1
z
e
=

.

BI TP CHNG III

3.1. Tớnh cỏc tớch phõn
1.
( 3 )
C
I z i dz
= +

,
C
cú phng trỡnh
2 , 4 1, 1 3
x t y t t

= =
.
2.
(2 )
C
I z z dz
=

,
C
cú phng trỡnh
2
, 2, 0 2
x t y t t
= = +
.
3.
2
| |
C
I z dz
=

,
C
cú phng trỡnh
2
1
, , 1 2
x t y t

t
= = <
.
4.
Re( )
C
I z dz
=


,
C
l ng trũn
| | 1
z
=
.
5.
C
I z dz
=


,
C
l ng trũn
| | 1
z
=
.

6.
2
1 2
3
( )
C
I dz
z i
z i




= +





+
+



,
C
l ng trũn
| | 1
z i
+ =

.

3.2. Tớnh cỏc tớch phõn
1.
2 3
( )
C
I x iy dz
= +

,
C
l on thng
a) ni t
1
z
=
n
z i
=
.
b) ni t
1
z i
= +
n
z i
=
.
2.

1
C
z
I dz
z
+
=

,
C
l
a) n
a phi ng trũn n v t
z i
=
n
z i
=
.
b) na trỏi ng trũn n v t
z i
=
n
z i
=
.

3.
2 2
( )

C
I x iy dz
=

,
C
cú im u
1
z
=
v im cui
1
z
=
, trong hai trng hp sau
Baứi taọp haứm bieỏn phửực Trang 7

a)
C
l na di ng trũn n v.
b)
C
l na trờn ng trũn n v.
4.
z
C
I e dz
=

, vi

C
l ng gp khỳc ni
0, 2, 1
i

+
.
5.
C
I zdz
=

theo ellip
2
2
1
4
x
y
+ =
t im
2
z
=
n im
z i
=
, chiu ngc kim ng h.
6.
1

C
I dz
z
=

theo ng trũn
| | 1
z
=
t im
1
z
=
n im
1
z
=
, trong hai trng hp
a) na mt phng trờn.
b) na mt phng di.
7.
Im( )
C
I z i dz
=


, vi biờn
C
gm ng trũn

| | 1
z
=
t
1
z
=
n
z i
=
v on thng t
z i
=

n
1
z
=
.
8.
z
C
I ze dz
=


, vi
C
l biờn ca hỡnh vuụng cú cỏc nh l
0

z
=
,
1, 1
z z i
= = +
v
z i
=
.

3.3. Tớnh tớch phõn
( )
C
I f z dz
=


, vi
C
l biờn ca tam giỏc cú cỏc nh l
0, 1
z z
= =
v
1
z i
= +
:
1.

( ) Re( )
f z z
=
2.
2
( )
f z z
=
3.
( ) 2 1
f z z
=
4.
2
( )
f z z
=


3.4. Tớnh tớch phõn
2
( 2)
C
I z z dz
= +

, vi
C
cú im u
z i

=
, im cui
1
z
=
, trong cỏc trng hp:
1.
C
l ng thng
1
x y
+ =
2.
C
l ng gp khỳc ni
, 1 , 1
i i
+

3.
C
l parabol
2
1
y x
=
4.
C
l ng trũn
| | 1

z
=
(chiu cựng kim ng h)

3.5. Tớnh tớch phõn
2
( )
C
I z dz
=

vi
C
l ng ni t im
0
z
=
n im
1
z i
= +
trong trng hp
1.
C
l on thng. 2.
C
l ng gp khỳc ni
0, 1, 1
i
+

.

3.6. Tớnh tớch phõn
| |
C
I z dz
=

, nu
C
l
1. on thng ni t im
z i
=
n im
z i
=
.
2. na trỏi ng trũn
| | 1
z
=
ni t im
z i
=
n im
z i
=
.
3. na phi ng trũn

| | 1
z
=
ni t im
z i
=
n im
z i
=
.

3.7. Tớnh tớch phõn
C
I zdz
=

t
1
z
=
n
z i
=
, theo mi ng sau
1. dc theo trc
Ox
n
O
, ri dc theo trc
Oy

n
i
.
2. dc theo ng thng
1
y x
=
.
3. d
c theo ng thng ng n
(1 )
i
+
ri dc theo ng ngang n
i
.

3.8. Tớnh tớch phõn
| |
C
I z dz
=


, vi
C
l ng trũn
| | 1
z i
=

.

Baứi taọp haứm bieỏn phửực Trang 8

3.9. Tớnh cỏc tớch phõn
1.
3
2
0
i
z dz
+

2.
/2
i
z
i
e dz


3.
2
sin
2
i
z
dz

+


4.
0
i
z
ze dz

5.
1
0
sin
i
z zdz
+

6.
1
2
1
( 1)
i
dz
z +



3.10. Tớnh cỏc tớch phõn
1.

cos
C
I zdz
=

, trong ú
2
{( , ) : , 0 1}
C x y x y y
= =

2.
2
1
C
I dz
z
=


,
C
l elip
2 2
1
( 1) ( 2) 1
4
x y
+ + =
.

3.
2
3
2 2
C
z
I dz
z z

=
+ +


,
: | | 1
C z
=

4.
2 1
(1 )
C
I z dz

= +


,
C
l elip

2 2
4 1
x y
+ =
.

3.11. Tớnh cỏc tớch phõn
1.
2
2
C
z
I dz
z i
=



a)
: | | 3
C z
=
b)
: | | 1
C z
=

2.
2
1

9
C
I dz
z
=
+


a)
: | 2 | 2
C z i
=
b)
1
: | 2 |
2
C z i
+ =

3.
2
2
3 4
C
z z i
I dz
z z
+
=
+

a)
: | | 2
C z
=
b)
3
: | 5 |
2
C z
+ =

4.
2
2
( 1 )
C
z
I dz
z z i
+
=



a)
: | | 1
C z
=

b)
: | 1 | 1
C z i
=

5.
3
1
( 4)
C
I dz
z z
=



a)
: | | 1
C z
=
b)
: | 2 | 1
C z
=

6.
3
( 1)
z
C

e
I dz
z z
=



a)
: | 2 | 1
C z
+ =
b)
1
: | |
2
C z
=

c)
1
: | 1 |
2
C z
=
d)
: | | 2
C z
=

3.12. Tớnh cỏc tớch phõn
1.
2 2
cos( )
C
iz
I dz
z

=
+


,
: | | 4
C z i
=
2.
2
sin( / 2)
1
C
z
I dz
z

=

,
: | | 3
C z i+ =
3.
2 2
, C : | | 2
( 1)
z
C
e
I dz z i
z z
= =
+


4.
2 2
, C : | | 2
( 1)( 3)
C
z
I dz z
z z
= =
+ +



5.

3 2
1
, C : | 2 | 4
( 1)
C
I dz z
z z
= =



6.
5
cos2
, C : | | 1
C
z
I dz z
z
= =





Baứi taọp haứm bieỏn phửực Trang 9

3.13. Tớnh cỏc tớch phõn
1.
2 2

2
sin
4
, : 2
1
C
z
I dz C x y x
z

= + =



.
2.
2
cos
1
C
z
I dz
z
=



,
C
l biờn ca tam giỏc cú cỏc nh

0
z
=
,
2 2
z i
=
v
2 2
z i
= +
.
3.
2
2
4
C
z
I dz
z
=
+


,
C
l biờn ca hỡnh vuụng cú cỏc nh l
2
z
=

,
2
z
=
,
2 4
z i
= +
v
2 4
z i
= +
.
4.
2 2
4
iz
C
e
I dz
z
=



,
C
cú phng trỡnh
2 , 0 2
it

z i e t

= +
.

3.14. Cho
0
t
>
v
C
l ng cong trn, kớn, bao quanh im
1
z
=
.
Chng minh rng:
2
3
1
2 2
( 1)
zt
t
C
ze t
dz t e
i
z

=





+



.

3.15. Cho
C
l na trờn ng trũn
| |
z R
=
t
z R
=
n
z R
=
.

Chng minh rng:
2 2 2 2
( 0, 0)
imz
C
e R
dz m R a
z a R a

> > >
+

.


BI TP CHNG IV

4.1. Xỏc nh xem cỏc dóy sau, dóy no hi t, dóy no phõn k
1.
2
1
in
in


+

2.
(1 )
1
n
n i
n


+



+


3.
n
n i
n


+





4.
1

2 arctan
n
e i n



+






4.2. Chng t cỏc dóy sau hi t bng cỏch s dng nh lý 4.1
1.
4 3
2
n i n
n i


+



+


2.
1

4
n
i




+
















4.3. Tỡm tng ca cỏc chui sau (nu chui hi t)
1.
1
( 1)
k
i

k k

=
+

2.
1
(1 )
k
k
i

=


3.
1
3
2
1 3
k
k
i

=

+



4.
1
1
1
4
2
k
k
i


=

5.
1
1
(1 )
k
k
k
i
i


=
+

6.
1
1 1
2 1 2
k
k i k i

=

+ + +




4.4. Tỡm bỏn kớnh v hỡnh trũn hi t ca cỏc chui
1.
2
1
( )
n
n
z i
n

=


2.
1
0
( 2 )
(1 )
n
n
n
z i

i

+
=



3.
1
( 1)
( 1 )
2
n
n
n
n
z i
n

=

+


4.
2
1
(2 )!
( 1)
( !)

n
n
n
z
n

=


5.
1
n
n
n
z
n

=

6.
1
1 2
( 1) ( 2 )
2
n
n n
n
i
z i

=

+



+








Baứi taọp haứm bieỏn phửực Trang 10

4.5. Khai trin Taylor cỏc hm s sau trong lõn cn im
a
v cho bit min khai trin c
1.
1
( )
1
f z
z
=

,
3

a i
=
2.
( )
z
f z e
=
,
a i

=

3.
1
( )f z
z
=
,
1
a
=
4.
( ) cos
f z z
=
,
/ 4
a

=

5.
1
( )
2
f z
z
=
+
,
1
a
=
,
a i
=
6.
1
( )
1
z
f z
z

=
+
,
0
a
=

,
1
a
=


4.6. Khai trin Taylor cỏc hm s sau trong lõn cn im
a
v tỡm bỏn kớnh hi t
1.
( )
( )( 2 )
i
f z
z i z i
=

,
0
a
=
2.
2
( )
2 3
z
f z
z z
=

,
0
a
=
,
2
a
=


4.7. Khai trin Laurent cỏc hm s trong min cho trc
1.
cos
( ) , | | 0
z
f z z
z
= >
2.
5
sin
( ) , | | 0
z z
f z z
z

= >

3.
2

1
( ) , | | 0
z
f z e z

= >
4.
2
2
( ) , | 1 | 0
( 1)
z
e
f z z
z
= >


5.
2
( ) sin , | | 0
f z z z
z
= >
6.
1
2
( ) , | | 0
z
f z z e z

= >

7.
2
1
( ) , 0 | 1 | 1
( 1)
f z z
z z
= < + <
+
8.
2 2
1
( ) , 0 | | 2
( 1)
f z z i
z
= < + <
+

4.8. Khai trin Laurent hm s
2
1
( )
( 1)
f z
z z
=

trong cỏc min
1.
0 | | 1
z
< <
2.
| | 1
z
>

4.9. Khai trin Laurent hm s
1
( )
( 3)
f z
z z
=

trong cỏc min
1.
0 | | 3
z
< <
2.
| | 3
z
>
3.
0 | 3 | 3
z

< <

4.
| 3 | 3
z
>
5.
1 | 4 | 4
z
< <
6.
1 | 1 | 4
z
< + <


4.10. Khai trin Laurent hm s ( )
( 1)( 2)
z
f z
z z
=
+
trong cỏc min
1.
1 | | 2
z
< <
2.
| 1 | 2

z
+ >
3.
0 | 1 | 3
z
< + <
4.
0 | 2 | 3
z
< <


4.11. Tỡm v phõn loi cỏc im bt thng cụ lp ca cỏc hm s
1.
3
2
( )
( 1) ( 1)
z
f z
z z z
+
=
+
2.
3
1 cos
( )
z
f z

z

= 3.
2 2
2
( )
( 1)
f z
z
=
+

4.
1
( )
z
f z ze
=
5.
5
2
( )
( 1)
z
f z
z
=

6.
1

( ) cosf z
z i
=
+

7.
2
3 1
( )
2 5
z
f z
z z

=
+ +
8.
3
1
( )
( 1)( 1)
z
f z
z z

=
+ +

9.
2 3

sin
( )
z
f z
z z
=

10.
6
cos cos2
( )
z z
f z
z

=
Baứi taọp haứm bieỏn phửực Trang 11

4.12. Tớnh thng d ca cỏc hm sau ti cỏc im bt thng cụ lp
1.
4 8
( )
2 1
z
f z
z
+
=

2.

2
( )
16
z
f z
z
=
+
3.
2
1
( )
1
f z
z
=

4.
2
1
( )
z
f z e

=

5.
3
cos
( )

z
f z
z
=
6.
4
3
( )
( 1)
z
f z
z
=
+
7.
2
1
( )
( )
f z
z z i
=

8.
2
( )
( 1)
z
e
f z

z
=


9.
3 4
1
( )f z
z z
=

10.
1
2
( )
z
f z z e
=
11.
3
sin
( )
z z
f z
z

= 12.
2
2 4 1
( )

( 1)( 2)( 3)
z z
f z
z z z
+
=
+ +

13.
2 2
( )
( 1)
z
f z
z
=
+
14.
2 3
cos
( )
( )
z
f z
z z

=

15.
1

( )
sin
f z
z z
= 16.
2
2
( ) ( 3) sin
3
f z z
z




= +





+



4.13. S dng thng d tớnh cỏc tớch phõn
1.
2
1
( 1)( 2)

C
I dz
z z
=
+


a)
1
: | |
2
C z
=
b)
3
: | |
2
C z
=
c)
: | | 3
C z
=

2.
2
1
( 2 )
C
z

I dz
z z i
+
=



a)
: | | 1
C z
=
b)
: | 2 | 1
C z i
=
c)
: | 2 | 3
C z i
=

3.
2
3 1/z
C
I z e dz

=


a)

: | | 5
C z
=
b)
: | | 2
C z i
+ =
c)
: | 3 | 1
C z
=

4.
1
sin
C
I dz
z z
=


a)
: | | 3
C z
=
b)
: | 2 | 1
C z i
=
c)

: | 2 | 4
C z i
=

5.
2 2
4
iz
C
e
I dz
z

=



,
: | | 2
C z i
=
.
6.
3 4
1
( 1)
C
I dz
z z
=

,
3
: | 2 |
2
C z
=
.
7.
2
( 1)( 1)
C
z
I dz
z z
=
+ +


,
2 2
: 16 4
C x y
+ =
.
8.
2 3
2 1

( 1)
C
z
I dz
z z

=
+


,
C
l biờn ca hỡnh ch nht xỏc nh bi
2, 1,
x x
= =

1
2
y

= v
1
y
=
.
9.
4/( 2)z
C
I e dz

=


,
: | 1 | 3
C z
=
.
10.
2 2
1
( 1) ( 3)
C
I dz
z z
=



,
2/3 2/3 2/3
: 2
C x y
+ =
.
11.
3 10
1
( 2)

C
I dz
z z
=



,
: | | 2
C z
=
.
12.
6
1
C
z
I dz
z
=
+


,
C
l na ng trũn xỏc nh bi
0
y
=
v

2
4
y x
=
.

4.14. a) S
dng khai trin c bn chng t
0
z
=
l khụng im cp 2 ca
1 cos
z

.
b) T kt qu cõu a) suy ra
0
z
=
l cc im cp 2 ca hm
( )
1 cos
z
e
f z
z
=

v t õy

f
cú khai trin
Laurent
2 1
0
2
( )
1 cos
z
c c
e
f z c
z z
z

= = + + +

, vi
0 | | 2
z

< <
.
Baứi taọp haứm bieỏn phửực Trang 12

S dng khai trin ca
z
e
v
1 cos

z

v ng nht cỏc h s t
2 1
0
2
(1 cos )
z
c c
e z c
z
z






= + + +







xỏc nh
2 1
,
c c

v
0
c
.
c) Tớnh tớch phõn
1 cos
z
C
e
I dz
z
=



,
: | | 2
C z i = .

4.15. S dng thng d tớnh cỏc tớch phõn
1.
2
0
1
5 4 sin
I dt
t

=

+

2.
0
5 3 cos
dt
I
t

=


3.
2
0
cos
2 sin
t
I dt
t

=
+


4.
2
0
sin
5 3 cos

t
I dt
t

=


5.
2
0
1
1 sin
I dt
t

=
+

6.
2
2
0
cos
3 sin
t
I dt
t

=

7.
2
0
1
cos 2 sin 3
I dt
t t

=
+ +

8.
2
0
cos2
5 4 cos
t
I dt
t

=




4.16. Chng minh rng
1.
2

2 3
0
1
, 1
( cos )
( 1)
a
I dt a
a t
a


= = >
+



2.
2
2
2 2
2
0
sin 2
( ), 0
cos
t
I dt a a b a b
a b t
b

= = > >
+



4.17. S dng thng d tớnh cỏc tớch phõn
1.
2
1
16
I dx
x
+

=
+

2.
2
1
2 2
I dx
x x
+

=
+

3.
2
1
6 25
I dx
x x
+

=
+ +


4.
2 2
1
( 1)( 4)
I dx
x x
+

=
+ +

5.
2 2
0
1
( 9)
I dx
x

+
=
+

6.
2
2 2
0
( 1)
x
I dx
x
+
=
+


7.
2 3
1
( 1)
I dx
x
+

=
+

8.
2

4
0
1
1
x
I dx
x
+
+
=
+

9.
6
0
1
1
I dx
x
+
=
+


10.
2
2 2
( 2 2)( 1)
x
I dx

x x x
+

=
+ + +

11.
2
cos
1
x
I dx
x
+

=
+

12.
2
sin 2
1
x x
I dx
x
+

=
+

13.
2
sin
2 2
x
I dx
x x
+

=
+ +

14.
2 2
0
cos
( 4)
x
I dx
x
+
=
+

15.
4
0
cos2
1

x
I dx
x
+
=
+


16.
4
0
sin
1
x x
I dx
x
+
=
+

17.
2 2
cos 4
( 1)( 9)
x
I dx
x x
+

=

+ +

18.
2 2
0
sin 3
( 1)( 4)
x x
I dx
x x
+
=
+ +








Baứi taọp haứm bieỏn phửực Trang 13

BI TP CHNG V

5.1. Cỏc hm sau cú phi l hm gc khụng? Nu khụng thỡ ti sao?
1.
1
0, 0
( )

( 3) , 0
t
f t
t t



<


=






2.
5
0, 0
( )
, 0
t
t
f t
e t


<

=







3.
2
0, 0
( )
, 0
t
t
f t
e t


<


=

4.
3 2
0, 0
( )
, 0
t
t
f t
t e t


<


=







5.2. Chng minh hm
1
( )f t
t
=
khụng cú bin i Laplace.

5.3. Tỡm bin i Laplace

1.
2 3
( ) 3
t t
f t e e t

= + +
2.
t
( ) sin 2 cos
2
f t t= 3.
( ) ( 2)
t
f t t t e
=

4.
3
( ) sin 2
t
f t e t

=
5.
( ) 3 cos 4
f t t t
=
6.
( ) sin2 2 cos 2t

f t t t
=

7.
2
( ) cos
f t t
=
8.
3
( ) 2 5
t t
f t t e te

= +

9.
( ) 2 sin 5 cos2 3
t
f t t t e t
= +
10.
3 2
( ) ( 3 5)
t
f t e t t

= +

5.4. Tỡm bin i Laplace
1.
2
sin
( )
t
f t
t
=
2.
2
1
( )
t
e
f t
t

=
3.
0
( ) s in2xdx
t
f t x=


4.
2
0
( ) cos xdx

t
f t =

5.
2 3
( ) ( 3).( 3)
t
f t u t t e

=
6.
2( 1)
( ) ( 1). cos 3( 1)
t
f t u t e t

=


5.5. Tỡm bin i Laplace
1.
, 0 4
( )
1, 4
t t
f t
t


<

=





2.
(2 ), 0 2
( )
0, 2
t t t
f t
t


<

=






3.
0, 0 1
2, 1
( )
1, 2

, 2
t
t
f t
t
t t





<



<


=


<







4.

cos ,
3 3
( )
0,
3
t t
f t
t


















=

<




5.
2
1, 0 1
( ) 3, 1 3
, 3
t
f t t
t t


<



= <







6.
1, 0 2

( ) 1, 2 3
0, 3
t
f t t
t


<



= <









5.6. Tỡm bin i Laplace ngc
1.
6
1
( )F s
s
=
2.
3

1
( )
( 5)
F s
s
=

3.
2
1
( )
4
F s
s
=
+

4.
2
1
( )
3
F s
s
=

5.
2 2
1
( )

( )
F s
s a b
=
+
6.
3
2
( )
5 6
s
e
F s
s s

=
+

Baứi taọp haứm bieỏn phửực Trang 14

7.
4
2
( )
( 4)
s
e
F s
s

=

8.
4
1
( )
1
F s
s
=

9.
2
4
( )
6 11
s
F s
s s
+
=
+ +

10.
3 2
6
( )F s
s s
=
+

11.
2
1
( )
2 5
F s
s s
=
+ +
12.
2 2
( )
( 1)
s
F s
s
=
+

13.
1
( )
( 1)( 3)
F s
s s
=
+ +
14.
2
( )

4 13
s
F s
s s
=
+ +
15.
2
1
( )
( 9)
F s
s s
=
+

16.
2
2 1
( )
( 4)
s
F s
s s

=
+
6.
2 2
1

( )
( 4)( 9)
F s
s s
=
+ +
18.
2
1
( )
( 2) ( 1)
s
F s
s s
+
=
+

19.
( /4)
2
( )
1
s
e
F s
s

=
+

20.
2
3
( )
( 1)
s
e
F s
s

=
+


5.7. Dựng thng d, tỡm bin i Laplace ngc
1.
2
( )
4 3
s
F s
s s
=
+
2.
2
( )
( 1)( 2)
s
F s

s s
=
+
3.
3
1
( )
( 1)
F s
s s
=


4.
3 1
( )
( 1)( 2)( 4)
s
F s
s s s

=

5.
2 2
1
( )
( 1)
F s
s

=
+
6.
3
( )
1
s
F s
s
=
+


5.8. Dựng bin i Laplace gii phng trỡnh vi phõn
1.
3
2 2 ; (0) 1
t
y y e y

+ = =

2.
; (0) 2
y y t y

= =

3.
2

2 ; (0) 3
t
y y e y

= =

4.
3 4 4 3; (0) 0, (0) 1
y y y t y y

+ = = =

5.
4 1 sin ; (0) 0, (0) 1
y y t y y

+ = = =

6.
3 2 3 ; (0) 1, (0) 1
t
y y y e y y

+ = = =

7.
4 5 ; (0) 1, (0) 0
t
y y y te y y

+ = = =

8.
2 5 8 ; ( ) 2, ( ) 12
t
y y y e y y




+ = = =

9.
2 3
6 9 ; (0) 2, (0) 17
t
y y y t e y y

+ = = =

10.
4 6 1 ; (0) 0, (0) 0
t
y y y e y y


+ + = + = =

11.
16 cos 4 ; (0) 0, (0) 1

y y t y y

+ = = =

12.
2
4 4 3 ; (0) 0, (0) 0
t
y y y e y y


+ + = = =

13.
4 sin 5 cos 2 ; (0) 1, (0) 2
y y t t y y

= + = =

14.
2
0; (0) , (0) , \ {0}
y y y A y B


+ = = =


15.
2

2 0; (0) 2, (0) 2
y ky k y y y k

+ = = =

16.
4 1; (0) (0) (0) 0
y y y y y

+ = = = =

17.
2
2 2 ; (0) 0, (0) 0, (0) 2
t
y y y x e y y y


+ = = = =

18.
(4)
0; (0) 1, (0) (0) (0) 0
y y y y y y

= = = = =

Baứi taọp haứm bieỏn phửực Trang 15

5.9. Dựng bin i Laplace gii phng trỡnh vi phõn

1.
1, 0 1
4 ; (0) (0) 0
0, 1
t
x x x x
t


<


= = =







2.
cos , 0
; (0) (0) 0
0,
t t
x x x x
t

<


+ = = =







3.
, 0 1
2 ; (0) (0) 0
0, 1
t t
x x x x
t


<


+ = = =

4.
2 ( ); (0) 0
x x f t y

+ = =
, vi
0, 0 1
( ) , 1 2
0, 2
t
t
f t e t
t



<



= <

5.
8 sin , 0
9 ; (0) 0, (0) 4
0,
t t
y y y y
t




<


+ = = =







6.
3 sin cos , 0 2
4 ; (0) 1, (0) 0
3 sin 2 cos2 , 2
t t t
y y y y
t t t

<


+ = = =







5.10. Dựng bin i Laplace gii h phng trỡnh vi phõn
1.
' 2 3
; (0) 8, (0) 3
' 2
x x y
x y
y y x


=

= =


=

2.
' 4 4 0
; (0) 3, (0) 15
' 2 6 0
x x y
x y
y x y


+ + =

= =


+ + =




3.
2
2
' 3 4 9
; (0) 2, (0) 0
' 2 3 3
t

t
x x y e
x y
y x y e


+ =


= =


+ =





4.
' 2 4 cos
; (0) (0) 0
' 2 sin
x x y t
x y
y x y t


=

= =

+ + =




5.
'
; (0) 1, (0) 0
'
x x y
x y
y x y


=

= =


= +




6.
4
' 2 4
; (0) 1, (0) 0

' 2
t
x x y e
x y
y x y


= + +


= =


= +





Ht.