我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!
5
13s
1/

 {}
n
α
^?
l ?$ ?T
l ? p
ì¥K
1
1i
1i
n
n
n
α
+
=
 2
i
1;
2
n
n
α

=+


3
i
(1) ;
1
n
n
n
α =? +
+
4  5
i/2n
n
e
π
α
=
i/2
1
n
n
e
n
π
α
=
3 1
2
22
1i1 2
i
1i1 1
n
nnn
nnn
α
+?
==+
++
?
2
12
lim 1,lim 0
11
nn
nn
→∞ →∞
2
=?=
++
n
# α
l ?
lim 1
n
n
α
→∞
=?
2
i2
1
2
5
nn
i
n
e
θ
α

=+ =


?
2
lim 0
5
n
i
n
e
θ?
→∞

=


#
n
α
l ? lim 0
n
n
α
→∞
=
3??
n
α ¥
L?
{ }
(1)
n
? ?#
n
α ? ?
4??
i/2
cos isin
22
n
n
nn
e
π
π π
α
==?  
L?a′?
 (? ?#
n
α ? ?
5
i/2
11 1
cos i sin
22
n
n
nn
e
nn
π
π π
α
==? ?
11
lim cos 0,lim sin 0
22
nn
nnπ π
→∞ →∞
= = 
#
n
α
l ? lim 0
n
n
α
→∞
=
2£
ü
0,| |<1,
,||>1,
lim
1,1,
||=1,1.
n
n
α
α
α
α
αα
→∞

=
=

?i
3
/
 )
¥ '
l ??D
l ??
1
1
i
n
n
n

=

 2
2
i
ln
n
n
n

=

 3
1
(6+5i)
8
n
n
n

=

 4
2
cosi
2
n
n
n

=

b
3 1? icos isin
22
n
nnπ π
=+
1
cos
2
n
n
n
π

=

D
1
sin
2
n
n
n
π

=

1
l ?¥?p[
L)

[
1
i
n
n
n

=

l ??
i1
n
nn
= #
1
i
n
n
n

=

? ?e)
Hq
l ?
1
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!
2D 1? ¨]"¥ZEi ?¨
11
(2
ln
n
nn
≥≥)
3y
(6+5i) 61
88
n
n
n

=

7
1
61
8
n
n

=




l ?#
1
(6+5i)
8
n
n
n

=

'
l ?
4y 7cosi chn= n
ch
lim 0
2
n
n
n
→∞
≠ #
2
cosi
2
n
n
n

=

? ?b
4/

aE
^?? ?$1
I
1$

1
B?
a)

¥
l ???
))
l ?

2
B?
a)
¥f

l ??
= V
μ ?

3
B?  ??¥f
B? V[ ¥
#×
=Z 7? Taylor)
b
0
z
0
z
3 
1? b ? 
l ??


=0n
n
z 1<z
=
l ??
l ??? 1=z
i?
l ?

2?b
a)
¥f

l ??
=13f
?
μ ?
(3)?b ? ( ) zzf =  ?
ü
 ??,?
 ??¥
#×
= (?
Z 7? Taylor )
b
5.
a)
?
l ?7(
0
2
n
n
n
cz

=

) 0=z 3=z ? ?$
3 ?
by ? ()
0
2
n
n
n
cz

=

0=z
l ?5 ? Abel ? ? 
l ? ??
220 =?≥R 7 2123 <=? '  
l ??3=z 2|2| <?z
=#)

l ?
±b
()
0
2
n
n
n
cz

=

3=z
6 p/

a )
¥
l ???

1
1
()
n
p
n
z
p
n

=

1??
 
2
1
!
n
n
n
n
z
n

=




 
3 
0
1)
nn
n
iz

=




4
1
i
n
n
n
ez
π

=

 
5
1
i
ch ( 1)
n
n
z
n

=




 
6
1
ln i
n
n
z
n

=




b
3 
1 1/lim lim 1
n p
n
n
nn
Ra
→∞ →∞
= ==
2
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!

2
1
1
1
(1 )
1/ lim lim lim 0
1
n
nn
nnn
nn
aa
n
R
aa
+
→∞ →∞ →∞
+
+
===
+
=;

3 1/lim lim1/ |1 i| 1/ 2
n
n
n
n
Ra
→∞
→∞
==+=

4 1/lim 1
n
n
n
Ra
→∞
==

5
1
1/lim 1/lim ch 1/lim cos 1
n n
n
n
nn
i
Ra
nn
→∞ →∞ →∞

== =


=

6 1/lim lim |lni |
n
n
nn
Ra
→∞ →∞
===∞;
7 ?T ¥
l ???1 R£
ü)
¥
l ???
0
n
n
n
cz

=

()
0
Re
n
n
n
cz

=

R≥ b
£
ü ? ? Rz <||
=¥ ?iB? z?X? '
l ?'
0
n
n
n
cz

=

0
n
n
n
cz

=

l ??
y Re
n
cc≤
n
V7 Re | || |
n
n
nn
cz c z≤ #??[)
¥1?
Y E
0
Re
n
n
n
cz

=

9
l ?' ()
0
Re
n
n
n
cz

=

Rz <||
= '
l ??
^ 
l ??? R≥ b
8£
ü ?T
1
lim
n
n
n
c
c
+
→∞
i
/
 ??
a)
μM]¥
l ??? ≠∞
n
n
cz


1
1
nn
c
z
n
+
+


1n
n
nc z

b
£
ü
!
1
lim
n
n
n
c
c
ρ
+
→∞
= 5
a)
¥
l ???1 1/
n
n
cz

| |ρ 
a)
1
1
nn
c
z
n
+
+

¥
l ???1
1
1
/( 1)
1/lim lim 1/ | |
/( 2)
nn
nn
nn
acn
R
acn
ρ
+
→∞ →∞
+
+
== =
+

a)
¥
l ???1
1n
n
nc z

1
1
1/lim lim 1/ | |
(1)
nn
nn
nn
anc
R ρ
+
→∞ →∞
+
== =
+

#[
 ??
a)
μM]¥
l ???b
9
!)
l ?7
0
n
n
c

=

0
n
n
c

=

? ?£
ü
0
n
n
n
cz

=

¥
l ???1 1b
3
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!
£
ü ?)
0
n
n
c

=

l ? ?
a)
0
n
n
n
cz

=

 1z = )
l ? ? Abel? ? ?
¥
l ???
0
n
n
n
cz

=

1R ≥  7
0
n
n
c

=

? ?? 
0
||
n
n
n
cz

=

||1z = )? ?#
0
n
n
n
cz

=

¥
l ?? ?
1R ≤ b
[ ¥
l ???1 1b
0
n
n
n
cz

=

10 ?T)
0
n
n
n
cz

=


¥
l ??¥ ??
B? ) ' 
l ? £
ü

l ? ?
?¥> u×
 '
l ?b
0
z
£
ü ? Abel? ? ?  
l ??
= '
l ? £ ??
 '
l ?'
Vb??
 ? |B?
0
n
n
n
cz

=

η ?
0
00
|||
nn
nn
ccη
∞∞
==
=
∑∑
|z
0
n
n
n


=

'
l ?#2
? ?b
11ü/
 òf
Z 7? z¥
a)
i·
ì¥
l ???b

1
3
1
1
z+
 
2
()
2
2
1
1
z+
 
3  
4 sh 
2
cos z z

5  
6  
7ch z
2
sin
2
ze
z 1
z
z
e
 
8
z?1
1
sin
3 
1? 1||,1
1
1
32
<+?+?=
+
zzzz
z
" #
() …+?+…+?+?=
+
nn
zzzz
z
3963
3
11
1
1
 1|| <z 
7
l ??? R=1

2y () …+?+…+?+?=
+
nn
zzzz
z
11
1
1
32
 1|| <z 
# () …+?+…++?=
+
nn
zzz
z
242
2
11
1
1
 1|| <z 
?y ′?
+
2
1
1
z
()
2
2
1
2
z
z
+
= 
()
…+?+?=′?
+
=
+
642
22
2
4321
1
1
2
1
1
1
zzz
zz
z
 1|| <z 
7 =1 R

3y,,
!6!4!2
1cos
642
∞<…+?+?= z
zzz
z# "+?+?=
!6!4!2
1cos
1284
2
zzz
z
4
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!
+∞<|| z 7 
l ???  +∞=R

4y,||,
!3!2
1,
2
sh
32
+∞<…++++=
=
z
zz
ze
ee
z
z
zz
,||,
!3!2
1
32
+∞<…+?+?=
z
zz
ze
z
#,||,
!5!3
sh
33
+∞<…+++= z
zz
zz 7
l ??? R =+∞

5
24
ch 1,| |,
2! 4!
zz
zz=+++… <+∞

6y,||,
!3!2
1
64
2
2
+∞<…++++= z
zz
ze
z
,||,
!5!3
sin
106
22
+∞<…++?= z
zz
zz
#
…++?
…++++=
!5!3
.
!3!2
1sin
106
2
64
22
2 zz
z
zz
zze
z
,||,
3
6
42
+∞<…+++= z
z
zz
7
l ??? ;+∞=R

7y
23
1,|
2! 3!
z
zz
ez z=++ + +… <+∞,
23 1
0
,| | 1,
1
n
n
z
zz z z z
z

+
=
=… =? <

#
12 13
23
1 00
1
0
()()
11
2! 3! 2! 3!
nn
z
n nn
z
n
zz
zz
ez z
∞∞
++

+ ==
=
=? +? + = + <
∑∑

"",|1z
7
l ??? R=1b

8y,
1
sin1cos
1
cos1sin
1
1sin
1
1
sin
z
z
z
z
z
z
z?
+
=?
+=
,1||,
1
0
132
<=…+++=


=
+
zzzzz
z
z
n
n
# ()()…+…+++?…+++=
3
3232
!3
1
1
sin zzzzzz
z
z
…+++=
32
6
5
zzz   1|| <z
( ) ()…+…+++?…+++?=
4
32
2
32
!4
1
2
1
1
1
cos zzzzzz
z
z
…+=
32
2
1
1 zz   1|| <z
#?
…++++?
…+=
3232
6
5
1cos
2
1
11sin
1
1
sin zzzzz
z
= () 1||,1sin1cos
6
5
1sin
2
1
1cos1cos1sin
32
<+?
+?
++ zzzz " 
7
l ??? R=1b
12 p/
 òf
·?? )¥ Taylor Z 7
T i·
ì¥
l ???
0
z
5
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!

1
1
1
+
z
z
 1
0
=z 
2
()()21 ++ zz
z
 2
0
=z

3
2
1
z
 1
0
=z 
4
z34
1
 i1
0
+=z

5 tan z
0
/4z π= 
6 arctan z
0
0z =
3 
1y ()
()
2
1
1
1
2
1
21
1
1
1
1
+
=
+?
=
+
z
z
z
z
z
z
# 1||,1
1
1
32
<+?+?=
+
zzzz
z
" b#
()
+?

+

+
=
+
""
1
1
2
2
1
1
2
1
2
1
1
2
1
1
1
n
n
zzzz
z
z
() "" +?

++?

=
n
n
zzz
2
1
1
2
1
2
1
1
2
()
()
n
n
n
n
z 1
2
1
1
1
=


=
 2|1| <?z
?
^
l ??? R=2b

2y
()() 1
1
2
2
1
2
2
4
2
1
21 +
+
=?
+
+
=
++ zzzzzz
z
#
()
4
2
1
1
4
1
24
1
2
1
+
=
+
=
+
z
zz
4|2|,
4
2
4
2
1
4
1
2
<?


+
= z
zz
"
()
3
2
1
1
3
1
23
1
1
1
+
=
+
=
+
z
zz
= 3|2|,
3
2
3
2
1
3
1
2
<?


+
z
zz
"
#e
T ()
+
=
"
2
222
2
2
1
2
2
1
4
2
z
z ( )
+
"
2
2
3
2
3
2
1
3
1 zz
=
()( ) ( )( )
∑∑

=

=


00
2
3
21
3
1
2
21
2
1
n
n
nn
n
n
nn
zz ( )
()
( )
()
∑∑

=
+

=
+

=
0
1
0
12
2
3
1
2
2
1
n
n
n
n
n
n
n
n
zz
() ()


=
++

=
0
112
2
3
1
2
1
1
n
n
nn
n
z  3|2| <?z 7 3=R b

3y ′?
=
zz
11
2
#
()
()()[ ]"+++++?=
+?
=
2
111
11
11
zz
zz
  1|1| <+z
# () ()"" ++++++=
1
2
1121
1
n
znz
z
()()


=
++=
0
11
n
n
zn  1|1| <+z 
7 R=1b
6
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!

4y
()[]i33i134
1
34
1
+
=
zz ()[]i13i31
1
+
=
z
()[]i1
i31
3
1
1
i31
1
+?
=
z
()[] ()[]
++
++?
+
= "
2
2
i1
i31
3
i1
i31
3
1
i31
1
zz 
? ()[]1i1
i31
3
<+?
z #
()
()[]
n
n
n
n
z
z
i1
i31
3
34
1
0
1
+?
=


=
+
 ()
3
10
3
i31
|i1| =
<+?z 
O
l ???
3
10
=R b

5y
3
5
2
tan,| |
315 2
z
zz z z
π
= ++ + <" ?
4
4
1tan( )
tan
1tan( )
z
z
z
π
π
+?
=

#
2
444
tan [1 tan( )](1 tan( ) tan ( ) )zz z z
πππ
=+? +?+?+"
"
23
8
12( )2( ) ( ),
4434
zz z
π ππ
+?+?+?+" O
l ???
4
R
π
= b

6y
2
1
(arctan ) '
1
z
z
=
+
?
24
2
1
1,
1
zz z
z
|1=?+? <
+
" #
21
2
2
00
00
1
arctan ( 1) ( 1)
12
n
zz
nn n
n
z
zdz zdz
zn
+∞∞
=
==
+ +
∑∑
∫∫
||1z < 
O
l ??? 1R = b
131
I
1  u× ||zR<
=3  O uW (,)R R? |
L
′¥f
()f z Z 7? ¥
a)
HZ 7
T¥"
?
^
L
$
z
3 ()f z Z 7? ¥
a )
HZ 7
T¥"
1z
()
(0)
!
n
n
f
c
n
= 7f
()f z  uW
(,)R R? |
L
′ V? 91
L
b#Z 7
T¥"
?
^
L
b
()
(0)
n
f
14£
ü
1
() cos( )fz z
z
= + [ ¥ò
aV¥
μZ 7
T?¥ò"
1 z
2
0
1
cos(2cos )cos,( 0,1,2,)
2
n
cndn
π
θθθ
π
==

"±b
£
ü
1
() cos( )fz z
z
= + ˉ
ü
= ?? 0z = ?3
[
= V
Z 7?
μ)
0||z<<+∞
1
cos( )
n
n
n
zc
z

=?∞
+=

z ?
7
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!
1
||
1
cos( )
1
,( 0,1,2,,0 )
2
n
n
zr
z
z
cdn
izπ
+
=
+
r= =±± <<+∞

"
v
D1£
ü
T0? ¥V
U
TM1?á
ì |
n
c 1r = i ?¨ˉs¥9

T V¤
ii
22
1i
00
||1
1
cos( )
11cos()1
cos(2cos )cos
2i 2 2
n nn
z
z
ee
z
cd d
ze
θθ
ππ
θ
ndθ θθ
ππ π
+
=
+
+
== =
∫∫ ∫v
θ
2
0
i
cos(2cos )sin
2
nd
π
θ θθ
π

"
2
0
1
cos(2cos )cos
2
nd
π
θ θθ
π

y
2
0
cos(2cos )sin nd
π
θ θθ

" cos(2cos )sin nd
π
π
θ θθ

7 cos(2cos )sin nθ θ 1 θ¥ 
f
b
15/
2
^?? ?$
¨é"E
23
1
z
zz z
z
= +++
" 
2
11
1
1
z
zzz
= ++ +
"
y1 0
11
zz
zz
+=


[
2
11
1
zz
++++"
23
0zz z+ ++=" b
3 ?? ? by 1é" E
¤?¥

T 
P
ì? ?¥ z ′¥S ??] 
s Y 1
 || yN?
MFb ||1z < 1z >
16ü/
 òf
·?¥?ì×
=Z 7? Laurent)
b

1
()()21
1
2
+ zz
 2||1 << z  
2
()
2
1
1
zz?
 1|1|0,1||0 <?<<< zz 

3
()()
+∞<?<<?<

|2|1,1|1|0,
21
1
zz
zz

4
1
1 z
e
 +∞<< ||1 z

5
2
1
(izz? )
[ i1??¥?ì×
= 
6
z?1
1
sin  +∞<?< |1|0 z

7
(1)(2)
,3 | | 4,4 | |
(3)(4)
zz
zz
zz

< <<<+


3 
1y
2
5
1
1
5
2
1
5
1
)2)(1(
1
222
+
+
+
+
=
+ zzz
z
zz
#
2
1
1
10
1
1
1
11
5
2
1
1
11
5
1
)2)(1(
1
2
2
2
22
z
z
z
z
z
z
zz
+
+
=
+
8
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!
() ()
∑∑∑

=

=

=
=
000
2
1
222
210
1
1
1
5
21
1
1
5
1
nn
n
n
n
n
n
n
z
z
zzz
z
() ()
∑∑∑

=

=
+

=
+
=
00
)1(2
0
12
210
11
1
5
21
1
5
1
n
n
n
n
n
n
n
n
n
z
zz
""++=
80402010
11
5
11
5
21
5
11
5
2
32
234
zzz
zzzz
2||1 << z 

2 1||0 << z
=
()
( )
2
2
2
1
1
1
1
"" +++++=
n
zzz
zzz
()( )"" ++++++=
n
znzz
z
1321
1
2
() "" ++++++=
1
132
1
n
znz
z
()


=
+=
1
2
n
n
zn

= 1|1|0 <?< z
()()()()
()( )


=

=
+?
=
0
222
11
1
1
11
1
1
1
1
1
n
nn
z
zzzzz
()( )
n
n
n
z 11
2
=


=


3 1|1|0 <?< z
=
() 1
1
11
1
1
1
2
1
)2)(1(
1

=
=
zzzzzz
()
()


=
=

=
0
1
1
1
1
1
11
1
n
n
z
z
zz
()


=
=
1
1
n
n
z
 +∞<?< |2|1 z
=
()12
1
2
1
1
1
2
1
)2)(1(
1
+?
=
=
zzzzzz
2
1
1
1
2
1
2z
1
+
=
z
z
()
()
n
n
n
zzz 2
1
1
2
1
2
1
0
=


=
()
()


=
+
+
+
=
0
1
1
2
1
1
2
1
n
n
n
zz
()
()


=
+
=
1
2
1
1
2
1
n
n
n
zz
()
()


=
=
2
1
2
1
1
n
n
n
z

4 +∞<< ||1 z
=y
+++?=?
+++
=
=
""
322
11111
1
1
1
1
1
1
1
zzzzzz
z
z
z
9
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!
# """" +?
+++
++++?
+++?=
3
32
2
3232
1
1
111
!3
1111
!2
1111
1
zzzzzzzzz
e
z

"++=
432
1
!4
11
!3
11
!2
11
1
zzzz

5
=y0| i|1z<?<
11
2
1
1
(1),| |1
(1 )
nn
n
nz z
z


=
=? <
+

#
2
1
(izz )?
"
22
1
i
i( i)(1 )
i
z
z
=
+
2
1
n+1
1
(i)
(1)
i
n
n
n
nz

=

 1|
=yi|z<?<+∞
2
1
(izz )?
"
32
1
i
(i)(1 )
i
z
z
+
#
2
1
(izz )?
"
-1
1
23
10
i(
(1) (1)
(-i) (-i)
nn
nn
zz
∞∞
1
+ +
==
+
=?
∑∑

6y ()
()
+∞<
+
=


=
+
0
12
||,
!12
1sin
n
n
n
z
n
z
z,#
()
()()


=
+
+
=
0
12
1
1
!12
1
1
1
1
sin
n
n
n
znz
()
()()
+∞<?<
+
=
+

=

|1|0,
1
1
!12
1
1
12
0
z
zn
n
n
n

3|
=y|4z<<
2
(1)(2) 1 1
(32)(
(3)(4) 3 4
zz
zz
zz z z

=?+?

)# 
2
3
4
(1)(2) 1 1
(32)(
( 3)( 4) (1 ) 4(1 )
z
z
zz
zz
zz z

= + +

)
"
2
11
00
3
(32)(
4
nn
nn
nn
z
zz
z
∞∞
++
==
+ +
∑∑
)"
1
01
3
12
24 3
nn
nn
nn
zz
∞∞
+
==?

∑∑

4|
=|z<<+∞
2
(1)(2) 1 1
(32)(
(3)(4) 4 3
zz
zz
zz z z

=?+?

)
2
34
(1)(2) 1 1
(32)(
(3)(4) (1) (1)
zz
zz
zz
zz z z

=?+?

)
"
2
11
00
43
(32)(
nn
nn
nn
zz
zz
∞∞
++
==
+?
∑∑
)
n
"
21 1
1
1 (32 23 )
nn
n
z


=
+


f
1
tan
z
??ì× 
0||zR<< 0 R< <+∞
=Z 7?
μ)
$
1
I
1$
10
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!
3 ?
Z?
μ)
by?ì× 0||z R< <
=
1
tan
z
?3b
 ?T 1
@1"k
2
1k < ¥
L
£
ü
2
0
sin
sin( 1)
12cos
n
n
kn
kk
θ
θ
θ

=
+=
+

 
2
0
cos
cos( 1)
12cos
n
n
k
kn
kk
θ
θ
θ

=
+=
+

b
£
ü
1
zk?
 ||
=13f
| Z?
μ)
μzk>
1
0
11
(1 )
n
n
n
k
k
zk z
z
z

+
=
=




T?
7
i
ze
θ
= 
ii1
0
1
()
n
n
n
k
ek e
θθ

+
=
=

()
2
0
cos isin
cos( 1) i sin( 1)
12cos
nn
n
k
kn k
kk
θθ
θ θ
θ

=

=+?+
+

' 
s ?
L?′?'¤2
b
 ?T 1?_?? ||C 3z =  ps ()
C
f zdz

¥′
! ()f z 1

1
(2zz+ )
  
2
(1)
z
zz
+
+
  
2
1
(1)zz+
  
(1)(2)
z
zz+ +
b
3
1
() 2i
C
f zdz cπ
=



1
2
0
1111111 11 2
(1),| | 2
(2)2 2 2 (1) 2
n
n
n
n
z
z
zz zz zz z z

+
=


=?=? =

++ +



>b
# ()
C
f zdz

"b

1
1
0
21 1
(2) (2) (1),||
(1) (1)
n
n
n
z
z
zz
zz zz z z

+
=
+

1= +? =+ >


++




# ()
C
f zdz

" 2iπ b

1
23 23 1
1
1
111
(1),| |1
(1) (1)
n
n
n
z
n
z
zz z z z

=
= =?
++

>b# ()
C
f zdz

"b
11
我要答案网 www.51daan.net
本文件是从网上收集,严禁用于商业用途!

12
00
11 (1)(1)2
,| | 2
(1)(2) (1) (1)
nnn
n
nn
zz
z
zz z z z z
∞∞
==


=?=?

++ + +

∑∑
>
# ()
C
f zdz

" 2iπ b

k ps ¥′ ? 1?ê? ||
2
()
n
n
C
zdz

=?

∫v
C 1z =
=¥ ?BH?üV
e?¥e?> wLb
3
2
2
11 1
,0 | | 1
1
n
n
zz
zz z

=?
=++ <<

b7 1?ê? ||C 1z =
=¥ ?BH?ü
Ve?¥e?> wL#
1
2
()2i2
n
n
C
zdz c iπ π

=?
==

∫v
b
12