复变函数课后习题答案(全)
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习题一答案
1.求下列复数的实部、虚部、模、幅角主值及共轭复数:
(1)
1
32i
+
(2)
(1)(2)
i
i i
--
(3)13
1
i
i i
-
-
(4)821
4
i i i
-+-
解:(1)
132
3213
i z
i
-
==
+
,
因此:
32 Re, Im
1313 z z
==-,
232
arg arctan,
31313
z z z i
==-=+
(2)
3
(1)(2)1310
i i i
z
i i i
-+
===
---
,
因此,
31
Re, Im
1010
z z
=-=,
131
arg arctan,
31010 z z z i
π
==-=--(3)
133335
122
i i i
z i
i i
--
=-=-+=
-
,
因此,
35
Re, Im
32
z z
==-,
535
,arg arctan,
232
i
z z z
+ ==-=
(4)821
41413
z i i i i i i
=-+-=-+-=-+
因此,Re1,Im3
z z
=-=,
arg arctan3,13
z z z i
π
==-=--
2. 将下列复数化为三角表达式和指数表达式: (1)i (2
)1-+ (3)(sin cos )r i θθ+
(4)(cos sin )r i θθ- (5)1cos sin (02)i θθθπ-+≤≤
解:(1)2
cos
sin
2
2
i
i
i e π
π
π
=+=
(2
)1-+2
3
222(cos sin )233
i i e πππ=+=
(3)(sin cos )r i θθ+()2
[cos()sin()]22i
r i re
π
θππ
θθ-=-+-=
(4)(cos sin )r i θ
θ-[cos()sin()]i r i re θθθ-=-+-=
(5)2
1cos sin 2sin 2sin cos 222
i i θ
θθ
θθ-+=+ 2
2sin [cos
sin
]2sin 22
22
i
i e
πθ
θπθ
πθ
θ
---=+=
3. 求下列各式的值:
(1
)5)i - (2)100100(1)(1)i i ++-
(3
)(1)(cos sin )
(1)(cos sin )
i i i θθθθ-+-- (4)
23(cos5sin 5)(cos3sin 3)i i ϕϕϕϕ+-
(5
(6
解:(1
)5)i -5[2(cos()sin())]66
i ππ
=-+-
5
552(cos()sin()))66
i i ππ
=-+-=-+
(2)100
100(1)
(1)i i ++-50505051(2)(2)2(2)2i i =+-=-=-
(3
)(1)(cos sin )
(1)(cos sin )
i i i θθθθ-+--
2[cos()sin()](cos sin)
33
)sin()][cos()sin()]
44
i i
i i
ππ
θθ
ππ
θθ
-+-+
=
-+--+-
)sin()](cos2sin2)
1212
i i
ππ
θθ
=-+-+
(2)
12
)sin(2)]
1212
i
i
π
θ
ππ
θθ-
=-+-=
(4)
2
3
(cos5sin5)
(cos3sin3)
i
i
ϕϕ
ϕϕ
+
-
cos10sin10
cos19sin19
cos(9)sin(9)
i
i
i
ϕϕ
ϕϕ
ϕϕ
+
==+
-+-
(5
=
11
cos(2)sin(2)
3232
k i k
ππ
ππ
=++
+
1
,0
22
1
,1
22
,2
i k
i k
i k
+=
⎪
⎪
⎪
=-+=
⎨
⎪
-=
⎪
⎪
⎩
(6
=
11
(2)sin(2)]
2424
k i k
ππ
ππ
=+++
8
8
,0
,1
i
i
e k
e k
π
π
=
=
⎪=
⎩
4.
设
12
,
z z i
==-试用三角形式表示
12
z z与1
2
z
z
解:
12
cos sin, 2[cos()sin()]
4466 z i z i
ππππ
=+=-+-,所以