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复变函数课后习题答案(全)(2020年10月整理).pdf


10
3
10 10
(3) z = 1 − 3i = −i + 3 − 3i = 3 − 5i ,
i 1−i
2
2
因此, Re z = 3 , Im z = − 5 ,
3
2
z = 34 , arg z = −arctan 5 , z = 3 + 5i
2
3
2
(4) z = −i8 + 4i21 − i = −1+ 4i − i = −1+ 3i
+
i sin ) 12

z1 = 1 [cos( + ) + i sin( + )] = 1 (cos 5 + i sin 5 )
z2 2
46
4 6 2 12
12
5. 解下列方程:
(1) (z + i)5 = 1
(2) z4 + a4 = 0 (a 0)
解:(1) z + i = 5 1, 由此
(3) (1− 3i)(cos + i sin ) (1− i)(cos − i sin )
2[cos(− ) + i sin(− )](cos + i sin )
=
3
3
2[cos(− ) + i sin(− )][cos(− ) + i sin(− )]
4
4
= 2[cos(− ) + i sin(− )](cos 2 + i sin 2 )
3
z=
5
1
−i
=
2 k i
e5

i

(k = 0,1, 2,3, 4)
(2) z = 4 −a4 = 4 a4 (cos + i sin )
= a[cos 1 ( + 2k ) + i sin 1 ( + 2k )] ,当 k = 0,1, 2,3时,对应的 4
4
4
个根分别为: a (1+ i), a (−1+ i), a (−1− i), a (1− i)
因此, Re z = −1, Im z = 3,
z = 10, arg z = − arctan 3, z = −1− 3i
2. 将下列复数化为三角表达式和指数表达式:
(1) i
(2) −1+ 3i
(3) r(sin + i cos )
(4)r(cos − i sin )
(5)1− cos + i sin (0 2 )
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 3 + 2i
(2)
i
(i −1)(i − 2)
(3) 1 − 3i i 1−i
(4) −i8 + 4i21 − i
解:(1) z = 1 = 3 − 2i , 3 + 2i 13
因此: Re z = 3 , Im z = − 2 ,
i
解:(1) i = cos + i sin = e 2
2
2
1
(2) −1+
3i
=
2(cos
2
+
i
sin
2
)
=
2 i
2e 3
3
3
(3) r(sin
+ i cos )
= r[cos(
− ) + i sin(
( − )i
− )] = re 2
2
2
(4) r(cos − i sin ) = r[cos(− ) + i sin(− )] = re−i
i
2e 8 ,
k k
=0 =1
4.

z1
=
1+ i 2
,
z2 =
3

i,
试用三角形式表示
z1 z2

z1 z2
解: z1 = cos 4 + i sin 4 , z2 = 2[cos(− 6 ) + i sin(− 6 )],所以
z1 z2
=
2[cos( 4

6
)
+
i sin( 4

6
)]
=
2(cos 12
(5)1− cos + i sin = 2sin2 + 2i sin cos
2
22
=
2sin
[cos

+ i sin

]=
2sin
− i
e2
2
2
2
2
3. 求下列各式的值:
(1) ( 3 − i)5
(2) (1+ i)100 + (1− i)100
(3) (1− 3i)(cos + i sin ) (1− i)(cos − i sin )
13
13
z = 1 , arg z = −arctan 2 , z = 3 + 2 i
13
3
13 13
(2) z =
i
= i = −3 + i ,
(i −1)(i − 2) 1− 3i 10
因此, Re z = − 3 , Im z = 1 ,
10
10
z = 1 , arg z = − arctan 1 , z = − 3 − 1 i
2
2
3 2
+
1 i, 2
k =0
=
cos
1 3
( 2
+
2k
)
+
i
sin
1 3
( 2
+
2k
)
=

3 + 1 i, 22
k =1
−i,
k=2
(6) 1+ i = 2(cos + i sin )
4
4
=
4
2[cos
1 2
( 4
+
2k
)
+
i sin
1 2
( 4
+
2k
)]
=
4
i
2e 8 ,
− 4
12
12
=
2[cos(2 − ) + i sin(2 − )] =
(2 − )i
2e 12
12
12
2
(4)
(cos (cos
5 3
+ −
i i
sin sin
5 3
)2 )3
= cos10 + i sin10 = cos19 + i sin19 cos(−9) + i sin(−9)
(5) 3 i = 3 cos + i sin
(4)
(cos (cos5 3Leabharlann + −i i
sin sin
5 3
)2 )3
(5) 3 i
(6) 1+ i
解:(1) ( 3 − i)5 = [2(cos(− ) + i sin(− ))]5
6
6
= 25 (cos(− 5 ) + i sin(− 5 )) = −16( 3 + i)
6
6
(2) (1+ i)100 + (1− i)100 = (2i)50 + (−2i)50 = −2(2)50 = −251
2
2
2
2
x+ y
6. 证明下列各题:(1)设 z = x + iy,则
z x+ y
2
证明:首先,显然有 z = x2 + y2 x + y ;
其次,因
x2 + y2 2 x y ,
固此有
2(x2 + y2 ) ( x + y )2,
从而 z = x2 + y2 x + y 。 2
(2)对任意复数 z1, z2 , 有 z1 + z2 2 = z1 2 + z2 2 + 2 Re(z1 z2 ) 证明:验证即可,首先左端 = (x1 + x2 )2 + ( y1 + y2 )2 , 而右端 = x12 + y12 + x22 + y22 + 2 Re[(x1 + iy1)(x2 − iy2 )] = x12 + y12 + x22 + y22 + 2(x1x2 + y1 y2 ) = (x1 + x2 )2 + ( y1 + y2 )2 ,
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