内容正文:
数学 (选修 1 - 2·人教 A 版)
3. B 由题意得 z = 1 + 2i,∴ iz = i(1 + 2i) = - 2 + i,故选 B.
4. 13 ∵ 5 - i
1 + i
= (5 - i)(1 - i)
(1 + i)(1 - i)
= 2 - 3i,
∴ 5 - i1 + i = |2 - 3i | = 13.
5. (1)( 1
2
+ 3
2
i)(4i - 6)
= 1
2
·4i + 1
2
·( - 6) + 3
2
i·4i + 3
2
i·( - 6)
= 2i - 3 - 6 - 9i = - 9 - 7i.
(2)(1 - i)(1 + 2i)
1 + i
= (1 - i)(1 + 2i)(1 - i)
(1 + i)(1 - i)
= ( - 2i)(1 + 2i)
2
= - i(1 + 2i) = 2 - i.
章末整合提升
专题突破·启智能
典例试做 1:(1)由题意可得 m(m - 1) = 0
m2 + 2m - 3 = 0{ ,
即
m = 0 或 m = 1
m = - 3 或 m = 1{ ,∴ m = 1.
所以当 m = 1 时,复数 z 为零.
(2)由题意可得 m(m - 1) = 0
m2 + 2m - 3≠0{ ,
解得
m = 0 或 m = 1
m≠ - 3 且 m≠1{ ,所以 m = 0.
所以 m = 0 时,z 为纯虚数.
(3)由题意可得 m(m - 1) = 2
m2 + 2m - 3 = 5{ ,
解得
m = 2 或 m = - 1
m = - 4 或 m = 2{ ,∴ m = 2.
所以当 m = 2 时,复数 z 为 2 + 5i.
典例试做 2:(1) 1
10
+ 7
10
i (2) D (1) ∵ 1 + 2i
3 + i3
= 1 + 2i
3 - i
=
(1 + 2i)(3 + i)
(3 - i)(3 + i)
= 3 - 2 + 7i
10
= 1
10
+ 7
10
i.
(2)∵ (1 + i
1 - i
)2 = 2i
- 2i
= - 1,( - 1
2
+ 3
2
i)3 = 1,
∴ (1 + i
1 - i
)4 + ( - 1
2
+ 3
2
i)18
= [(1 + i
1 - i
)2 ]2 + [( - 1
2
+ 3
2
i)3 ]6 = 2.
典例试做 3:D ∵ 点 Z(3,1)对应的复数为 z,
∴ z = 3 + i, z
1 + i
= 3 + i
1 + i
= (3 + i)(1 - i)
(1 + i)(1 - i)
= 4 - 2i
2
= 2 - i,
该复数对应的点的坐标是(2, - 1),即 H 点.
典例试做 4:|1 - i + z | = | cosθ + isinθ + 1 - i |
= (cosθ + 1)2 + (sinθ - 1)2
= 2(cosθ - sinθ) + 3 = 2 2cos(θ + π
4
) + 3,
当 θ = 7π
4
时,|1 - i + z | max = 2 + 1;
当 θ = 3π
4
时,|1 - i + z | min = 2 - 1.
典例试做 5:D 设 z = x + yi(x、y∈R),则z = x - yi.
由 z + z = 4,z z = 8,得
x + yi + x - yi = 4
(x + yi)(x - yi) = 8{ ,即
x = 2
x2 + y2 = 8{ ,解得
x = 2
y = ± 2{ .
∴ z
z
= x - yi
x + yi
= x
2 - y2 - 2xyi
x2 + y2
= ± i.
典例试做 6:D ∵ zi z- 1 1 = zi + z = 4 + 2i,
∴ z = 4 + 2i
1 + i
= 3 - i.
典例试做 7:2 3 令 z1 = 2( cosθ + isinθ),z2 = 2 ( cosα +
isinα),
∴ z1 + z2 = 2(cosθ + cosα) + 2(sinθ + sinα)i,
又∵ z1 + z2 = 3 + i,∴
2(cosθ + cosα) = 3 ①
2(sinθ + sinα) = 1 ②{ ,
①2 + ②2 得,4(2 + 2cosθcosα + 2sinθsinα) = 4,
∴ cosθcosα + sinθsinα = - 1
2
,
∴ | z1 - z2 | = |2(cosθ - cosα) + 2(sinθ - sinα)i |
= 4(cosθ - cosα)2 + 4(sinθ - sinα)