3134bet:高考数学一轮复习 第11章 算法、复数、推理与证明 11.2 数系的扩充与复数的引入习题课件 文_图文

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课后作业夯关 11.2 数系的扩充与复数的引入

[基础送分 提速狂刷练]

一、选择题

1.(2018·湖南长沙四县联考)i是虚数单位,若复数z满

足zi=-1+i,则复数z的实部与虚部的和是( )

A.0 B.1 C.2 D.3

解析 复数z满足zi=-1+i,可得z=

-1+i i



?-1+i?i i·i

=1+i.故复数z的实部与虚部的和是1+1=2,故

选C.

2.(2018·湖北优质高中联考)已知复数z=1+i(i是虚数

单位),则2z-z2的共轭复数是(

)

A.-1+3i B.1+3i C.1-3i D.-1-3i

解析 2z-z2=1+2 i-(1+i)2=?1+2?1i?-?1-i? i?-2i=1-i- 2i=1-3i,其共轭复数是1+3i,故选B.

3.(2017·河南洛阳模拟)设复数z满足 -z =|1-i|+i(i为 虚数单位),则复数z=( )
A. 2-i B. 2+i C.1 D.-1-2i 解析 复数z满足 -z =|1-i|+i= 2 +i,则复数z= 2 -i.故选A.

4.(2018·广东测试)若z=(a- 2)+ai为纯虚数,其中a

∈R,则1a++ai7i=(

)

A.i B.1 C.-i D.-1 解析 ∵z为纯虚数,∴?????aa≠-0,2=0, ∴a= 2,



a+i7 1+ai



2-i 1+ 2i



? 2-i??1- ?1+ 2i??1-

2i? 2i?

=-33i

=-i.故选

C.

5.(2018·安徽江南十校联考)若复数z满足z(1-i)=|1- i|+i,则z的实部为( )
解A.析22-由1 z(1B-. i2)=-|11-Ci|+.i1,得Dz.=22+12-+1ii= ??12-+i?i??1?1++i?i?= 22-1+ 22+1i,z的实部为 22-1,故选A.

6.(2017·安徽十校联考)若z=22+-ii,则|z|=(

)

1 A.5

B.1

C.5

D.25

解析 解法一:z=22- +ii=??22- +ii????22- -ii??=35-45i,故|z|=

1.故选B.

解法二:|z|=???22- +ii???=||22- +ii||= 55=1.故选B.

7.(2017·河南百校联盟模拟)已知复数z的共轭复数为 -z ,若 ????32z+-2z ???? (1-2 2 i)=5- 2 i(i为虚数单位),则在复 平面内,复数z对应的点位于( )
A.第一象限 B.第二象限 C.第三象限 D.第四象限

解析

依题意,设z=a+bi(a,b∈R),则

3z 2



-z 2

=2a

+bi,故2a+bi=15--2 22ii=1+ 2i,

故a=

1 2

,b=

2 ,则在复平面内,复数z对应的点为

????12,

?
2??,位于第一象限.故选A. ?

8.(2018·新乡、许昌、平顶山调研)复数z1,z2满足z1

=m+(4-m2)i,z2=2cosθ+(λ+3sinθ)i(m,λ,θ∈R),并

且z1=z2,则λ的取值范围是( )

A.???-1,1???

B.????-196,1????

C.????-196,7 ????

D.????196,7 ????

解析 由复数相等的充要条件,可得

??m=2cosθ, ???4-m2=λ+3sinθ,

化简得4-4cos2θ=λ+3sinθ,由此可得λ=-4cos2θ-

3sinθ+4=-4(1-sin2θ)-3sinθ+4=4sin2θ-3sinθ=

4

????sinθ-83????

2-

9 16

,因为sinθ∈[-1,1],所以λ∈

????-196,7????

.故

选C.

9.对于复数z1,z2,若(z1-i)z2=1,则称z1是z2的“错

位共轭”复数,则复数

3 2



1 2

i的“错位共轭”复数为

()

A.- 63-12i B.- 23+32i

C. 63+12i

D. 23+32i

解析

?
由(z-i)?? ?

23-12i????=1,可得z-i=

1= 23-12i

23+12

i,所以z= 23+32i.故选D.

10.已知z=a+bi(a,b∈R,i是虚数单位),z1,z2∈ C,定义:D(z)=||z||=|a|+|b|,D(z1,z2)=||z1-z2||,给出 下列命题:
(1)对任意z∈C,都有D(z)>0; (2)若 z 是复数z的共轭复数,则D( z )=D(z)恒成立; (3)若D(z1)=D(z2)(z1,z2∈C),则z1=z2;

(4)对任意z1,z2,z3∈C,结论D(z1,z3)≤D(z1,z2)+

D(z2,z3)恒成立.

其中真命题为( )

A.(1)(2)(3)(4) B.(2)(3)(4)

C.(2)(4)

D.(2)(3)

解析 对于(1),由定义知当z=0时,D(z)=0,故(1)错 误,排除A;对于(2),由于共轭复数的实部相等而虚部互 为相反数,所以D( z )=D(z)恒成立,故(2)正确;对于(3), 两个复数的实部与虚部的绝对值之和相等并不能得到实部 与虚部分别相等,所以两个复数也不一定相等,故(3)错 误,排除B,D,故选C.

二、填空题 11.(2017·江苏高考)已知复数z=(1+i)(1+2i),其中i 是虚数单位,则z的模是____1_0___.
解析 解法一:∵z=(1+i)(1+2i)=1+2i+i-2=-1 +3i,
∴|z|= ?-1?2+32= 10. 解法二:|z|=|1+i||1+2i| = 2× 5= 10.

12.(2016·天津高考)已知a,b∈R,i是虚数单位.若

(1+i)(1-bi)=a,则ab的值为____2____. 解析 由(1+i)(1-bi)=a得1+b+(1-b)i=a,则

?? b+1=a, ???1-b=0,

解得????? ab= =21, ,

所以ab=2.

13.(2016·北京高考)设a∈R.若复数(1+i)(a+i)在复平 面内对应的点位于实轴上,则a=__-__1____.
解析 (1+i)(a+i)=(a-1)+(a+1)i, ∵a∈R,该复数在复平面内对应的点位于实轴上, ∴a+1=0,∴a=-1.

14.若虚数z同时满足下列两个条件:①z+

5 z

是实数;

②z+3的实部与虚部互为相反数.则z=_-__1_-__2_i或__-__2_-__i_.

解析 设z=a+bi(a,b∈R,b≠0), 则z+5z=a+bi+a+5 bi =a????1+a2+5 b2????+b????1-a2+5 b2????i.

又z+3=a+3+bi实部与虚部互为相反数,z+

5 z

是实

数,根据题意有???b????1-a2+5 b2????=0, ??a+3=-b,

因为b≠0,所以

?? a2+b2=5, ???a=-b-3,

解得

?? a=-1, ???b=-2



?? a=-2, ???b=-1.
所以z=-1-2i或z=-2-i.

三、解答题

15.(2017·徐汇区校级模拟)已知z是复数,z+2i与

z 2-i

均为实数(i为虚数单位),且复数(z+ai)2在复平面上对应点

在第一象限.

(1)求z的值;

(2)求实数a的取值范围.

解 (1)设z=x+yi(x,y∈R), 又z+2i=x+(y+2)i为实数,∴y+2=0,解得y=-2. ∴2-z i=x2--2ii=??x2--2ii????22++ii??=?2x+2?+5 ?x-4?i, ∵2-z i为实数,∴x-5 4=0,解得x=4. ∴z=4-2i.

(2)∵复数(z+ai)2=[4+(a-2)i]2=16-(a-2)2+8(a- 2)i=(12+4a-a2)+(8a-16)i,
∴?????182a+ -416a->0a,2>0, 解得2<a<6, 即实数a的取值范围是(2,6).

16.(2017·孝感期末)已知复数z=(m-1)+(2m+1)i(m ∈R).
(1)若z为纯虚数,求实数m的值; (2)若z在复平面内的对应点位于第二象限,求实数m的 取值范围及|z|的最小值. 解 (1)∵z=(m-1)+(2m+1)i(m∈R)为纯虚数, ∴m-1=0且2m+1≠0,∴m=1. (2)z在复平面内的对应点为(m-1,2m+1).

由题意得?????m2m-+1<1>0,0, ∴-21<m<1,

即实数m的取值范围是????-12,1????. 而|z|= ?m-1?2+?2m+1?2 =

5m2+2m+2 =

5????m+51????2+59, 当m=-15∈????-12,1????时,|z|min=

95=3

5

5 .


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