复数2022高考题.pdf

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1、第 1 页共 4 页2010 年全国各地高考数学真题分章节分类汇编复数一、选择题:1（2010 年高考山东卷理科2）已知2(,R)aibi a bi2aibii（a,bR），其中 i 为虚数单位，则 a+b=(A)-1(B)1(C)2(D)3【答案】B【解析】由a+2i=b+ii得a+2i=bi-1,所以由复数相等的意义知:a=-1,b=2,所以a+b=1,故选 B.【命题意图】本题考查复数相等的意义、复数的基本运算，属保分题。2（2010 年高考全国卷I 理科 1）复数3223ii(A)i(B)i (C)12-13 (D)12+13【答案】A【解析】32(32)(23)694623(23)(

2、23)13iiiiiiiii.【命题意图】本小题主要考查复数的基本运算,重点考查分母实数化的转化技巧.3(2010 年高考湖北卷理科1)若为虚数单位，图中复平面内点z 表示复数z，则表示复数1zi的点是A.E B.F C.G D.H【答案】D【解析】由图知z=2+i，所以1zi21ii(2)(1-i)422(1+)(1-i)2iiii，故选 D。4（2010 年高考福建卷理科9）对于复数a,b,c,d,若集合S=a,b,c,d具有性质“对任意x,yS,必有xyS”,则当22a=1b=1c=b时,b+c+d等于()A.1 B.-1 C.0 D.【答案】B【解析】由题意，可取a=1,b=-1,c=

3、i,d=-i，所以b+c+d=-1+i+-i1，选 B。【命题意图】本题属创新题，考查复数与集合的基础知识。5（2010 年高考安徽卷理科1）是虚数单位，33iiA、13412iB、13412iC、1326iD、1326i1.B 第 2 页共 4 页【解析】(33)3313391241233iiiiii，选 B.【规律总结】33ii为分式形式的复数问题，化简时通常分子与分母同时乘以分母的共轭复数3i，然后利用复数的代数运算，结合21i得结论.6.(2010 年高考天津卷理科1)i 是虚数单位，复数1312ii=（A）1+i（B）5+5i（C）-5-5i（D）-1-i【答案】A【解析】1312i

4、i(13)(12)5ii5515ii，故选 A。【命题意图】本小题考查复数的基本运算，属保分题。7（2010 年高考广东卷理科2）若复数 z1=1+i，z2=3-i，则 z1z2=（）A4+2 iB.2+i C.2+2 i D.3【答案】A【解析】12(1)(3)1 3 1 1(3 1)42zziiii。8（2010 年高考四川卷理科1）i 是虚数单位，计算ii2i3（A）1（B）1（C）i（D）解析：由复数性质知：i2 1故 i i2i3i(1)(i)1答案：A 9.(2)(2010年全国高考宁夏卷2）已知复数23(13)izi，z是 z的共轭复数，则zz=A.14B.12C.1 D.2【答

5、案】A 解析：2333(3)(2 2 3)3144(13)1 2 332 2 3(22 3)(2 2 3)iiiiiziiiiii，所以22311()()444z z另解：2223(3)31113144(13)(31)(31)(31)3ii iiziiiiiiiii，下略10 （2010年 高 考 陕 西 卷 理 科2）复 数1izi在 复 平 面 上 对 应 的 点 位 于（A）（A）第一象限（B）第二象限（C）第三象限（D）第四象限【答案】A【解析】iiiiiiz21211112，复数在复平面上对应的点位于第一象限.故选.11（2010 年高考江西卷理科1）已知()(1)xiiy，则实数，

6、分别为A1x，1yB1x，2yC1x，1yD1x，2y文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I

7、2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文

8、档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6

9、E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T1

10、0 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R

11、7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N

12、2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:

13、CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10第 3 页共 4 页【答案】D12（2010 年高考浙江卷5）对任意复数z=x+yi (x,yR),i为虚数单位，则下列结论正确的是【答案】D 13（2010 年高考辽宁卷理科2）设 a,b 为实数，若复数11+2iiabi，则（A）31,22ab(B)3,1ab(C)13,22ab(D)1,3ab【答案】A 14(2010 年高考全国 2 卷理数 1）复数231ii（A）34i（B）34i（C）34i（D）34i【答案】A【命题意图】本试题主要考查复数的运算.【解析】231ii22(3)(1)(1 2)342iii

14、i.二、填空题：1（2010 年高考江苏卷试题2）设复数 z 满足 z(2-3i)=6+4i（其中 i 为虚数单位），则 z 的模为 _.【答案】2 解析 考查复数运算、模的性质。z(2-3i)=2(3+2 i),2-3i 与 3+2 i 的模相等，z 的模为 2。2（2010 年高考北京卷理科9）在复平面内，复数21ii对应的点的坐标为。【答案】（-1，1）【解析】因为21ii=2(1)12iii，故复数21ii对应的点的坐标为（-1，1）。3（2010 年高考上海市理科2）若复数12zi（为虚数单位），则z zz。【答案】6-2i【解析】因为12zi,所以1412z zzi6-2i.4.(

15、2010年高考重庆市理科11)已知复数1zi，则2zz_【答案】-2i 文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10

16、文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R

17、6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T

18、10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1

19、R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5

20、N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码

21、:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N

22、7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10第 4 页共 4 页解析：iiiii211112.5（2010 年上海市春季高考3)计算：21ii（为虚数单位）答案：1i解析：22(1)2211(1)(1)2iiiiiiii。文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10

23、ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B

24、10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E

25、4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC

26、2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T1

27、0文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10

28、R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10文档编码:CC2I2D5N2E4 HZ1R7B10N8T10 ZL10R6E6N7T10