数系的扩充与复数的引入单元测
试题
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姓名:___________班级:____________一、选择题(本题含有12个小题,每小题5分,共60分)1.是复数为纯虚数的()A.充分条件B.必要条件C.充要条件D.非充分非必要条件2.设,则在复平面内对应的点位于()A.第一象限B.第二象限C.第三象限D.第四象限3.()A.B.C.D.4.复数z满足,那么=()A.2+iB.2-iC.1+2iD.1-2i如果复数的实部与虚部互为相反数,那么实数b等于()A.EQ\R(2)B.EQ\F(2,3)C.2D.-EQ\F(2,3)6.集合{Z︱Z=},用列举法
表
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示该集合,这个集合是()A.{0,2,-2}B.{0,2}C.{0,2,-2,2}D.{0,2,-2,2,-2}7.设O是原点,向量对应的复数分别为,那么向量对应的复数是()复数,则在复平面内的点位于第()象限。A.一B.二C.三D.四9.复数不是纯虚数,则有()设i为虚数单位,则的值为()A.4B.-4C.4iD.-4i对于两个复数,,有下列四个结论:①;②;③;④,其中正确的结论的个数为()A.1B.2C.3D.412.1,,EMBEDEquation.KSEE3\*MERGEFORMAT是某等比数列的连续三项,则的值分别为()A.B.C.D.二、填空题(本题含有4个小题,每小题5分,共20分)13.设(为虚数单位),则z=;|z|=.14.复数的实部为,虚部为15.已知复数z与(z+2)2-8i均是纯虚数,则z=16.设,,复数和在复平面内对应点分别为A、B,O为原点,则的面积为____三.解答题(本题含有6个小题,其中第17题10分,其余每小题12分,共70分)17.18.已知复数z=(2+)EMBEDEquation.3EMBEDEquation.DSMT4).当实数m取什么值时,复数z是:(1)零;(2)虚数;(3)纯虚数;(4)复平面内第二、四象限角平分线上的点对应的复数。19.已知关于的方程组有实根,求的值。20.已知若,求的值。21.复数,求实数使。(其中为的共轭复数)若,求的最大值和最小值.参考答案1.解析:B2.解析:D点拨:。3.解析:B点拨:原式=EMBEDEquation.DSMT4=4.解析:B点拨:化简得EMBEDEquation.DSMT45.解析:D点拨:EMBEDEquation.DSMT4EMBEDEquation.DSMT4,由因为实部与虚部互为相反数,即,解得。6.解析:A点拨:根据成周期性变化可知。7.解析:B点拨:EMBEDEquation.DSMT4EMBEDEquation.DSMT48.解析:D点拨:EMBEDEquation.DSMT49.解析:D10.解析:B点拨:=-411.解析:B12.解析:C13.解析:,点拨:EMBEDEquation.DSMT4EMBEDEquation.DSMT414.解析:1,点拨:EMBEDEquation.DSMT4EMBEDEquation.DSMT415.解析:点拨:设代入解得,故16.解析:1点拨:17.解:EMBEDEquation.DSMT4EMBEDEquation.DSMT4.19.解:由第一个等式得,将上述结果代入第二个等式中得:21.解析:由,可知,代入得:EMBEDEquation.DSMT4,即EMBEDEquation.DSMT4EMBEDEquation.DSMT4则,解得或.22.解:(可利用复数运算几何意义化归为几何问题)而|z|则表示该圆上的点到原点O的距离..由平面几何知识可知,使圆上的点到原点距离取最大(最小)值的点在直线OC与圆的交点处。本章知识点归纳
总结
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