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li826236:2019高中数学人教A版选修2-2习题:第三章数系的扩充与复数的引入 3.2.1

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3.2 3.2.1 复数代数形式的四则运算 课时过关· 能力提升 基础巩固 复数代数形式的加、减运算及其几何意义 1(6-2i)-(3i+1)等于( ) A.3-3i C.7+i B.5-5i D.5+5i 解析(6-2i)-(3i+1)=(6-1)+(-2-3)i=5-5i. 故选 B. 答案 B 2 若复数 z 满足 z+i-3=3-i,则 z 等于( A.0 解析∵z+i-3=3-i, B.2i C.6 ) D.6-2i ∴z=(3-i)-(i-3)=(3+3)+(-i-i)=6-2i, 故选 D. 答案 D 3 在复平面内,已知点 A 对应的复数为 2+3i,向量 A.1+5i C.-3-i 解析因为 答案 B 4 若 z1=2+i,z2=3+ai(a∈R),且 z1+z2 所对应的点在实轴上,则 a 的值为( A.3 B.2 C.1 D.-1 ) B.3+i D.1+i ,所以 对应的复数为(2+3i)-(-1+2i)=(2+1)+(3-2)i=3+i.故选 B. 对应的复数为-1+2i,则向量 对应的复数为( ) 解析 z1+z2=2+i+3+ai=(2+3)+(1+a)i=5+(1+a)i.∵z1+z2 所对应的点在实轴上, ∴1+a=0.∴a=-1. 答案 D 5 若在复平面内的?ABCD 中, A.2+14i B.1+7i 对应复数 6+8i, 对应复数-4+6i,则 对应的复数是( ) 1 C.2-14i 解析设 则有 故 答案 D D.-1-7i 对应的复数分别为 z1 与 z2, 得 2z2=2+14i,z2=1+7i, 对应的复数是-1-7i. 6 已知复数 z1=3+2i,z2=1-3i,则复数 z=z1-z2 在复平面内对应的点 Z 位于复平面内的第 答案一 7 已知 z1=m2-3m+m2i,z2=4+(5m+6)i(m∈R).若 z1-z2=0,则 m= 解析∵z1-z2=(m2-3m+m2i)-[4+(5m+6)i] =(m2-3m-4)+(m2-5m-6)i=0, . 象限. ∴ 答案-1 - - ∴m=-1. 8 已知 z1= a+(a+1)i,z2=-3 b+(b+2)i(a,b∈R).若 z1-z2=4 ,则 a+b= . 解析 z1-z2= a+(a+1)i-[-3 b+(b+2)i] = +(a-b-1)i=4 , 由复数相等的条件,知 - 解得 答案 3 9 若|z-1|=1,试说明复数 z 对应点的轨迹. 分析解答本题可根据复数的减法和模的几何意义求解. 解根据复数的减法和模的几何意义,知|z-1|=1 表示复数 z 对应的点到点(1,0)的距离为 1, 所以复数 z 对应的点的轨迹是以点(1,0)为圆心,以 1 为半径的圆. 故 a+b=3. 能力提升 1 已知复数 z1= i,复数 z2=cos 60°+isin 60°,则 z1+z2 等于( ) A.1 B.-1 2 C. 答案 A i D. i 2 已知 z1=3-4i,z2=-5+2i,z1,z2 对应的点分别为 P1,P2,则 A.-8+6i C.8+6i B.8-6i D.-2-2i 对应的复数为( ) 解析由复数减法的几何意义知: 答案 B 对应的复数为 z1-z2=3-4i-(-5+2i)=(3+5)+(-4-2)i=8-6i,故选 B. 3 已知 A,B 分别是复数 z1,z2 在复平面内对应的点,O 是坐标原点.若|z1+z2|=|z1-z2|,则△AOB 一定是( A.等腰三角形 C.等边三角形 B.直角三角形 D.等腰直角三角形 ) 解析因为|z1+z2|=|z1-z2|,所以由复数加减运算的几何意义知,以 OA,OB 为邻边的平行四边形是矩形,故△AOB 是直角三 角形. 答案 B ★ A. C.5 4 已知 z∈C,|z-2|=1,则|z+2+5i|的最大值和最小值分别是( +1 和 和 -1 B.3 和 1 D. 和3 ) 解析由|z-2|=1 知 z 对应的点在以(2,0)为圆心,半径为 1 的圆上,而|z+2+5i|=|z-(-2-5i)|表示 z 对应的点到点(-2,-5)的距离. 而圆心(2,0)与(-2,-5)间的距离为 答案 A ★ 5 已知|z1|=1,|z2|=1,|z1+z2|= ,则|z1-z2|= . ,则向量 对应 z1+z2, 对应 z1-z2. ,故最大值为 +1,最小值为 -1 . 解析在平面直角坐标系内以原点 O 为起点作出 z1,z2 对应的向量 由题意知| |=1,| |=1,| |= ,可得∠OZ1Z=120°, 所以∠Z2OZ1=60°,即△Z2OZ1 是等边三角形. 所以在△Z2OZ1 中,| 答案 1 6 已知集合 A={z1||z1+1|≤1,z1∈C},B={z2|z2=z1+i+m,z1∈A,m∈R}. (1)当 A∩B=?时,求实数 m 的取值范围; (2)是否存在实数 m,使得 A∩B=A? 解因为|z1+1|≤1,所以 z1 所对应的点构成的集合 A 是以(-1,0)为圆心,以 1 为半径的圆面(圆周及其内部).又 z2=z1+i+m, 所以 z1=z2-i-m. 所以|z2-i-m+1|≤1,即|z2-[(m-1)+i]|≤1. 所以 z2 所对应的点的集合 B 是以点(m-1,1)为圆心,1 为半径的圆面(圆周及其内部). |=1,即|z1-z2|=1. 3 (1)若 A∩B=?,说明上述两圆外离,其圆心距 d= m<}. - >2,解得 m 的取值范围是{m|m∈R,且 m> 或 (2)若 A∩B=A,因为两圆半径相等,所以两圆重合,但由圆心的坐标(-1,0)及(m-1,1)可知它们不可能重合,所以不存在 实数 m,使 A∩B=A. ★ 7 在复平面内,复数 z1 对应的点在连接 1+i 和 1

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