2.1复数的加法与减法课时过关·能力提升1.(5-i)-(3-i)-5i等于()A.5iB.2-5iC.2+5iD.2答案:B2.设f(z)=z,z1=4-3i,z2=1+2i,则f(z1+z2)=()A.1-3iB.-2+11iC.-2+iD.5-i解析:f(z1+z2)=z1+z2=(4-3i)+(1+2i)=4+1+(-3+2)i=5-i.答案:D3.若z=3-4i,则复数z-|z|+(1-i)在复平面内的对应点在()A.第一象限B.第二象限C.第三象限D.第四象限解析:∵z=3-4i,∴z-|z|+(1-i)=3-4i−√32+(-4)2+1−i=(3-5+1)+(-4-1)i=-1-5i.答案:C4.已知z=11-20i,则1-2i-z等于()A.z-1B.z+1C.-10+18iD.10-18i1解析:∵z=11-20i,∴1-2i-z=1-2i-(11-20i)=(1-11)+(-2+20)i=-10+18i.答案:C5.实数x,y满足(1+i)x+(1-i)y=2,则xy的值是()A.1B.2C.-2D.-1解析:原式变形,得(x+y)+(x-y)i=2,由实部和虚部对应相等,得{x+y=2,x-y=0,解得{x=1,y=1,即xy=1.答案:A6.设m∈R,复数z=(2m2+3i)+(m-m2i)+(-1+2mi),若z为纯虚数,则m等于()A.-1B.3C.12D.1或2解析:∵z=(2m2+m-1)+(3-m2+2m)i为纯虚数,∴{2m2+m-1=0,3-m2+2m≠0,解得m¿12.答案:C7.若|z-1|¿√5,且z−2i为实数,则z=¿¿解析:∵z-2i为实数,∴设z-2i=x(x∈R),则z=x+2i,代入|z-1|¿√5,得|x-1+2i|¿√5,即(x-1)2+4=5,∴x=0或x=2.∴z=2i或z=2+2i.答案:2i或2+2i8.计算(-i2+2i)+(i+|i|)-(1+2i)=.解析:原式=1+2i+i+1-1-2i=1+i.2答案:1+i9.★方程|z|=1+3i-z,复数z=.解析:由|z|=1+3i-z,得z的虚部是3.设z=m+3i(m∈R),代入方程,得√m2+9=1+3i−m−3i,所以√m2+9=1−m,所以m=-4,所以z=-4+3i.答案:-4+3i10.已知z1=(3x+y)+(y-4x)i,z2=(4y-2x)-(5x+3y)i(x,y∈R).若z1-z2=13-2i.求z1,z2.解:z1-z2=(3x+y)+(y-4x)i-[(4y-2x)-(5x+3y)i]=[(3x+y)-(4y-2x)]+[(y-4x)+(5x+3y)]i=(5x-3y)+(x+4y)i,∵z1-z2=13-2i,∴(5x-3y)+(x+4y)i=13-2i,∴{5x-3y=13,x+4y=-2,解得{x=2,y=-1.∴z1=(3×2-1)+(-1-4×2)i=5-9i,z2=[4×(-1)-2×2]-[5×2+3×(-1)]i=-8-7i.11.已知复数z1=cosα+isinα,z2=cosβ+isinβ,|z1-z2|¿2√55,cos求(α−β)的值.解:因为z1-z2=cosα-cosβ+(sinα-sinβ)i,3且|z1-z2|¿2√55,所以√(cosα-cosβ)2+(sinα-sinβ)2=2√55,即√2-2cos(α-β)=2√55,解得cos(α-β)¿35.12.★已知|z|=2,试求z+3-4i对应的点的轨迹.解法一设ω=z+3-4i,则z=ω-3+4i,因为|z|=2,所以|ω-3+4i|=2.故ω的对应点的轨迹为以点(3,-4)为圆心,2为半径的圆.解法二设ω=z+3-4i,ω=x+yi(x,y∈R),z=a+bi(a,b∈R),所以x+yi=a+3+(b-4)i,所以{x=a+3,y=b-4,即{a=x-3,b=y+4.由题意,知a2+b2=4,所以(x-3)2+(y+4)2=4.故ω的对应点的轨迹为以点(3,-4)为圆心,2为半径的圆.45
