1、2.1复数的加法与减法2.2复数的乘法与除法课后训练案巩固提升一、A组1.复数2+i1-2i的虚部为()A.0B.1C.-1D.2解析:2+i1-2i=(2+i)(1+2i)(1-2i)(1+2i)=2+i+4i-25=i,虚部为1.答案:B2.若a,bR,设z1=2+bi,z2=a+i,当z1+z2=0时,复数a+bi为()A.1+iB.2+iC.3D.-2-i解析:z1+z2=(2+bi)+(a+i)=(2+a)+(b+1)i=0,2+a=0,b+1=0.a=-2,b=-1.a+bi=-2-i.答案:D3.设f(z)=|z|,z1=3+4i,z2=-2-i,则f(z1-z2)=()A.10
2、B.55C.2D.52解析:z1-z2=5+5i,f(z1-z2)=f(5+5i)=|5+5i|=52.答案:D4.设复数z=a+bi(a,bR),若z1+i=2-i成立,则点P(a,b)在()A.第一象限B.第二象限C.第三象限D.第四象限解析:z1+i=2-i,z=(2-i)(1+i)=3+i.a=3,b=1.点P(a,b)在第一象限.答案:A5.复数z的共轭复数为z,i为虚数单位,且z+2iz=5+4i,则z=()A.1+2iB.1-2iC.-1+2iD.-1-2i解析:设z=a+bi,则z=a-bi,a+bi+2i(a-bi)=5+4i,即(a+2b)+(b+2a)i=5+4i,a+2
3、b=5,b+2a=4,解得a=1,b=2.z=1+2i.答案:A6.已知z1=32a+(a+1)i,z2=-33b+(b+2)i(a,bR),若z1-z2=43,则a+b=.解析:z1-z2=32a+(a+1)i-33b+(b+2)i=32a+33b+(a-b-1)i=43,32a+33b=43,a-b-1=0.a=2,b=1.a+b=3.答案:37.已知复数z=3+i(1-3i)2,z是z的共轭复数,则zz=.解析:z=3+i(1-3i)2=3+i-2-23i=3+i-2(1+3i)=(3+i)(1-3i)-2(1+3i)(1-3i)=23-2i-8=-34+14i,z=-34-14i.zz
4、=-34+14i-34-14i=316+116=14.答案:148.若x,yR,且x1-i-y1-2i=51-3i,则x=,y=.解析:x1-i-y1-2i=51-3i,x(1-2i)-y(1-i)(1-i)(1-2i)=5(1+3i)(1-3i)(1+3i).(x-y)+(y-2x)i-1-3i=1+3i2.(x-y)+(y-2x)i=-(1+3i)22=4-3i.x-y=4,y-2x=-3.x=-1,y=-5.答案:-1-59.计算:(1)-12+32i(2-i)(3+i);(2)(2+2i)2(4+5i)(5-4i)(1-i).解:(1)-12+32i(2-i)(3+i)=-12+32i
5、(7-i)=3-72+73+12i.(2)(2+2i)2(4+5i)(5-4i)(1-i)=4i(4+5i)5-4-9i=-20+16i1-9i=-4(5-4i)(1+9i)82=-4(41+41i)82=-2-2i.10.导学号18334049设复数z满足4z+2z=33+i,w=sin -icos ,求复数z及|z-w|的取值范围.解:设z=a+bi(a,bR),并将其代入条件中,得4(a+bi)+2(a-bi)=33+i,即6a+2bi=33+i.6a=33,2b=1,解得a=32,b=12.z=32+12i.|z-w|=32+12i-(sin-icos)=32-sin+12+cosi=
6、32-sin2+12+cos2=2-3sin+cos=2-2sin-6.-1sin-61,0|z-w|2.故所求的z=32+12i,|z-w|的取值范围是0,2.二、B组1.如果一个复数与它的模的和为5+3i,那么这个复数是()A.115B.3iC.115+3iD.115+23i解析:设z=x+yi(x,yR),则|z|=x2+y2.x+yi+x2+y2=5+3i,x+x2+y2=5,y=3,解得x=115,y=3.z=115+3i.答案:C2.导学号18334050若z2+z+1=0,则z2 014+z2 015+z2 017+z2 018的值为()A.2B.-2C.-12+32iD.-12
7、32i解析:z2+z+1=0,两边同乘(z-1),得z3-1=0,z3=1(z1),则z4=z,z2 014=(z3)671z=z,于是原式=z2 014(1+z+z3+z4)=z(1+z+1+z)=z(2+2z)=2(z+z2)=-2.答案:B3.已知复数z=x+yi(x,yR),且|z-2|=3,则yx的最大值为.解析:由|z-2|=3,知点z的轨迹是以点(2,0)为圆心,半径为3的圆,yx表示圆上的点与原点连线的斜率,结合图形易知,当直线与圆相切时取最值.答案:34.设zC,若|z|=1,且zi.(1)证明z1+z2必是实数;(2)求z1+z2对应点的轨迹.(1)证明:设z=a+bi(a
8、,bR),则a2+b2=1(a0).于是z1+z2=a+bi1+(a+bi)2=a+bi1+a2-b2+2abi=a+bi2a2+2abi=2a3+2ab24a4+4a2b2=12aR.因此z1+z2必是实数.(2)解:由(1)知zz2+1=12a(a0).a2+b2=1,-1a0或0a1.12a-12或12a12,即zz2+1对应的点的轨迹是x轴上除去-12,12这个区间的所有点的两条射线.5.已知z为虚数,z+9z-2为实数,若z-2为纯虚数,求虚数z及|z|.解:z为虚数且z-2为纯虚数,可设z=2+bi(bR,b0).又z+9z-2=2+bi+9bi=2+bi-9bi=2+b-9bi为实数,b-9b=0,b=3.z=23i.故|z|=13.
