第四章平面向量、数系的扩充与复数的引入4.4数系的扩充与复数的引入真题演练文1.(2015·广东卷)若复数z=i(3-2i)(i是虚数单位),则=()A.2-3iB.2+3iC.3+2iD.3-2i解析:i(3-2i)=3i-2i2=2+3i,所以z=2+3i,所以=2-3i,故选A.答案:A2.(2013·江苏卷)设z=(2-i)2(i为虚数单位),则复数z的模为________.解析:∵z=(2-i)2=3-4i,∴|z|==5.答案:53.(2011·江苏卷)设复数z满足i(z+1)=-3+2i(i为虚数单位),则z的实部是________.解析:方法1:∵i(z+1)=-3+2i.∴z=-1=-(-3i-2)-1=1+3i,故z的实部是1.方法2:令z=a+bi(a,b∈R),由i(z+1)=-3+2i得i[(a+1)+bi]=-3+2i,-b+(a+1)i=-3+2i,∴b=3,a=1,故z的实部是1.答案:14.(2015·北京卷)复数i(2-i)=()A.1+2iB.1-2iC.-1+2iD.-1-2i解析:i(2-i)=2i-i2=1+2i,故选A.答案:A5.(2015·湖南卷)已知=1+i(i为虚数单位),则复数z=()A.1+iB.1-iC.-1+iD.-1-i解析:z====-1-i.答案:D6.(2011·上海卷)已知复数z1满足(z1-2)(1+i)=1-i(i为虚数单位),复数z2的虚部为2,且z1·z2是实数,求z2.解:∵(z1-2)(1+i)=1-i,∴z1=2-i.设z2=a+2i,a∈R,z1·z2=(2-i)(a+2i)=(2a+2)+(4-a)i.∵z1·z2∈R,∴a=4,∴z2=4+2i.