1、7.1.1数系的扩充和复数的概念课后训练巩固提升1.以3i-2的虚部为实部,以3i2+2i的实部为虚部的复数是()A.3-3iB.3+iC.-2+2iD.2+2i解析:3i-2的虚部为3,3i2+2i=-3+2i的实部为-3,故选A.答案:A2.若z=(x2-1)+(x-1)i为纯虚数,则实数x的值为()A.-1B.0C.1D.-1或1解析:z为纯虚数,x2-1=0,x-10,解得x=-1.答案:A3.已知复数z=a2-(2-b)i的实部和虚部分别是2和3,则实数a,b的值分别是()A.2,1B.2,5C.2,5D.2,1解析:由题意得a2=2,-(2-b)=3,a=2,b=5.故选C.答案:
2、C4.下列复数中,满足方程x2+2=0的是()A.1B.iC.2iD.2i解析:因为x2=-2=2i2,所以x=2i.答案:C5.若复数cos +isin和sin +icos相等,则的值为()A.4B.4或54C.2k+4(kZ)D.k+4(kZ)解析:由复数相等可得cos=sin,sin=cos,tan=1,=k+4(kZ),故选D.答案:D6.若复数(m2-3m-4)+(m2-5m-6)i是虚数,则实数m满足.解析:m2-3m-4+(m2-5m-6)i是虚数,m2-5m-60,m-1且m6.答案:m-1且m67.若a-2i=bi+1(a,bR),则b+ai=.解析:根据复数相等可得a=1,
3、b=-2,b+ai=-2+i.答案:-2+i8.若x-2+(y-1)i0(x,yR),则x的取值范围是.解析:由题意知y-1=0,x-20,y=1,x2.答案:(2,+)9.设复数z=log2(m2-3m-3)+ilog2(3-m),mR,若复数z是纯虚数,求m的值.解:由题意得m2-3m-30,3-m0,log2(m2-3m-3)=0,log2(3-m)0,解得m=-1.10.已知M=1,(m2-2m)+(m2+m-2)i,P=-1,1,4i,若MP=P,求实数m的值.解:MP=P,MP,即(m2-2m)+(m2+m-2)i=-1或(m2-2m)+(m2+m-2)i=4i.由(m2-2m)+(m2+m-2)i=-1,得m2-2m=-1,m2+m-2=0,解得m=1;由(m2-2m)+(m2+m-2)i=4i,得m2-2m=0,m2+m-2=4,解得m=2.综上可知,m=1或m=2.
