1、3.2.3复数的除法1复数1+ii=()A.-1-iB.1-iC.-1+iD.-i解析:1+ii=(1+i)ii2=-i-i2=1-i.答案:B2复数3-i1+i2=()A.-3-4iB.-3+4iC.3-4iD.3+4i解析:3-i1+i2=(3-i)2(1+i)2=8-6i2i=8i+62i2=-3-4i.答案:A3复数z满足(z-i)(2-i)=5,则z=()A.-2-2iB.-2+2iC.2-2iD.2+2i解析:由题意可得,z-i=52-i=5(2+i)(2-i)(2+i)=2+i,故z=2+2i.答案:D4定义运算abcd=ad-bc,则符合条件1-1zzi=4+2i的复数z为()
2、A.3-iB.1+3iC.3+iD.1-3i解析:由定义1-1zzi=zi+z,得zi+z=4+2i,解得z=4+2i1+i=3-i.答案:A5若1+i1-in+1-i1+in=2,则n的值可能为()A.4B.5C.6D.7解析:1+i1-i=i,1-i1+i=-i,in+(-i)n=2,n=4k,0,n=4k+1,-2,n=4k+2,0,n=4k+3,kN+,n的值可能为4.答案:A6已知复数z满足(1+2i)z=4+3i,那么z=.解析:由(1+2i)z=4+3i,得z=4+3i1+2i=2-i,故z=2+i.答案:2+i7若x,yR,且x1-i-y1-2i=51-3i,则x=,y=.解析
3、: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,解得x=-1,y=-5.答案:-1-5 8已知复数z1=2+3i,z2=3+2i(2+i)2,则z1z2等于.解析:z1z2=(2-3i)(2+i)23+2i=(2-3i)(3-2i)(4+i2+4i)(3+2i)(3-2i)=(6+6i2-9i-4i)(3+4i)9+4=-13i(3+4i)13=-3i-4i2=4-3i.答案:4-
4、3i9已知f(z)=z2-z+1z2+z+1,求f(2+3i)与f(1-i)的值.解f(z)=z2-z+1z2+z+1,f(2+3i)=(2+3i)2-(2+3i)+1(2+3i)2+(2+3i)+1=-6+9i-2+15i=147229+72229i,f(1-i)=(1-i)2-(1-i)+1(1-i)2+(1-i)+1=-2i-1+i+1-2i+1-i+1=-i2-3i=313-213i. 10设zC,若|z|=1,且zi.(1)证明:z1+z2必是实数;(2)求z1+z2对应的点的轨迹.分析:设z=a+bi(a,bR),先将z1+z2进行化简再求解.(1)证明设z=a+bi(a,bR),则a2+b2=1(a0).z1+z2=a+bi1+(a+bi)2=a+bi1+a2-b2+2abi=a+bi2a2+2abi=2a3+2ab24a4+4a2b2=12aR.(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这个区间内的所有点的两条射线.