答案:3/4+i

8.已知复数z满足|z|=1+3i-z,化简("(" 1+i")" ^2 "(" 3+4i")" ^2)/2z.

解设z=a+bi(a,b∈R),∵|z|=1+3i-z,

　　∴√(a^2+b^2 )-1-3i+a+bi=0.

　　∴{■(√(a^2+b^2 )+a"-" 1=0"," @b"-" 3=0"," )┤解得{■(a="-" 4"," @b=3"," )┤∴z=-4+3i,

　　∴("(" 1+i")" ^2 "(" 3+4i")" ^2)/2z=(2i"(-" 7+24i")" )/(2"(-" 4+3i")" )=(24+7i)/(4"-" 3i)=3+4i.

9.已知复数z的实部为正数,|z|=√2,z2的虚部为2.

(1)求复数z;

(2)设z,z2,z-z2在复平面内对应的点分别为A,B,C,求△ABC的面积.

解(1)设z=a+bi(a,b∈R),则由条件|z|=√2可得a2+b2=2　①.

　　因为z2=a2-b2+2abi,所以其虚部为2ab=2　②.

　　联立①②,解得a=b=1或a=b=-1.

　　又复数z的实部为正数,所以a>0,所以a=b=1,于是z=1+i.

　　(2)由(1)可知z=1+i,则z2=2i,z-z2=1-i,所以A(1,1),B(0,2),C(1,-1),由此可得S△ABC=1,所以△ABC的面积为1.

10.导学号88184072设O为坐标原点,已知向量(OZ_1 ) ⃗,(OZ_2 ) ⃗分别对应复数z1,z2,且z1=3/(a+5)+(10-a2)i,z2=2/(1"-" a)+(2a-5)i,a∈R.若¯(z_1 )+z2可以与任意实数比较大小,求(OZ_1 ) ⃗·(OZ_2 ) ⃗的值.

解由题意,得¯(z_1 )=3/(a+5)-(10-a2)i,

　　则¯(z_1 )+z2=3/(a+5)-(10-a2)i+2/(1"-" a)+(2a-5)i=(3/(a+5)+2/(1"-" a))+(a2+2a-15)i.

　　∵¯(z_1 )+z2可以与任意实数比较大小,

　　∴¯(z_1 )+z2是实数,

　　∴a2+2a-15=0,解得a=-5或a=3,

　　又a+5≠0,∴a=3,∴z1=3/8+i,z2=-1+i.

　　∴(OZ_1 ) ⃗=(3/8 "," 1),(OZ_2 ) ⃗=(-1,1),

　　∴(OZ_1 ) ⃗·(OZ_2 ) ⃗=3/8×(-1)+1×1=5/8.

B组

1.设z1,z2是复数,则下列命题中的假命题是(　　)

A.若|z1-z2|=0,则¯(z_1 )=¯(z_2 )

B.若z1=¯(z_2 ),则¯(z_1 )=z2

C.若|z1|=|z2|,则z1·¯(z_1 )=z2·¯(z_2 )

D.若|z1|=|z2|,则z_1^2=z_2^2

解析:设z1=a+bi,z2=c+di,a,b,c,d∈R.

　　若|z1-z2|=0,则z1-z2=(a-c)+(b-d)i=0,所以a=c,b=d,所以¯(z_1 )=¯(z_2 ),所以A正确;

　　若z1=¯(z_2 ),则a=c,b=-d,所以¯(z_1 )=z2,故B正确;

若|z1|=|z2|,则a2+b2=c2+d2,所以z1·¯(z_1 )=z2·¯(z_2 ),故C正确;