∴{■(a+2b=5"," @b+2a=4"," )┤解得{■(a=1"," @b=2"." )┤∴z=1+2i.

答案:A

6.已知z1=√3/2a+(a+1)i,z2=-3√3b+(b+2)i(a,b∈R),若z1-z2=4√3,则a+b=　　　　.

解析:∵z1-z2=√3/2a+(a+1)i-[-3√3b+(b+2)i]=√3/2a+3√3b+(a-b-1)i=4√3,

　　∴{■(√3/2 a+3√3 b=4√3 "," @a"-" b"-" 1=0"." )┤∴{■(a=2"," @b=1"." )┤∴a+b=3.

答案:3

7.已知复数z=(√3+i)/("(" 1"-" √3 i")" ^2 ),¯z是z的共轭复数,则z·¯z=　　　　　.

解析:∵z=(√3+i)/("(" 1"-" √3 i")" ^2 )=(√3+i)/("-" 2"-" 2√3 i)=(√3+i)/("-" 2"(" 1+√3 i")" )

　　=("(" √3+i")(" 1"-" √3 i")" )/("-" 2"(" 1+√3 i")(" 1"-" √3 i")" )=(2√3 "-" 2i)/("-" 8)=-√3/4+1/4i,

　　∴¯z=-√3/4-1/4i.

　　∴z·¯z=("-" √3/4+1/4 i)("-" √3/4 "-" 1/4 i)

　　=3/16+1/16=1/4.

答案:1/4

8.若x,y∈R,且x/(1"-" i)-y/(1"-" 2i)=5/(1"-" 3i),则x=　　　　,y=　　　　.

解析:∵x/(1"-" i)-y/(1"-" 2i)=5/(1"-" 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+3i)/2.

　　∴(x-y)+(y-2x)i=("-(" 1+3i")" ^2)/2=4-3i.

　　∴{■(x"-" y=4"," @y"-" 2x="-" 3"." )┤∴{■(x="-" 1"," @y="-" 5"." )┤

答案:-1　-5

9.计算:(1)("-" 1/2+√3/2 i)(2-i)(3+i);

(2)("(" √2+√2 i")" ^2 "(" 4+5i")" )/("(" 5"-" 4i")(" 1"-" i")" ).

解(1)("-" 1/2+√3/2 i)(2-i)(3+i)