解析：∵(1+i)/(1"-" i)=i,(1"-" i)/(1+i)=-i,

　　∴in+(-i)n={■(2",　" n=4k"," @0",　" n=4k+1"," @"-" 2",　" n=4k+2"," @0",　" n=4k+3"," )┤k∈N+,

　　∴n的值可能为4.

答案：A

6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+3i)/(1+2i)=2-i,故z=2+i.

答案：2+i

7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"," )┤解得{■(x="-" 1"," @y="-" 5"." )┤

答案：-1　-5

8已知复数¯(z_1 )=2+3i,z2=(3+2i)/("(" 2+i")" ^2 ),则 z_1/z_2 等于　　　　.

解析：z_1/z_2 =("(" 2"-" 3i")(" 2+i")" ^2)/(3+2i)

　　=("(" 2"-" 3i")(" 3"-" 2i")(" 4+i^2+4i")" )/("(" 3+2i")(" 3"-" 2i")" )

　　=("(" 6+6i^2 "-" 9i"-" 4i")(" 3+4i")" )/(9+4)

　　=("-" 13i"(" 3+4i")" )/13=-3i-4i2=4-3i.

答案：4-3i