=("-" 1/2+√3/2 i)(7-i)=(√3 "-" 7)/2+(7√3+1)/2i.

　　(2)("(" √2+√2 i")" ^2 "(" 4+5i")" )/("(" 5"-" 4i")(" 1"-" i")" )=(4i"(" 4+5i")" )/(5"-" 4"-" 9i)=("-" 20+16i)/(1"-" 9i)

　　=("-" 4"(" 5"-" 4i")(" 1+9i")" )/82

　　=("-" 4"(" 41+41i")" )/82=-2-2i.

10.设复数z满足4z+2¯z=3√3+i,w=sin θ-icos θ,求复数z及|z-w|的取值范围.

解设z=a+bi(a,b∈R),并将其代入条件中,得

　　4(a+bi)+2(a-bi)=3√3+i,即6a+2bi=3√3+i.

　　∴{■(6a=3√3 "," @2b=1"," )┤解得{■(a=√3/2 "," @b=1/2 "." )┤

　　∴z=√3/2+1/2i.

　　|z-w|=|(√3/2+1/2 i)"-(" sinθ"-" icosθ")" |

　　=|(√3/2 "-" sinθ)+(1/2+cosθ)i|

　　=√((√3/2 "-" sinθ)^2+(1/2+cosθ)^2 )

　　=√(2"-" √3 sinθ+cosθ)=√(2"-" 2sin(θ"-" π/6) ).

　　∵-1≤sin(θ"-" π/6)≤1,

　　∴0≤|z-w|≤2.

　　故所求的z=√3/2+1/2i,|z-w|的取值范围是[0,2].

B组

1.如果一个复数与它的模的和为5+√3i,那么这个复数是0(　　)

A.11/5 B.√3i C.11/5+√3i D.11/5+2√3i

解析:设z=x+yi(x,y∈R),则|z|=√(x^2+y^2 ).

　　∵x+yi+√(x^2+y^2 )=5+√3i,

　　∴{■(x+√(x^2+y^2 )=5"," @y=√3 "," )┤解得{■(x=11/5 "," @y=√3 "." )┤∴z=11/5+√3i.

答案:C

2.导学号88184057若z2+z+1=0,则z2 014+z2 015+z2 017+z2 018的值为(　　)

A.2 B.-2

C.-1/2+√3/2i D.-1/2±√3/2i