解析∵a=("-" 3"-" i)/(1+2i)=("(-" 3"-" i")(" 1"-" 2i")" )/5=-1+i,

　　∴a4=[(-1+i)2]2=(-2i)2=-4.

答案-4

7已知复数 =a+3i(i为虚数单位,a>0),若 2是纯虚数,则a=　　　　　.

解析 2=(a+3i)2=a2+6ai+9i2=(a2-9)+6ai,

　　由题易知{■(a^2 "-" 9=0"," @6a≠0"," @a>0"," )┤解得a=3.

答案3

8若复数 满足 (2-3i)=6+4i(i为虚数单位),则 的模为　　　　　.

解析由 (2-3i)=6+4i,得 =(6+4i)/(2"-" 3i)=("(" 6+4i")(" 2+3i")" )/("(" 2"-" 3i")(" 2+3i")" )=2i,所以| |=2.

答案2

9设复数 满足| |=1,且(3+4i)· 是纯虚数,求¯z.

分析利用复数问题实数化的思想,设 =a+bi(a,b∈R),利用已知条件建立关于a,b的方程组,求解即可.

解设 =a+bi(a,b∈R),由| |=1,得√(a^2+b^2 )=1.

　　由题意,得(3+4i)· =(3+4i)(a+bi)=3a-4b+(4a+3b)i是纯虚数,则{■(3a"-" 4b=0"," @4a+3b≠0"." )┤

　　由{■(√(a^2+b^2 )=1"," @3a"-" 4b=0"," @4a+3b≠0"," )┤

　　解得{■(a=4/5 "," @b=3/5)┤或{■(a="-" 4/5 "," @b="-" 3/5 "." )┤

　　所以 =4/5+3/5i或 =-4/5-3/5i.

　　故¯z=4/5-3/5i或¯z=-4/5+3/5i.

能力提升

1若复数 满足( -i)(2-i)=5,则 等于(　　)

A.-2-2i B.-2+2i

C.2-2i D.2+2i

解析由题意可得, -i=5/(2"-" i)=(5"(" 2+i")" )/("(" 2"-" i")(" 2+i")" )=2+i,

　　所以 =2+2i.

答案D

2("(" 1+i")" ^3)/("(" 1"-" i")" ^2 )=(　　)