9已知f(z)=(z^2 "-" z+1)/(z^2+z+1),求f(2+3i)与f(1-i)的值.
解∵f(z)=(z^2 "-" z+1)/(z^2+z+1),
∴f(2+3i)=("(" 2+3i")" ^2 "-(" 2+3i")" +1)/("(" 2+3i")" ^2+"(" 2+3i")" +1)
=("-" 6+9i)/("-" 2+15i)=147/229+72/229 i,
f(1-i)=("(" 1"-" i")" ^2 "-(" 1"-" i")" +1)/("(" 1"-" i")" ^2+"(" 1"-" i")" +1)=("-" 2i"-" 1+i+1)/("-" 2i+1"-" i+1)
=("-" i)/(2"-" 3i)=3/13-2/13 i.
10设z∈C,若|z|=1,且z≠±i.
(1)证明:z/(1+z^2 ) 必是实数;
(2)求 z/(1+z^2 ) 对应的点的轨迹.
分析:设z=a+bi(a,b∈R),先将 z/(1+z^2 ) 进行化简再求解.
(1)证明设z=a+bi(a,b∈R),则a2+b2=1(a≠0).
∴z/(1+z^2 )=(a+bi)/(1+"(" a+bi")" ^2 )=(a+bi)/(1+a^2 "-" b^2+2abi)
=(a+bi)/(2a^2+2abi)=(2a^3+2ab^2)/(4a^4+4a^2 b^2 )=1/2a∈R.
(2)解由(1),知 z/(z^2+1)=1/2a(a≠0).
∵a2+b2=1,
∴-1≤a<0或0 ∴1/2a≤-1/2 或 1/2a≥1/2, 即 z/(z^2+1) 对应的点的轨迹是x轴上除去("-" 1/2 "," 1/2)这个区间内的所有点的两条射线.