Question

已知虚数z的虚部系数等于根号3/2,且满足lzl=1,求z/(z的平方+1)的值

Answer 1

设：z=a+(√3/2)i

|z|=√[a^2 + (√3/2)^2] =1

a=±1/2

z=1/2 + (√3/2)i 或 z=-1/2 + (√3/2)i

z / (z^2 + 1) =[1/2 + (√3/2)i] / {[1/2 + (√3/2)i]^2 + 1} = [1/2 + (√3/2)i] / [1/4 + (√3/2)i - 3/4 + 1]

=[1/2 + (√3/2)i] / [1/2 + (√3/2)i]=1

或 z / (z^2 + 1) = -1