设点z是单位圆x^2+y^2=1上的动点,复数w是复数z的函数w=1/(1+z) ^2试求w的轨

设z=cosθ+isinθ,则
w=1/(1+cosθ+isinθ)^2
=1/{2cos(θ/2)[cos(θ/2)+isin(θ/2)]}^2
=1/{4[cos(θ/2)]^2*(cosθ+isinθ)}
=(cosθ-isinθ)/(2+2cosθ)=x+yi,
∴x=cosθ/(2+2cosθ),①
y=-sinθ/(2+2cosθ),②
由①,2x+2xcosθ=cosθ,
∴cosθ=2x/(1-2x),
代入②^2,y^2=(1-cosθ)/[4(1+cosθ)]=(1-4x)/4=-(x-1/4),
∴所求轨迹是抛物线y^2=-(x-1/4).