由|z|=1设z=cosθ+isinθ(θ∈[0,2π)
由z^5+z=1得cos5θ+cosθ+i(sin5θ+sinθ)=1
于是cos5θ+cosθ=1,sin5θ+sinθ=0
由sin5θ+sinθ=0得θ=0,π/3,2π/3,π,4π/3,5π/3或θ=π/4,3π/4,5π/4,7π/4
当θ=π/3,5π/3时cos5θ+cosθ=1成立
于是z=1/2+√3/2i或z=1/2-√3/2i