Let equation xn+an−1xn−1+an−2xn−2+⋯+a1x+a0=0 with real coefficients satisfy 0<a0≤a1≤a2≤⋯≤an−1≤1. Suppose that λ (|λ|>1) is a complex root of the equation, prove that λn+1=1.
λn=−(an−1λn−1+an−2λn−2+⋯+a0)λn+1=−(an−1λn+an−2λn−1+⋯+a0λ)λn+1−λn=−an−1λn+(an−1−an−2)λn−1+⋯+(a1−a0)λ+a0⟹|λ|n+1=|(1−an−1)λn+(an−1−an−2)λn−1+⋯+(a1−a0)λ+a0|≤(1−an−1)|λ|n+(an−1−an−2)|λ|n−1+⋯+(a1−a0)|λ|+a0≤(1−an−1)|λ|n+(an−1−an−2)|λ|n+⋯+(a1−a0)|λ|n+a0|λ|n=|λ|n⟹|λ|≤1⟹|λ|=1. Thus, the triangle inequality holds and a0=a1=⋯=an−1=1⟹λn+1=1