一、设 n 是自然数.求证:方程z~(n+1)-z~n-1=0有模为1的复根的充分必要条件是 n+2可被6整除.证必要性.设 w 是方程 z~(n+1)-z~n-1=0的一个模为1的复根.则 w~(n+1)-w~n-1=0.所以 w~n(w-1)=1,从而|w|~n|w-1|=1.因为|w|=1,由此得出|w-1|=1.既然|w|=1,故 w 在复平面的单位圆上.而单位圆上满足|w-1|=1的点 w 只能是 First, let n be a natural number. Prove that: The equation z~(n+1)-z~n-1=0 is a sufficient and necessary condition for a complex root with a modulus of 1, that is, n+2 can be divisible by 6. Necessity. w is a modulo 1 complex root of the equation z~(n+1)-z~n-1=0. Then w~(n+1)-w~n-1=0. So w~n(w -1)=1, so that |w|~n|w-1|=1. Since |w|=1, it results in |w-1|=1. Since |w|=1, then w is in The unit circle on the plane. The point w that satisfies |w-1|=1 on the unit circle can only be