设n≥２，证明一切n次单位根的和等于零。
热心网友
设n≥２，x^n=1在复数范围内有n个根：x(k)=cos[2kπ/n]+isin[2kπ/n]，k=0,1,2,。。。,n-1cos0+cos[2π/n]+cos[4π/n]+。。。+cos[2(n-1)π/n]={cos0+cos[2π/n]+cos[4π/n]+。。。+cos[2(n-1)π/n]}sin[π/n]/sin[π/n]=(1/2){[sin(π/n)-sin(-π/n)]+[sin(3π/n)-sin(π/n)]+。。。+[sin(2n-1)π/n-sin(2n-3)π/n]}/sin(π/n)=(1/2){sin[(2n-1)π/n]-sin(-π/n)}/sin(π/n)=(1/2){sin[2π-π/n]-sin(-π/n)}/sin(π/n)=0sin0+sin[2π/n]+sin[4π/n]+。。。+sin[2(n-1)π/n]={sin0+sin[2π/n]+sin[4π/n]+。。。+sin[2(n-1)π/n]}sin[π/n]/sin[π/n]=(1/2){[cos(-π/n)-cos(π/n)]+[cos(π/n)-sin(3π/n)]+。。。+[cos(2n-3)π/n-cos(2n-1)π/n]}/sin(π/n)=(1/2){-cos[(2n-1)π/n]+cos(-π/n)}/sin(π/n)=(1/2){-cos[2π-π/n]+cos(-π/n)}/sin(π/n)=0∴x1+x2+。。。+x(n-1)={cos0+cos[2π/n]+cos[4π/n]+。。。+cos[2(n-1)π/n]}++i{sin0+sin[2π/n]+sin[4π/n]+。。。+sin[2(n-1)π/n]}=0证毕。