z^n-1=0                (1)

(1)有n個解，都在單位圓上。角度是0(2π/n),1(2π/n),...,(n-1)(2π/n)

寫成指數形式是

zk=e^(ik(2π/n))。k=0,1,2,...(n-1)

z^n - 1 = (z-z0)(z-z1)(z-z2)...(z-z(n-1))   (2)

1+z+z^2+...+z^(n-1) = (z-z1)(z-z2)...(z-z(n-1))

令z=1，

n = (1-z1)(1-z2)...(1-z(n-1))    (3)

1-z^k= 1-(cos(k(2π/n))+i sin(k(2π/n)))=

1+2sin^2(k(π/n))-1-i 2sin(k(π/n))cos(k(π/n))=

2sin(k(π/n))(sin(k(π/n))-icos(k(π/n))

|1-z^k|= 2sin(k(π/n))

代入(3)得

n = 2^(n-1)sin(π/n)*sin(2π/n)*sin(3π/n)*...*sin[(n-1)π]/n)