X4+1因式分解

令 x⁴= -1 = -1 + i×0 = cos(2k+1)π + isin(2k+1)π = e^[i(2k+1)π]
∴ x = e^[¼i(2k+1)π]
k=0,x.= e^[¼iπ] = cos(¼π) + isin(¼π) = √2/2 + i√2/2
k=1,x₁= e^[¾iπ] = cos(¾π) + isin(¾π) = -√2/2 + i√2/2
k=2,x₂= e^[5iπ/4] = cos(5π/4) + isin(5π/4) = -√2/2 - i√2/2
k=3,x₃= e^[7iπ/4] = cos(7π/4) + isin(7π/4) = √2/2 - i√2/2
k=4,x₄= e^[9iπ/4] = cos(9π/4) + isin(9π/4) = √2/2 + i√2/2
.
∴ x⁴+ 1 = [x-(√2/2+i√2/2)][x-(-√2/2+i√2/2)][x-(-√2/2-i√2/2)][x-(√2/2-i√2/2)]
= (x-√2/2-i√2/2)(x+√2/2-i√2/2)(x+√2/2+i√2/2)(x-√2/2+i√2/2)