在实数域无解!只能在复数域求解:
Z⁴ + 1 = 0 ----------------------- (1)
设:Z = R(cos a + j sin a) ------- (2)
Z⁴ = -1 = (cos π + j sin π) ------ (3)
Z = cos[(π+2kπ)/n + j sin[(π+2kπ)/n] ---------- (4) //: n=4,k=0,1,2,3
Z₀ = cos(π/4)+jsin(π/4) =√2(1+j)/2 --- (5) //: 验证:Z₀^4=-1
Z₁ = cos (3π/4)+jsin(3π/4) ------------ (6)
Z₂ = cos (5π/4)+jsin(5π/4) ------------ (7)
Z₃ = cos(7π/4)+jsin(7π/4) -------------(8)
(5,6,7,8)是4个复根。