求证x4次方十4=(x-1-i)(x-1+i)(x+1-i)(x+1+i)并写出在复数范围内-1的4个四次方根

Question

求证x4次方十4=(x-1-i)(x-1+i)(x+1-i)(x+1+i)并写出在复数范围内-1的4个四次方根

Answer 1

x^4+4
=x^4+4x^2+4-4x^2
=(x^2+2)^2-(2x)^2
=(x^2+2x+2)(x^2-2x+2)
=[(x+1)^2+1][(x-1)^2+1]
=(x+1+i)(x+1-i)(x-1+i)(x-1-i)
得证
由z^4=-1,得：(√2z)^4+4=0
故√2z=1+i,1-i,-1+i,-1-i
得：z=(1+i)/√2,(1-i)/√2,(-1+i)/√2,(-1-i)/√2