Question

x4+1=0在复数范围内有几个解

Answer 1

z^4=-1 设：z=r(cosθ+isinθ)则：z^4=r^4(cos4θ+isin4θ)因为：-1=1(cos(π+2kπ)+isin(π+2kπ))根据复数相等,实部和虚部都相等得：r=14θ=π+2kπz=r(cosθ+isinθ)=[cos((π+2kπ)/4)+isin((π+2kπ)...

x=a+bi(a、b为实数)
x^4=(a+bi)^4
=[a^2+2abi+(bi)^2]^2
= (a^2+2abi-b^2)^2
=a^4+(2abi)^2+b^4+4a^3bi-2a^2b^2-4ab^3i
=a^4-4a^2b^2+b^4-2a^2b^2+(4a^3b-4ab^3)i
=(a^4-6a^2b^2+b^4)+(4a^3-4ab^3)i
=1
a^4-6a^2b^2+b^2=1，4a^3b-4ab^3=0，4ab(a^2-b^2)=0，a=0或b=0或a=b(舍去)或a=-b(舍去)。
a=0,b=+-1或b=0,a=+-1。
所以，x=i或x=-i或x=1或x=-1。

Answer 2

x^4+1=0
x^4+2x^2+1-2x^2=0
(x^2+1)^2-2x^2=0
(x^2+√2x+1)(x^2-√2x+1)=0
x1=(-√2-√2i)/2、x2=(-√2+√2i)/2、x3=(√2-√2i)/2、x4=(√2+√2i)/2