复变函数问题z^4+i=0如何解

Question

复变函数问题 z^4+i=0
如何解

Answer 1

z^4=-i
z^4=e^i(-π/2)
z=e^i(-π/2+2kπ)/4, k=0,1,2,3
即：
z1=e^i(-π/8)=cos(π/8)-isin(π/8)
z2=e^i(3π/8)=cos(3π/8)+isin(3π/8)
z3=e^i(7π/8)=-cos(π/8)+isin(π/8)
z4=e^i(11π/8)=-cos(3π/8)-isin(3π/8)
更进一步的话,有：
sin(π/8)=√[1-cos(π/4)]/2=[√(2-√2)]/2
cos(π/8)=√[1+cos(π/4)]/2=[√(2+√2)]/2
sin(3π/8)=cos(π/8)
cos(3π/8)=sin(π/8)