因为e^(ix)=cosx+isinx
所以e^(iπ/2)=cos(π/2)+isin(π/2),即i=e^(iπ/2)
于是√i=√[e^(iπ/2)]=[e^(iπ/2)]^(1/2)
=e^(iπ/4)=cos(π/4)+isin(π/4)
=(√2/2)+(√2/2)i
=(√2/2)(1+i)
因为e^(ix)=cosx+isinx
所以e^(iπ/2)=cos(π/2)+isin(π/2),即i=e^(iπ/2)
于是√i=√[e^(iπ/2)]=[e^(iπ/2)]^(1/2)
=e^(iπ/4)=cos(π/4)+isin(π/4)
=(√2/2)+(√2/2)i
=(√2/2)(1+i)