i=cos(π/2)+isin(π/2)=cos(2kπ+π/2)+isin(2kπ+π/2)所以√i=cos(kπ+π/4)+isin(kπ+π/4)=√2/2+√2/2i或-√2/2-√2/2i
i为虚数，√i=？=√2(1/2+1/2i)
好像不能开根号吧。。。
