1-i=√2(1/√2-i/√2)=√2(cos(-π/4)+isin(-π/4))=√2(cos(2kπ-π/4)+isin(2kπ-π/4))所以平方根为：2^(1/4)*(cos(kπ-π/8)+isin(kπ-π/8))=2^(1/4)*(cos(π/8)-isin(π/8))或2^(1/4)(cos(7π/8)+isin(7π/8))...