Question

1+i的1-i次方是多少?

Answer 1

(1+i)^(1-i)的值有无穷多个|1+i|=√2,arg(1+i)=π/4Ln(1+i)=ln|1+i|+i[arg(1+i)+2kπ]=ln(√2)+i(π/4+2kπ)（k∈Z）(1+i)^(1-i)=exp[(1-i)*Ln(1+i)]=exp{(1-i)[ln(√2)+i(π/4+2kπ)]}（k∈Z）取k=0,即得(1+i)^(1-i...