(1+i)=√2(√2/2+i√2/2) 　　=√2[cos(π/4)+isin(π/4)] 　　则(1+i)^(1/4)=2^(1/8)[cos((π/4+2kπ)/4)+isin((π/4+2kπ)/4)],k=0,1,2,3 　　k=0时：2^(1/8)[cos(π/16)+isin(π/16)] 　　k=1时：2^(1/8)[cos(9π/16)+isin(9π/16)] 　　k=2时：2^(1/8)[cos(17π/16)+isin(17π/16)] 　　k=3时：2^(1/8)[cos(25π/16)+isin(25π/16)] 　　希望可以帮到你,如果解决了问题,请点下面的"选为满意回答"按钮,