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(1+i)的1/4次是多少
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(1+i)的1/4次是多少
▼优质解答
答案和解析
(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,
=√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,
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