1+i开立方 求详解

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1+i开立方 求详解
1+i开立方 求详解

1+i开立方 求详解
1+i=(√2)[cos(π/4)+isin(π/4)]
z=∛(1+i)=(∛√2){cos[(2kπ+π/4)/3]+isin[2kπ+π/4)/3]}(k=0,1,2)
当k=0时,zo=[2^(1/6)][cos(π/12)+isin(π/12)]
当k=1时,z₁=[2^(1/6)][cos(3π/4)+isin(3π/4)]=[2^(1/6)][-cos(π/4)+isin(π/4)]
当k=2时,z₂=[2^(1/6)][cos(17π/12)+isin(17π/12)]=[2^(1/6)][-cos(5π/12)-isin(5π/12)]