解答
z6=i
解答
z=42(1+3)+i42(−1+3),z=42(−1+3)+i42(1+3),z=−22+i22,z=42(−1−3)+i42(1−3),z=42(1−3)+i42(−1−3),z=22−i22
求解步骤
z6=i
For zn=athe solutions are zk=n∣a∣(cos(narg(a)+2kπ)+isin(narg(a)+2kπ)),
k=0,1,…,n−1
对于 n=6,a=i∣a∣=1
arg(a)=2π
z=61(cos(62π+2⋅0π)+isin(62π+2⋅0π)),z=61(cos(62π+2⋅1π)+isin(62π+2⋅1π)),z=61(cos(62π+2⋅2π)+isin(62π+2⋅2π)),z=61(cos(62π+2⋅3π)+isin(62π+2⋅3π)),z=61(cos(62π+2⋅4π)+isin(62π+2⋅4π)),z=61(cos(62π+2⋅5π)+isin(62π+2⋅5π))
化简 61(cos(62π+2⋅0π)+isin(62π+2⋅0π)):42(1+3)+i42(−1+3)
化简 61(cos(62π+2⋅1π)+isin(62π+2⋅1π)):42(−1+3)+i42(1+3)
化简 61(cos(62π+2⋅2π)+isin(62π+2⋅2π)):−22+i22
化简 61(cos(62π+2⋅3π)+isin(62π+2⋅3π)):42(−1−3)+i42(1−3)
化简 61(cos(62π+2⋅4π)+isin(62π+2⋅4π)):42(1−3)+i42(−1−3)
化简 61(cos(62π+2⋅5π)+isin(62π+2⋅5π)):22−i22
z=42(1+3)+i42(−1+3),z=42(−1+3)+i42(1+3),z=−22+i22,z=42(−1−3)+i42(1−3),z=42(1−3)+i42(−1−3),z=22−i22