解答
z4=163−16i
解答
z=22434+2+6−i22434−2−6,z=22434−2−6+i22434+2+6,z=−22434+2+6+i22434−2−6,z=−22434−2−6−i22434+2+6
求解步骤
z4=163−16i
For zn=athe solutions are zk=n∣a∣(cos(narg(a)+2kπ)+isin(narg(a)+2kπ)),
k=0,1,…,n−1
对于 n=4,a=163−16i∣a∣=32
arg(a)=−6π
z=432(cos(4−6π+2⋅0π)+isin(4−6π+2⋅0π)),z=432(cos(4−6π+2⋅1π)+isin(4−6π+2⋅1π)),z=432(cos(4−6π+2⋅2π)+isin(4−6π+2⋅2π)),z=432(cos(4−6π+2⋅3π)+isin(4−6π+2⋅3π))
化简 432(cos(4−6π+2⋅0π)+isin(4−6π+2⋅0π)):22434+2+6−i22434−2−6
化简 432(cos(4−6π+2⋅1π)+isin(4−6π+2⋅1π)):22434−2−6+i22434+2+6
化简 432(cos(4−6π+2⋅2π)+isin(4−6π+2⋅2π)):−22434+2+6+i22434−2−6
化简 432(cos(4−6π+2⋅3π)+isin(4−6π+2⋅3π)):−22434−2−6−i22434+2+6
z=22434+2+6−i22434−2−6,z=22434−2−6+i22434+2+6,z=−22434+2+6+i22434−2−6,z=−22434−2−6−i22434+2+6