解答
z4=−4−4i
解答
z=22852+2−2−i22852−2−2,z=22852−2−2+i22852+2−2,z=−22852+2−2+i22852−2−2,z=−22852−2−2−i22852+2−2
求解步骤
z4=−4−4i
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=−4−4i∣a∣=42
arg(a)=4π−π
z=442(cos(44π−π+2⋅0π)+isin(44π−π+2⋅0π)),z=442(cos(44π−π+2⋅1π)+isin(44π−π+2⋅1π)),z=442(cos(44π−π+2⋅2π)+isin(44π−π+2⋅2π)),z=442(cos(44π−π+2⋅3π)+isin(44π−π+2⋅3π))
化简 442(cos(44π−π+2⋅0π)+isin(44π−π+2⋅0π)):22852+2−2−i22852−2−2
化简 442(cos(44π−π+2⋅1π)+isin(44π−π+2⋅1π)):22852−2−2+i22852+2−2
化简 442(cos(44π−π+2⋅2π)+isin(44π−π+2⋅2π)):−22852+2−2+i22852−2−2
化简 442(cos(44π−π+2⋅3π)+isin(44π−π+2⋅3π)):−22852−2−2−i22852+2−2
z=22852+2−2−i22852−2−2,z=22852−2−2+i22852+2−2,z=−22852+2−2+i22852−2−2,z=−22852−2−2−i22852+2−2