解答
z5=2−2i
解答
z=421074+10+25−i421074−10+25,z=421074−10−25+i421074+10−25,z=−22107+i22107,z=−421074+10−25−i421074−10−25,z=421074−10+25−i421074+10+25
求解步骤
z5=2−2i
For zn=athe solutions are zk=n∣a∣(cos(narg(a)+2kπ)+isin(narg(a)+2kπ)),
k=0,1,…,n−1
对于 n=5,a=2−2i∣a∣=2
arg(a)=−4π
z=52(cos(5−4π+2⋅0π)+isin(5−4π+2⋅0π)),z=52(cos(5−4π+2⋅1π)+isin(5−4π+2⋅1π)),z=52(cos(5−4π+2⋅2π)+isin(5−4π+2⋅2π)),z=52(cos(5−4π+2⋅3π)+isin(5−4π+2⋅3π)),z=52(cos(5−4π+2⋅4π)+isin(5−4π+2⋅4π))
化简 52(cos(5−4π+2⋅0π)+isin(5−4π+2⋅0π)):421074+10+25−i421074−10+25
化简 52(cos(5−4π+2⋅1π)+isin(5−4π+2⋅1π)):421074−10−25+i421074+10−25
化简 52(cos(5−4π+2⋅2π)+isin(5−4π+2⋅2π)):−22107+i22107
化简 52(cos(5−4π+2⋅3π)+isin(5−4π+2⋅3π)):−421074+10−25−i421074−10−25
化简 52(cos(5−4π+2⋅4π)+isin(5−4π+2⋅4π)):421074−10+25−i421074+10+25
z=421074+10+25−i421074−10+25,z=421074−10−25+i421074+10−25,z=−22107+i22107,z=−421074+10−25−i421074−10−25,z=421074−10+25−i421074+10+25