解
z4=3+2i
解
z=813cos(4arctan(32))+813isin(4arctan(32)),z=813cos(4arctan(32)+2π)+813isin(4arctan(32)+2π),z=813cos(4arctan(32)+4π)+813isin(4arctan(32)+4π),z=813cos(4arctan(32)+6π)+813isin(4arctan(32)+6π)
解答ステップ
z4=3+2i
zn=aの場合, 解は zk=n∣a∣(cos(narg(a)+2kπ)+isin(narg(a)+2kπ)),
k=0,1,…,n−1
以下のため: n=4,a=3+2i∣a∣=13
arg(a)=arctan(32)
z=413(cos(4arctan(32)+2⋅0π)+isin(4arctan(32)+2⋅0π)),z=413(cos(4arctan(32)+2⋅1π)+isin(4arctan(32)+2⋅1π)),z=413(cos(4arctan(32)+2⋅2π)+isin(4arctan(32)+2⋅2π)),z=413(cos(4arctan(32)+2⋅3π)+isin(4arctan(32)+2⋅3π))
簡素化 413(cos(4arctan(32)+2⋅0π)+isin(4arctan(32)+2⋅0π)):813cos(4arctan(32))+813isin(4arctan(32))
簡素化 413(cos(4arctan(32)+2⋅1π)+isin(4arctan(32)+2⋅1π)):813cos(4arctan(32)+2π)+813isin(4arctan(32)+2π)
簡素化 413(cos(4arctan(32)+2⋅2π)+isin(4arctan(32)+2⋅2π)):813cos(4arctan(32)+4π)+813isin(4arctan(32)+4π)
簡素化 413(cos(4arctan(32)+2⋅3π)+isin(4arctan(32)+2⋅3π)):813cos(4arctan(32)+6π)+813isin(4arctan(32)+6π)
z=813cos(4arctan(32))+813isin(4arctan(32)),z=813cos(4arctan(32)+2π)+813isin(4arctan(32)+2π),z=813cos(4arctan(32)+4π)+813isin(4arctan(32)+4π),z=813cos(4arctan(32)+6π)+813isin(4arctan(32)+6π)