解
z4=−1+i
解
z=2822+2−2+i2822−2−2,z=−2822−2−2+i2822+2−2,z=−2822+2−2−i2822−2−2,z=2822−2−2−i2822+2−2
解答ステップ
z4=−1+i
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=−1+i∣a∣=2
arg(a)=−4π+π
z=42(cos(4−4π+π+2⋅0π)+isin(4−4π+π+2⋅0π)),z=42(cos(4−4π+π+2⋅1π)+isin(4−4π+π+2⋅1π)),z=42(cos(4−4π+π+2⋅2π)+isin(4−4π+π+2⋅2π)),z=42(cos(4−4π+π+2⋅3π)+isin(4−4π+π+2⋅3π))
簡素化 42(cos(4−4π+π+2⋅0π)+isin(4−4π+π+2⋅0π)):2822+2−2+i2822−2−2
簡素化 42(cos(4−4π+π+2⋅1π)+isin(4−4π+π+2⋅1π)):−2822−2−2+i2822+2−2
簡素化 42(cos(4−4π+π+2⋅2π)+isin(4−4π+π+2⋅2π)):−2822+2−2−i2822−2−2
簡素化 42(cos(4−4π+π+2⋅3π)+isin(4−4π+π+2⋅3π)):2822−2−2−i2822+2−2
z=2822+2−2+i2822−2−2,z=−2822−2−2+i2822+2−2,z=−2822+2−2−i2822−2−2,z=2822−2−2−i2822+2−2