解
z5=−3i
解
z=45325+5−i45323−5,z=45310−25+i453+535,z=−45310−25+i453+535,z=−45325+5−i45323−5,z=−53i
解答ステップ
z5=−3i
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=5,a=−3i∣a∣=3
arg(a)=−2π
z=53(cos(5−2π+2⋅0π)+isin(5−2π+2⋅0π)),z=53(cos(5−2π+2⋅1π)+isin(5−2π+2⋅1π)),z=53(cos(5−2π+2⋅2π)+isin(5−2π+2⋅2π)),z=53(cos(5−2π+2⋅3π)+isin(5−2π+2⋅3π)),z=53(cos(5−2π+2⋅4π)+isin(5−2π+2⋅4π))
簡素化 53(cos(5−2π+2⋅0π)+isin(5−2π+2⋅0π)):45325+5−i45323−5
簡素化 53(cos(5−2π+2⋅1π)+isin(5−2π+2⋅1π)):45310−25+i453+535
簡素化 53(cos(5−2π+2⋅2π)+isin(5−2π+2⋅2π)):−45310−25+i453+535
簡素化 53(cos(5−2π+2⋅3π)+isin(5−2π+2⋅3π)):−45325+5−i45323−5
簡素化 53(cos(5−2π+2⋅4π)+isin(5−2π+2⋅4π)):−53i
z=45325+5−i45323−5,z=45310−25+i453+535,z=−45310−25+i453+535,z=−45325+5−i45323−5,z=−53i