解答
i⋅z5=3
解答
z=45325+5−i45323−5,z=45310−25+i453+535,z=−45310−25+i453+535,z=−45325+5−i45323−5,z=−53i
求解步骤
iz5=3
两边除以 i
z5=−3i
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=−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))
化简 53(cos(5−2π+2⋅0π)+isin(5−2π+2⋅0π)):45325+5−i45323−5
化简 53(cos(5−2π+2⋅1π)+isin(5−2π+2⋅1π)):45310−25+i453+535
化简 53(cos(5−2π+2⋅2π)+isin(5−2π+2⋅2π)):−45310−25+i453+535
化简 53(cos(5−2π+2⋅3π)+isin(5−2π+2⋅3π)):−45325+5−i45323−5
化简 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