Solve the equation `z^7=1`

Explanation

`z^7=1=cos 0+ sin 0`
`=cos(0+2kpi+isin 2kpi)`
`=cos 2kpi+isin 2kpi`
`:.z=(cos 2kpi+isin 2kpi)^(1/7)`
`=cos (2kpi)/7)+isin(2kpi)/7)`
`"where"k=0,1,2,3,4,5,6`