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consider:
i^m/n has n number of roots
so z = 1^1/3 has 3 roots.
It can also be written as z^3 - 1 =0 this is a cubic equation just as a quadratic one so roots of this cubic quation gives
the cube roots of unity.
1 = cos0 + isin0 =cos2k(pie) + isin2k(pie)
1^1/3 =e^i2kpie/3 where k = 0, 1, 2
so roots are :- 1, e^i2pie/3, e^i4pie/3
let z be a cube root of 1
=>z3 =1
=>z3 -1 =0
=>(z-1)(z2+z+1) =0
=>z=1 or z2+z+1=0
so the cube roots of unity are 1 and two conjugate complex nos.
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