`(costheta+isintheta)(cos2theta+isin2theta)....(cosntheta+isinntheta) = 1`
we know, `costheta+isintheta = e^itheta`
So, our equation becomes,
`=>(e^(itheta))(e^(i2theta)).....(e^(i ntheta)) = 1`
`=>e^(itheta(1+2+3...n))= 1`
`=>e^(itheta((n(n+1))/2)) = 1`
`=>cos((n(n+1))/2)theta+isin((n(n+1))/2)theta = 1`
Comparing real and imaginary parts,
`=>cos((n(n+1))/2)theta = 1`
`=>((n(n+1))/2)theta = 2mpi` ...... (here, `m in (1,2,3...n)`)
`=>theta = (4mpi)/(n(n+1))`
