Complex Numbers :Indian Statistical Institute B.Math & B.Stat

Let \(\omega\) be the complex cube root of unity. Find the cardinality of the set $S$ where \(S = \{(1+\omega+\omega^2+\dots+\omega^n)^m | m,n = 1,2,3,......\}\) $$$$ $n$ must be of the form \( 3k, 3k+1\) or \(3k+2\) where \(k \in N\) $$$$ when $n$ is multiple of $3$,\(1+\omega+\omega^2+\dots+\omega^n = 0\) whence \((1+\omega+\omega^2+\dots+\omega^n)^m = 0\) \(\forall\) \(n,m \in N\) $$$$ when $n$ is of the form $3k+1$,\(1+\omega+\omega^2+\dots+\omega^n = 1\) whence \((1+\omega+\omega^2+\dots+\omega^n)^m = 1^m=1\) \(\forall\) \(n,m \in N\) $$$$ when $n$ is of the form $3k+2$,\(1+\omega+\omega^2+\dots+\omega^n = 1+\omega(**)\) whence \((1+\omega+\omega^2+\dots+\omega^n)^m = (1+\omega)^m =(-\omega)^{2m}\) \(\forall\) \(n,m \in N\) $$$$ In this case the possible values are possible values of \((-\omega)^{2m}\) which are \(-1,1,\omega,-\omega,\omega^2,-\omega^2\), note m varies over the set of Natural numbers $$$$ Combining the three cases we see that \( S = \{0,-1,1,\omega,-\omega,\omega^2,-\omega^2\} \), therefore $|S|=7$. \((**)\) To understand these first observe that $\omega^n=1,\omega,\omega^2$ for any natural number $n$. So when $n$ is of the for $3k+2$ we can couple the three consecutive recurring terms $1,\omega,\omega^2$ to get the sum as $0$. After that we are left with two more terms $1$ and $\omega$, since the series \(1+\omega+\omega^2+\omega^3+\omega^4+\omega^5\dots+\omega^n\) has repeated terms after every 3 terms (similar to the first three terms). Similarly for the othere cases the same argument follows. This concludes the result .