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 .