Let \(\omega\) be the complex cube root of unity. Find the cardinality of the set \(S\) where \(S = \{(1+\omega+\omega^2+\dots+\omega^n)^m \mid m,n = 1,2,3,\dots\}\)
Note that the sum \(1+\omega+\omega^2+\dots+\omega^n\) contains \(n+1\) terms. The power \(n\) must be of the form \(3k\), \(3k+1\) or \(3k+2\) where \(k \in \mathbb{N} \cup \{0\}\).
When \(n\) is of the form \(3k+2\), there are \(3k+3\) terms (a multiple of 3). Because \(1+\omega+\omega^2 = 0\), the sum evaluates to \(0\). Whence \((1+\omega+\omega^2+\dots+\omega^n)^m = 0^m = 0\) \(\forall m \in \mathbb{N}\).
When \(n\) is a multiple of 3 (i.e., \(3k\)), there are \(3k+1\) terms. The sum evaluates to \(1\). Whence \((1+\omega+\omega^2+\dots+\omega^n)^m = 1^m = 1\) \(\forall m \in \mathbb{N}\).
When \(n\) is of the form \(3k+1\), there are \(3k+2\) terms. The sum evaluates to \(1+\omega = -\omega^2\). Whence \((1+\omega+\omega^2+\dots+\omega^n)^m = (-\omega^2)^m = (-1)^m \omega^{2m}\) \(\forall m \in \mathbb{N}\).
In this final case, as \(m\) varies over the natural numbers, the expression \((-1)^m \omega^{2m}\) generates 6 distinct values: \(-\omega^2, \omega, -1, \omega^2, -\omega, 1\).
Combining the three cases, we see that \( S = \{0, -1, 1, \omega, -\omega, \omega^2, -\omega^2\} \), therefore \(|S|=7\).
No comments:
Post a Comment