MHB Challenge problem #5 [Olinguito]

[Olinguito]

If $A$ and $B$ are nonempty sets of complex numbers, define
$$A\circ B\ =\ \{z_1z_2:z_1\in A,\,z_2\in B\}.$$
Further define $A^{[1]}=A$ and recursively $A^{[n]}=A^{[n-1]}\circ A$ for $n>1$.

Let $\zeta_n=\{z\in\mathbb C:z^n=1\}$. Given a fixed integer $n\geqslant2$ and any positive integer $r$, find the sum of all the elements in $\zeta_n^{[r]}$.

 

[castor28]

[sp]
As $\zeta_n^{[1]}$ is closed under multiplication, $\zeta_n^{[r]}\subset\zeta_n^{[1]}$. On the other hand, as $1\in\zeta_n^{[1]}$, $\zeta_n^{[1]}\subset\zeta_n^{[r]}$.
The conclusion is that $\zeta_n^{[r]}=\zeta_n^{[1]}$ for all $r$. The sum of the elements of that set is the sum of the roots of $z^n-1$, which is $0$.
[/sp]

 

[Olinguito]

Spoiler

Nice work. (Yes)

Alternatively, note that
$$A\circ B\ =\ \bigcup_{z\in A}\,\{z\}\circ B$$
and $\{\omega\}\circ\zeta_n=\zeta_n$ where $\omega$ is any $n$th root of unity.