Summing powers of complex root of unities gives 0

Question

I have a question regarding a proof. Let $z_N$ denote the complex N'th root of unity, from which we have the identities

$(z_N)^n=1$

$\sum_{i=0}^{N-1}{(z_N)^i}=0$

Now let $N=r\cdot t$ and let $H^\bot$ be the set of multiples of $t$ in $\mathbb{Z}_n$.

For any $x \notin H^\bot$, show the following holds: $\sum_{i=0}^{t-1}{z_N^{rxi}}=0$

Can you help me?

Answer 1

Hint. Prove that $z_N^{rx}$ is a $t$th root of unity besides $1$.

Answer 2

If $\displaystyle Z_N^{rx}-1\ne0,$

$\displaystyle\sum_{i=0}^{t-1}{z_N^{rxi}}=\sum_{i=0}^{t-1}(Z_N^{rx})^i=\frac{(Z_N^{rx})^t-1}{Z_N^{rx}-1}$ $\displaystyle=\frac{(Z_N^{rt})^x-1}{Z_N^{rx}-1}$

But, $Z_N^{rt}=1$