For a special $\rho \in \mathbb{C}$ $1+ \rho^k+\rho^{2k}+\rho^{3k}+...+\rho^{(n-1)k}=1$ if $n$ divides $k$.

Question

Let $\rho=cos(\frac{2\pi}{n})+i sen(\frac{2\pi}{n})$ such $z=1, z= \rho, z= \rho ^2,...,z= \rho^{n-1}$ are all the diferent solutions of $z^n=1$, then for $k \in \mathbb{N}$ we have $$1+ \rho^k+\rho^{2k}+\rho^{3k}+...+\rho^{(n-1)k}=1$$ if $n$ divides $k$.

My idea for the proof goes this way: If $n|k$ then there is a $m \in \mathbb{Z}$ such $nm=k$, so $$1+ \rho^k+\rho^{2k}+\rho^{3k}+...+\rho^{(n-1)k}\\ =1+\rho^{nm}+\rho^{2nm}+....+\rho^{(n-1)nm}\\ =1+(\rho^m)^{n}+(\rho^{2m})^n+...+(\rho^{(n-1)m})^{n}\\ =1+1+............................+1\\ =n.$$

Is my reasoning right? Can anyone help me ending the proof of the statement? I would apreciate it.

Answer 1

Your working depends on $\rho^m$ for any $m$ is a root for $z^n=1$.

Alternatively, I can use the fact that $\rho^n=1$ and get

\begin{align} &1+ \rho^k+\rho^{2k}+\rho^{3k}+ \ldots +\rho^{(n-1)k}\\ &=1+\rho^{nm}+\rho^{2nm}+ \ldots +\rho^{(n-1)nm}\\ &=1+(\rho^\color{blue}n)^{m}+(\rho^\color{blue}n)^{2m}+ \ldots +(\rho^{n})^{(n-1)m}\\ &=\sum_{k=0}^{n-1} \left( \rho^n\right)^{km}\\ &=\sum_{k=0}^{n-1} \left( 1\right)^{km}\\ &=\sum_{k=0}^{n-1} \left( \rho^n\right)^{km}\\ &=\sum_{k=0}^{n-1} 1\\ &=n. \end{align}

Remark:

If $n \not \mid k$, try to use the formula for geometric series to evaluate the sum.