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**Capillarian** Let $\displaystyle \omega$ be a real or complex number satisfying $\displaystyle \omega^q = 1$, where $\displaystyle q$ is a natural number. Show that $\displaystyle \sum_{\omega} \omega^k$ has the value $\displaystyle q$ if $\displaystyle k$ is divisible by $\displaystyle q$ and has the value 0 otherwise.

Slightly annoying that I cannot prove this. The case $\displaystyle k$ is divisible by $\displaystyle q$ is obvious. I can only prove the sum is equal to 0 when $\displaystyle (k,q)=1$...