# Thread: Sum over roots of unity

1. ## Sum over roots of unity

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 $\displaystyle (k,q)=1$...

2. Originally Posted by 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 $\displaystyle (k,q)=1$...
.

Divide with residue $\displaystyle k=xq+r\,,\,\,r=0\,\,or\,\,|r|<q$ , so

$\displaystyle \displaystyle{\sum\limits_ww^k=\sum\limits_w(w^q)^ xw^r=\sum\limits_ww^r$ .

As you said, if k is divisible by q then r = 0 and the sum is trivially = q, otherwise:

Take a q-th root of unity $\displaystyle \sigma$ s.t. $\displaystyle \sigma^r\neq 1$ (why must there be such a root?), so:

$\displaystyle \displaystyle{\sigma^r\sum\limits_ww^r=\sum\limits _w\sigma^rw^r=\sum\limits_w(\sigma w)^r=\sum\limits_ww^r$ ,

the last equality fowllowing because... ,and then

$\displaystyle \displaystyle{\sigma^r\sum\limits_ww^r=\sum\limits _ww^r\Longrightarrow (\sigma^r-1)\sum\limits_ww^r=0$ , and we're done.

Tonio

Pd. The above proof assumes you know a little group theory and that the q-th roots

of unity are a (cyclic) a group under complex multiplication.