I want to prove the following identity: $$1+\omega^{k\hspace{0.1cm}mod \hspace{ 0.1cm}d}+\omega^{(2k)\hspace{0.1cm}mod \hspace{ 0.1cm}d}+...+\omega^{[(d-1)*k]\hspace{0.1cm}mod \hspace{ 0.1cm}d}=d*\delta_{k,0}$$ where $\omega$ is $d^{th}$ root of unity and $0\leq k\leq d-1$.

By inspection I found the identity to be true but, I am still not able to prove it rigorously.

I tried to write a proof considering two cases. One in which $k$ divides $d$ and in the other one it doesn't. When $k$ doesn't divide $d$ then, the above identity reduces to: $$1++\omega+\omega^2+...+\omega^{d-1}$$ which is zero. But, when $k$ divides $d$ I am not able to prove the same. Is my approach correct? I would like to get some hints on how to approach towards proving the identity. I don't expect any complete answers as I would like to try it out myself.

  • $\begingroup$ What is $$\omega^{k\hspace{0.1cm}mod \hspace{ 0.1cm}d}$$ exactly? Do you mean $$\omega^k$$ instead? // What is an identity without equal sign? $\endgroup$ – Did Aug 20 '16 at 14:36
  • $\begingroup$ What is $\delta _{k,0}$? $\endgroup$ – MereMortal47 Aug 20 '16 at 14:46
  • 1
    $\begingroup$ @Asemismaiel: It's a Kronecker delta, which is $1$ if $k=0$ and $0$ otherwise. $\endgroup$ – Henning Makholm Aug 20 '16 at 14:49

As Did points out in comments, taking mod $d$ in the exponents is redundant exactly because $\omega$ is a $d$th root of unity -- we have $$ \omega^{ad+b} = (\omega^d)^a\omega^b = 1^a\omega ^b = \omega^b = \omega^{(ad+b)\bmod d} $$ for all $a\in\mathbb Z$ and $b\in\{0,1,\ldots,d-1\}$.

Your sum is therefore the same as $$ 1 + \omega^k + \cdots + \omega^{(d-1)k} = 1 + (\omega^k)^1 + \cdots + (\omega^k)^{d-1} $$ which is a geometric series with common factor $\omega^k$, so the sum is $$ \frac{1-(\omega^k)^d}{1-\omega^k} = \frac{1-(\omega^d)^k}{1-\omega^k} = \frac{1-1^k}{1-\omega^k} = 0$$ as long as $\omega^k\ne1$.

Beware that $\omega^k$ may be $1$ even for nonzero $k$ if $\omega$ is not a primitive $d$th root of unity.


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