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Let $\zeta\not= 1$ a $n$-th root of unity (I have doubts with this word in English but in Spanish is $n$-esima). Prove that for $k \in \Bbb N$ \begin{equation} 1+\zeta^{k}+\zeta^{2k}+\dots+\zeta^{(n-1)k}= \begin{cases} 0&\text{if $n$ not divide $k$,} \\ n &\text{if $n$ divide $k$.} \end{cases} \end{equation}

I'm not really good in English and I don't know, if someone can understand this question. But well, I tried to use different teorems about complex numbers and at the same time I used teorems about a $n$-th root (in Spanish $n$-esima). If someone could help me I would really appreciate it.

Masacroso
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TheAsteroide
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  • Your English is quite good! Writing $n$-th or $n$th is fine. Small corrections: With a natural number, we use the singular form of the verb, so "if $n$ divides $k$'' and ''if $n$ does not divide $k$". Also, in English, theorems. – Sammy Black May 11 '22 at 21:38
  • Well i tried to use something like it but I was not sure, but, thank you. – TheAsteroide May 11 '22 at 22:27

2 Answers2

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You have that

$$ 1+\zeta^k+\zeta ^{2k}+\ldots +\zeta ^{(n-1)k}=\sum_{j=0}^{n-1}\zeta ^{jk}=\sum_{j=0}^{n-1}(\zeta ^{k})^j=\frac{1-(\zeta ^k)^n}{1-\zeta^k }=\frac{1-1}{1-\zeta ^k}=0 $$

when $\zeta ^k\neq 1$, what is equivalent to say that when $n\nmid k$ (that is, $\zeta ^m=1$ if and only if $m$ is a multiple of $n$). By the other hand, if $\zeta ^k=1$ then $\zeta ^{jk}=1$ for any chosen $j\in \mathbb{N}$, so

$$ \sum_{j=0}^{n-1}\zeta ^{jk}=\sum_{j=0}^{n-1}1=n $$

Masacroso
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I'm going to assume you mean a primitive $n$th root of unity, as otherwise the result is false (consider $\zeta=-1$ as a $4$th root of unity with $k=2$).

Lemma. If $\zeta$ is a primitive $p$th root of unity for some integer $p\geq 2$ (meaning powers of $\zeta$ span all the roots of $x^p-1$), then $1+\zeta+\zeta^2+\dots+\zeta^{p-1}=0$.

Proof: Vieta's formula on $x^p-1$.

With this in hand, the key observation is that the map $x\mapsto x^k$ maps primitive $n$th roots of unity to primitive $n/\gcd(n,k)$'th roots of unity. This is obvious from the definition of a primitive root of unity (if we need $n$ powers to get to $1$ and raising to the $k$ gets us $\gcd(n,k)$ of the way there (since relatively prime factors just permute things), we still have $n/\gcd(n,k)$ of the way to go).

This observation plus the lemma quickly yields the result: if $\gcd(n,k)\neq n$, then by the lemma the desired sum is $0$. Otherwise, each term in the sum is equal to $n$, so the desired sum is $n$.

Connor Gordon
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