MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to prove that if $\theta _1,\theta _2,\theta _3$ be the arguments of the primitive roots of unity, $\sum \cos p\theta = 0$ when $p$ is a positive integer less than $\dfrac {n} {abc\ldots k}$, where $a, b, c, \ldots, k$ are the different constituent primes of n and when $p=\dfrac {n} {abc\cdots k}$, $\sum \cos p\theta = \dfrac {\left( -\right)^\mu n} {abc\cdots k}$, where $\mu$ is the number of the constituent primes.

Any help would be much appreciated.

Here is where i got stuck i was hoping for a direct proof starting with the LHS of $\sum \cos p\theta = 0$.

Solution attempt

Well as we know the number of primitive roots(roots the power of which give all the roots) is the number of integers(including unity) less than n and prime to it. Hence we get this number from Euler's totient or phi function.

In the first part of the problem we are asked to assume $0 < p < \dfrac {n} {abc\ldots k}$.

So applying Taylor's series expansion to LHS we get $\sum _{i=1}^{i=\varphi \left( n\right) }\cos p\theta _{i} = \sum _{i=1}^{i=\varphi \left( n\right) }\left( 1-\dfrac {\left( p\theta _{i}\right) ^{2}} {2!}+\dfrac {\left( p\theta _{i}\right) ^{4}} {4!}-...\right) $

Combining terms from various expansions we get $\sum _{i=1}^{i=\varphi \left( n\right) }\cos p\theta _{i} = \varphi \left( n\right) -\dfrac {p^{2}} {2!}\sum _{i=1}^{i=\varphi \left( n\right) }\theta _{i}^{2} + \dfrac {p^{4}} {24}\sum _{i=1}^{i=\varphi \left( n\right) }\theta _{i}^{4} - ...$

I am unsure how to proceed from here although the pattern of alternating signs required for the second result are starting to appear. Any clues or pointers about results related to arguments of primitive roots, would be highly appreciated.

share|cite|improve this question
Also posted to…, as OP should have told us. – Gerry Myerson Feb 20 '12 at 21:55
I could only understand the notation after reading David Speyer's answer. Also, the choice of tags is most misleading. – Vladimir Dotsenko Feb 21 '12 at 7:29
up vote 8 down vote accepted

This problem seems to use deliberately confusing notation. Let $Z_n$ be the set of primitive $n$-th roots of unity. Since $\cos p \theta$ is just the real part of $e^{i p \theta}$, the problem is to evaluate $\mathrm{Re} \sum_{\zeta \in Z_n} \zeta^p$. The real part is just a read herring, since the sum is real anyway, so our job is to evaluate $\sum_{\zeta \in Z_n} \zeta^p$.

I'm not sure whether you want a full solution or just a hint. For a starting hint, here is a sketch of the case $p=1$. Our goal is to show that $$\sum_{\zeta \in Z^n} \zeta = \begin{cases} 0 & n\ \mathrm{not\ squarefree} \\ (-1)^{\mu} & n \ \mathrm{squarefree} \\ \end{cases}$$ This can be written in a single line using the Mobius function: $$\sum_{\zeta \in Z_n} \zeta = \mu(n).$$

Let $M_n$ be the set of all $n$-th roots of unity. We have $\sum_{\zeta \in M_n} \zeta=0$ for $n>1$ and $\sum_{\zeta \in M_1} \zeta =1$. Now use inclusion-exclusion.

Once you've done that, the case of higher $p$ is not much more difficult.

share|cite|improve this answer
I am impressed that you were able to figure out the notation... – Igor Rivin Feb 20 '12 at 20:26
@David thank you very much for the hint. I 'll attempt a solution with it in mind and post back if is run into another problem. – Comic Book Guy Feb 20 '12 at 22:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.