Compute $$\log_2\left(\prod_{a=1}^{2015}\prod_{b=1}^{2015}\left(1+e^{2\pi iab/2015}\right)\right)$$ Here $i$ is the imaginary unit (that is, $i^2=-1$).
Can someone explain me this line from the first solution [by Kent merrifield] posted here
( https://artofproblemsolving.com/community/c7h1171025p5624333) :
If $a$ is relatively prime to $n$, then the numbers $\zeta^{ab}$ for $1\le b\le n$ encompass all $n$ of possible powers of $\zeta,$ and hence $\prod_{b=1}^n(1+\zeta^{b})=2.$
Thanks in advance !
You must be assuming $n$ is odd, and $\zeta$ is a primitive $n$-th root of $1$. Then $\zeta^a$ is also a primitive $n$-root of $1$. We have a factorisation $$X^n-1=\prod_{b=0}^{n-1}(X-\zeta^{ab}).$$ Setting $X=-1$ gives $$-2=\prod_{b=0}^{n-1}(-1-\zeta^{ab})=(-1)^n\prod_{b=0}^{n-1}(1+\zeta^{ab}).$$