Problem:
Compute $$\log_2\left(\prod_{a=1}^{2015} \prod_{b=1}^{2015} \left(1 + e^{2\pi iab/2015}\right)\right)$$
Solution:
We claim that the answer is $\boxed{13725}$.
For a fixed value of $a$, let $w=e^{2\pi iab/2015}$ be a primitive $m^{\text{th}}$ root of unity. Since $\frac{2\pi iab}{2015} = \frac{2\pi i \frac{a}{\gcd(a, 2015)}b}{\frac{2015}{\gcd(a, 2015)}}$ in its lowest form, we know that $m=\frac{2015}{\gcd(a, 2015)}$.
Then, we have
$$\prod_{b=1}^{2015} \left(1 + e^{2\pi iab/2015}\right) = [(1+w)(1+w^2)(1+w^3)\dots(1+w^m)]^{\gcd(a, 2015)} \qquad (\star),$$ where the power of $\gcd(a, 2015)$ is beacuse the cycle $1+w, 1+w^2, 1+w^3, \dots, 1+w^m$, from $b=1$ to $b=2015$, repeats $\frac{2015}{m} = \gcd(a, 2015)$ times.
It is well-known that $(x-w)(x-w^2)(x-w^3)\dots(x-w^m)=x^m-1$ for a primitive $m^{\text{th}}$ root of unity, so plugging in $x=-1$, it becomes clear that $$(1+w)(1+w^2)(1+w^3)\dots(1+w^m)=2$$ for odd $m$.
Using this fact along with $(\star)$, we see that the problem statement is equivalent to $$\log_2 \left(\prod_{a=1}^{2015} 2^{\gcd(a,2015)}\right) = \sum_{a=1}^{2015} \gcd(a, 2015)$$.
Notice that computing this sum of GCDs is equivalent to computing $$(5+\underbrace{1+\dots+1}_{4 \text{ ones}})(13+\underbrace{1+\dots+1}_{12 \text{ ones}})(31+\underbrace{1+\dots+1}_{30 \text{ ones}})$$
$$= 9 \cdot 25 \cdot 61 = 13725$$ as desired. $\square$
Comments
Post a Comment