In my research, I have come across the following problem:
Let $n \in \mathbb{N}$ and $r \geq 0$ be given. For $\theta \in [0,2 \pi)$, define $$f(\theta) = \biggl| \frac{\sum_{j=0}^{n-1} e^{2 \pi ij/n} e^{r \Re (e^{2 \pi i j/n + i\theta})}}{\sum_{j=0}^{n-1} e^{r \Re (e^{2 \pi i j/n + i\theta})}}\biggr|.$$ Is it then the case that $f(\theta)$ is maximal at $\theta = 0$?
Mathematica suggests that this is at least true for small $n$. Also, it is easy to check that $f(\theta)$ is $2 \pi/n$-periodic, and that $f(2\pi/n-\theta) = f(\theta)$.
With some work, one can verify that for any $ \theta_0 \in [0,2\pi) $, one has $$ f(\theta)^2 = \biggl( \frac{\sum_{j} \cos \bigl( 2 \pi j/n+\theta_0 \bigr) e^{r \cos (2 \pi j/n + \theta) }}{\sum_{j} e^{r \cos (2 \pi j/n + \theta) }} \biggr)^2 + \biggl( \frac{\sum_{j} \sin \bigl( 2 \pi j/n+\theta_0\bigr) e^{r \cos (2 \pi j/n + \theta) }}{\sum_{j} e^{r \cos (2 \pi j/n + \theta) }} \biggr)^2. $$ Putting $\theta_0 = \theta$ makes the expression look nice and symmetric, while choosing $\theta_0 = 0$ makes it easier to differentiate. Differentiating the expression, one finds local extrema at $\theta = 0, \pi/n, 2 \pi/n, 3\pi/n, \ldots$, but it is not clear to me neither that these are all local extreme points(?) not which of these are local maxima(?).
