Let $a + b i$ be a complex number whose absolute value is greater than $1$ and whose argument is not a rational multiple of $\pi$ .
For $n = 1, 2, 3, \cdots$ define $f(n) =| (a + b i )^n + (a - b i )^n |$ where $|\cdot|$ denotes the absolute value.
It is easy to show that $\limsup f(n) = \infty$ .
Is it also true that $\liminf f(n) = \infty$ ?
Thought this might be useful, but it's too long for a comment:
Denote $z = a + bi = re^{i \theta}$. We note that $$ f(n) = |z^n + (\overline z)^{n}| = r^n \left|e^{i n \theta} - e^{-in\theta} \right| = 2 r^n \sin (n\theta) $$ So, proving that $\liminf_{n \to \infty} f(n)\leq K$ amounts to showing that we can find arbitrarily large $n$ such that $$ -\sin^{-1}\left(\frac K2 r^{-n}\right) \leq[n \theta] \leq \sin^{-1} \left(\frac K2 r^{-n}\right) $$ Where $[n \theta]$ is the the value of $n \theta - 2 \pi k$ for $k \in \mathbb Z$ that falls between $-\pi$ and $\pi$. Note that $\sin^{-1} x \approx x$ for sufficiently small $x$.
This does not seem like it can be directly deduced from equidistribution.