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Let $\alpha,\lambda,\mu$ be nonzero complex numbers with $|\alpha| \gt 1$, and let $u_n=|\lambda \alpha^n + \mu \bar{\alpha}^n|$.

My question: Can anyone determine the set of all so-called adherence values of $(u_n)$, i.e. the values in $(0,+\infty)$ that are the limit of some subsequence of $(u_n)$.

What I tried: Write the polar decomposition of $\alpha$ : $\alpha =r e^{i\theta}$ with $r>0$ and $\theta\in{\mathbb R}$. When $\frac{\theta}{2\pi}$ is rational, it is easy to see that $v_n=\frac{u_n}{r^n}$ is periodic, so that the only possible adherence values for $u_n$ are either $0$ or $+\infty$. I have no ideas of how to proceed when $\frac{\theta}{2\pi}$ is irrational, however.

Gabriel Romon
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Ewan Delanoy
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    A remark: If WLOG $|\lambda|>|\mu|$, $|\lambda e^{in\theta}+\mu e^{-in\theta}|\geq |\lambda|-|\mu|>0$, hence $u_n\to \infty$. If $|\lambda|=|\mu|$, write $\lambda = \rho e^{ia}$ and $\mu = \rho e^{ib}$ to get $u_n = 2\rho r^n |\cos(n\theta+\frac{a-b}2)|$ – Gabriel Romon May 07 '18 at 09:12
  • In the simple case where $\theta =1$ and $a=b$, you may adapt this argument and get something like $|\cos(n)|\gg \frac{1}{n^\alpha}$, thus $u_n\to \infty$. – Gabriel Romon May 07 '18 at 10:01

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