'm readig a paper of recurrences sequences and i could not understand why there are conjugated units that are mentioned in the underlined lines in the image. I think you can not use Dirichlet's unit theorem, and it has to do with p-adic analysis. If someone has a clue or an answer, I would appreciate it very much.
Choose $a\in \Bbb Z$ such that $|a|_v=|\alpha_i^3|_v$ for all finite places $v$ (why can that be done?*). It follows that $\eta_i := \alpha_i^3/a$ are units in $\mathcal{O}_K$.
To see they are conjugate (by the Galois action, this has nothing to do with complex conjugation a priori), just note that the $\alpha_i$ are conjugate, hence so are their cubes, and $a$ is fixed by the Galois action since it is in $\Bbb Z$.
*As asked for in the comment, further hints why such an $a \in \Bbb Z$ exists:
First imagine we were in $\Bbb Q$, so the valuations are the usual $p$-adic ones. Say for finitely many $p$, I prescribe a certain $p$-adic value $\le 1$, then can you find an element $a\in \Bbb Z$ so that for each $p$, $|a|_p$ has the value I prescribed? To give a totally explicit example, say I want $|a|_2 = 1/2, |a|_5 = 1/125$ and $|a|_{17} = 1/17$ (and all other $|a|_p =1$), can you find such an $a \in \Bbb Z$?
Now once you get that, the only problem is that we are in $\mathcal{O}_K$, and we are prescribed values for $|a|_v$, but want values for $|a|_p$. Even worse, a priori there could be several valuations (finite places) $v$ lying above some primes $p$. But no, the sentence before the one you highlighted says something about total ramification ... so if for each integer prime $p$, there is exactly one $v$ above it in $\mathcal{O}_K$, then the $|a|_v$ completely and uniquely determine $|a|_p$ ... and the above method still works.