Let $K$ be a number field with $d \in \mathcal{O}_K \setminus \{0\}$, $\sqrt{d} \notin \mathcal{O}_K$. Consider the group of units $\mathcal{O}_K[\sqrt{d}]^\times$ and denote conjugation of elements of $\mathcal{O}_K[\sqrt{d}]$ by $\bar{.}$. I am reading a paper (see lemma 5 here) that considers the following subgroup of $\mathcal{O}_K[\sqrt{d}]^\times$: $$U = \left\{u \in \mathcal{O}_K[\sqrt{d}]^\times \middle| \left(\frac{u + \bar{u}}{2}, \frac{u - \bar{u}}{2\sqrt{d}}\right) \in \mathcal{O}_K^2 \right\}.$$
My question is: do we not have that $U = \mathcal{O}_K[\sqrt{d}]^\times$ always?
If $u = x + y\sqrt{d}$ then $\left(\frac{u + \bar{u}}{2}, \frac{u - \bar{u}}{2\sqrt{d}}\right) = (x, y)$ so the condition is simply that $(x, y) \in \mathcal{O}_K^2.$ My understanding is that $\mathcal{O}_K[\sqrt{d}]^\times$ is the invertible elements of $\mathcal{O}_K[\sqrt{d}]$ which are of the form $x + y\sqrt{d}$ for $(x, y) \in \mathcal{O}_K^2$ by definition, so this subgroup definition is redundant - we have $U = \mathcal{O}_K[\sqrt{d}]^\times$ always.
In the event we consider elements of $\mathcal{O}_L$ for $L = K(\sqrt{d})$ then I know that we may not have $\mathcal{O}_L = \mathcal{O}_K[\sqrt{d}]$ in which case I can see why we would look at this particular subgroup, but this doesn't appear to be the case here.
Your definition of $U$ is not what is written in the paper: each time you write $\overline{u}$, it should be $u^{-1}$ instead, which is clearly not the same.