Denote $\mathbb{Q}[\sqrt{n}i]=\{a+b\sqrt{n}i | a,b \in \mathbb{Q}\} \subset \mathbb{C}$ and $\mathbb{Z}[\sqrt{n}i]=\{a+b\sqrt{n} | a,b \in \mathbb{Z}\} \subset \mathbb{C}$.
The above two rings are integral domains (as subrings of fields).
Recall that a normal domain is an integral domain which is integrally closed in its field of fractions.
In the answers to this question, it is shown there that $\mathbb Z[\sqrt{n}]$ is integrally closed in $\mathbb Q(\sqrt{n})$ ($n\in\mathbb Z$, $n\neq1$) if and only if $n$ is square free and $n$ is not congruent to $1$ mod $4$ (or $n$ is a perfect square, in that case we have $\mathbb Z$ and $\mathbb Q$).
Is the above claim still valid for $\mathbb{Z}[\sqrt{n}i]$ and $\mathbb{Q}[\sqrt{n}i]$?
Edit: After receiving an answer to one of my 'old' versions of this question, if I am not wrong, $\mathbb{Q}[\sqrt{n}i]$ is a field, so the answer is that, trivially, for all $n \in \mathbb{N}$, $\mathbb{Q}[\sqrt{n}i]$ is integrally closed.
It remains to deal with $\mathbb{Z}[\sqrt{n}i]$ (if I am not wrong, its field of fractions is $\mathbb{Q}[\sqrt{n}i]$).
Does the same answer as in the above mentioned question still hold if we replace $\mathbb{Z}[\sqrt{n}]$ by $\mathbb{Z}[\sqrt{n}i]$? (I suspect that yes).
Any hints and comments are welcome!
