Suppose we have a function $f : U \to \mathbb{R}$, where $U = (0,1)^n \subset \mathbb{R}^n$ is the open box, and that $f(x_1,x_2,\cdots,x_n)$ is separately real analytic in each $x_i$.
Does there exist an extension of $f$ to a an open connected domain $D \subset \mathbb{C}^n$ such that $f$ is complex analytic in $D$? Is this extension unique?
By complex analytic, I mean that $f(z)$ coincides with its Taylor expansion.
Yes, more generally given a connected domain $U\subset \mathbb R^n$ and a real-analytic function $f:U\to \mathbb R$ there exists an extension $F:D\to \mathbb C$ of $f$ to to a connected open domain $D\subset \mathbb C^n$ and that extension is unique if $D$ is fixed.
The proof is very simple: given $p\in U$ the function $f$ has a development as a power series $\sum a_I (x-p)^I$ ($I=$ multiindex) converging for $|x_j-p_j|\lt \epsilon$ and you can extend this development to the polydisc $P_p\subset \mathbb C^n$ defined by $|z_j-p_j|\lt \epsilon$.
You thus obtain a holomorphic function $F_p\in \mathcal O(P_p)$ and these functions coincide on intersections $P_p\cap P_q$ , thus yielding the required extension $F:D=\bigcup _{p\in U} D_p\to \mathbb C$ of $f:U\to \mathbb R$.
(Connectedness is ensured by considering the connected component of $D$ containing $U$)