A complex valued function $f,$ defined on an open set $E$ in the plane $\mathbb R^{2}$, is said to be real analytic on $E$, if to every point $(s_{0}, t_{0}) \in E,$ there corresponds an expansion of the form $$F(s, t)= \sum_{m,n=0}^{\infty} a_{mn} \, (s-s_{0})^{m} \, (t-t_{0})^{n}, \hskip.1in a_{mn} \in \mathbb C$$ which converges absolutely for all $(s,t)$ in some neighbourhood of $(s_{0}, t_{0}).$
Assume that $f:E\subset \mathbb R^2 \to \mathbb C$ is real analytic at $(s_0, t_0)\in E.$ We put $$f(x)= \text{Re}f(x) + i\text{Im}f(x), x\in E$$ where $\text{Re}f(x)$ and $\text{Im} f(x)$ denotes the real and imaginary part of $f(x)\in \mathbb C$ respectively.
Can we say that $\text{Re}f:E \to \mathbb R $ is real analytic at $(s_0, t_0)$?
Yes. To see this, note that $\overline{F(s,t)}$ is also analytic, and thus so is $\operatorname{Re}(F(s,t))=(F(s,t)+\overline{F(s,t)})/2$.
The same argument works in any dimension, not just two.
Bear in mind that while $F$ extends to an analytic function on a region of $\mathbf C^2$, the real and imaginary parts of this extension will usually not be analytic (unless $F$ is constant).