To begin with, the definition of a rational function $R(x)$ can be found in Wiki. Suppose that $R(x)$ is defined in a subset $D \subseteq \mathbb{R}^n$. Then my question is:
Is any rational function $R(x)$ a real analytic function in its definition domain $D$ ?
Supplement: It is easy to verify that $R(x)$ is a real analytic function in its definition domain $\mathbb{R}^n$ if $R(x)$ is a polynomial.
Edit $1$:
Motivated by Daniel Fischer's comments, I have completed a proof of my problem by the methods of complex analysis. Though an approach of real analysis is my original ideal.
Proof: Since $R(z)$ is a complex function definded in its definition domain $D' \subseteq \mathbb{C}^n$, we need to prove that $R(x)$ is real analytic in its definition domain $D \triangleq D' \bigcap \mathbb{R}^n$. Fist, we need prove a result:
Suppose that $f(z)$ is a complex analytic function in an open subset $\Omega$ of $\mathbb{C}^n$ and it satisfies $f\left( \Omega \bigcap \mathbb{R}^n \right) \subseteq \mathbb{R}^n$. Then $f(x)$ is real analytic on $\Omega \bigcap \mathbb{R}^n$.
To prove this, we choose an arbitrary $x \in \Omega \bigcap \mathbb{R}^n$ ( Note that $\Omega \bigcap \mathbb{R}^n$ is an open subset of $\mathbb{R}^n$ ). Then there is an open neighborhood $U$ of $x$ and $U \subseteq \Omega$ such that $f(z) = \sum_{\alpha} b_{\alpha}(z-x)^{\alpha}$ for $z \in U$. Hence, we have $f(y) = \sum_{\alpha} b_{\alpha}(y-x)^{\alpha}$ for $y \in U \bigcap \mathbb{R}^n$. The coefficient $b_{\alpha}$ can be written as $b_{\alpha} = c_{\alpha} + i d_{\alpha}$. Thus $i \sum_{\alpha} d_{\alpha} (y-x)^{\alpha} = 0$, and $\sum_{\alpha} d_{\alpha} (y-x)^{\alpha} = 0$. That is, $\sum_{\alpha} d_{\alpha} (y-x)^{\alpha}$ is a zero polynomial in the open subset $U \bigcap \mathbb{R}^n$ of $\mathbb{R}^n$. Hence $d_{\alpha} = 0$. Therefore $f(x)$ is real analytic on $\Omega \bigcap \mathbb{R}^n$.
Next, we need to prove that $R(z)$ is a complex analytic function in its definition domain $D'$. Fortunately, I have found a theorem of Hartogs:
Suppose that $f(z)$ is a function on a region $\Omega$ of $\mathbb{C}^n$. Denote $g^j (z^j) \triangleq f(z^1, \cdots, z^j, \cdots, z^n)$. For each $j \in \{ 1, \cdots, n \}$, if $g^j (z^j)$ is an analytic function on an open subset $\Omega^j \triangleq \{ z^j : z \in \Omega \}$ of $\mathbb{C}$, then $f(z)$ is complex analytic on $\Omega$.
By Hartogs' theorem, we can easily get that $R(z)$ is complex analytic on $D'$. Therefore, we complete the proof. $\square$
Now, my question is this: How can I prove it only by the methods of real analysis ?
