Meromorphic function written as a quotient of entire functions with real Taylor series coefficients Let $f$ be a meromorphic function on $\mathbb C$ with $f(z)\in \mathbb R$ for all $z\in \mathbb R$, except possibly at the poles of $f$. Show that there exists  entire functions $$g(z)=\sum_{i=0}^{\infty} a_i z^i \quad\text{and}\quad h(z)= \sum_{i=0}^{\infty} b_i z^i ,$$ such that $f(z)=g(z)/h(z)$, and $a_i,b_i\in \mathbb R$ for all $i$.
I know from the Weierstrass factorization theorem that there are entire functions $g,h$ such that $f=g/h$. I also know that if $f(z) = \sum_{i=0}^{\infty} a_i z^i$ is an entire function with $f(z)\in \mathbb R$ for every $z\in \mathbb R$, then $a_i\in \mathbb R$ for all $i$. Thus, it remains to show that $g$ and $h$ maps the real line to itself. But this seems impossible to show. Perhaps I am missing something obvious?
 A: One can give a proof using Weierstrass factors but we can give a direct proof from your observation that $f(z)=g(z)/h(z)$, $g,h$ entire; it is obvious that $g,h$ have real Taylor coefficients (at zero) iff they preserve the real axis since $n!a_n=g^n(0)$ and one can take all the derivatives at zero (or any real for that matter) to be the usual real ones of the restriction $g: \mathbb{R} \to \mathbb{R}$ if $g$ preserves the real axis
We can assume $g,h$ have no common zeroes as we factor them out, so the zeroes of $g$ are the zeroes of $f$ and the zeroes of $h$ are the poles of $f$. The hypothesis implies that all non-real such come in conjugate pairs with same multiplicities for both $g,h$ respectively.
Let $g^*(z)=\bar {g(\bar z)}$ the conjugate entire functions of $g$ (note that $g$ preserves the real axis iff $g=g^*$ since then they coincide on the real axis hence everywhere) and $h^*$ same for $h$; the above observation implies that $g^*$ has same zeroes with same multiplicities as $g$, same for $h^*,h$ and obviously $G(z)=g(z)g^*(z)$ preserves the reals since if $z=x$ real, $G(x)=g(x)\bar g(x)=|g(x)|^2$, while all its zeroes occur with even multipicities, same for $H$. Also since $G,H$ are not identical zero, there is $G(a)=a_1>0, H(b)=b_1>0$ for some real values $a,b$. 
Since $G,H$ have all zeroes with even multiplicity, both have unique entire square roots $G_1,H_1$ determined by the condition $G_1(a)=\sqrt a_1>0, H_1(b)=\sqrt b_1>0$. If $x$ real,  $G_1(x)$ is a square root of the non-negative number $|g(x)|^2$ (could be plus or minus, do not know apriori of course except for $a$ by choice), so it is real, and same with $H_1$, so $G_1,H_1$ preserve the real axis, so they have real Taylor coefficients at zero.
But since $f^*=f$ it follows that $f^2=\frac{G}{H}$, so $f(z)=\pm\frac{G_1(z)}{H_1(z)}$ and for each non-zero, non-pole of $f$, the sign $\pm$ is preserved in a small neighborhood. This means it is locally constant outside the poles and zeroes of $f$, and since those are isolated, it is constant everywhere, so indeed $f(z)=c\frac{G_1(z)}{H_1(z)}$ for $c$ fixed $\pm 1$. Done!
