= Solution
Cauchy's formula gives $|f'(z_0)|\le M_R/R$ for an <entire function> bounded by $M$ on every radius-$R$ circle. If $f$ is globally bounded, letting $R\to\infty$ gives $f'=0$, proving Liouville's theorem.
Set $h(w)=f(1/w)$. Its <limit> at zero makes the singularity removable, so
$$
h(w)=a+c_1w+c_2w^2+\cdots.
$$
Consequently $f(z)=a+c_1/z+O(z^{-2})$ and $z^2f'(z)\to-c_1=:b$.
The <argument principle> on a large circle gives number of zeros minus poles equal to the winding number of $g$, which is zero because $g\to1$. Thus $m=n$. The <function>
$$
G(z)=g(z)\frac{\prod_i(z-p_i)}{\prod_j(z-q_j)}
$$
has removable singularities everywhere and tends to one at infinity. Liouville gives $G=1$, hence
$$
\boxed{g(z)=\frac{\prod_j(z-q_j)}{\prod_i(z-p_i)}.}
$$
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