 Solution

Cauchy's formula gives $|f^{\prime }(z_0)|\le M_R/R$ for an entire function bounded by $M$ on every radius-$R$ circle. If $f$ is globally bounded, letting $R\to \infty$ gives $f^{\prime }=0$, proving Liouville's theorem.

Set $h(w)=f(1/w)$. Its limit at zero makes the singularity removable, soConsequently $f(z)=a+c_1/z+O(z^{-2})$ and $z^2{f}^{\prime }(z)\to -c_1=:b$.

The argument principle on a large circle gives number of zeros minus poles equal to the winding number of $g$, which is zero because $g\to 1$. Thus $m=n$. The functionhas removable singularities everywhere and tends to one at infinity. Liouville gives $G=1$, hence