 Solution

Put $g=1/f$. The hypothesis $|f(z)|\to \infty$ gives $g(z)\to 0$ as $z\to a$, so defining $g(a)=0$ makes $g$ continuous on $U$ and analytic there by the stated assumption. Its zero at $a$ has some finite order $m\ge 1$, and henceThereforehas a pole of order $m$: its Laurent series has $c_{-m}\ne 0$ and $c_n=0$ for $n<-m$.

Now let $f$ be entire and tend to infinity at infinity. The functiontends to infinity as $z\to 0$, so the preceding argument says that $h$ has a pole at zero. If the Taylor series of $f$ is $f(w)={\sum }_{n\ge 0}{a}_n{w}^n$, thenA pole has only finitely many negative powers, so $a_n=0$ for all sufficiently large $n$. Thus $f$ is a polynomial.