= Solution
Put $g=1/f$. The hypothesis $|f(z)|\to\infty$ gives $g(z)\to0$ 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 $m\geq1$, and hence
$$
g(z)=(z-a)^m q(z),
\qquad q(a)\ne0.
$$
Therefore
$$
f(z)=(z-a)^{-m}\frac1{q(z)}
$$
has a pole of order $m$: its Laurent series has $c_{-m}\ne0$ and $c_n=0$ for $n<-m$.
Now let $f$ be entire and tend to infinity at infinity. The function
$$
h(z)=f(1/z)
$$
tends to infinity as $z\to0$, so the preceding argument says that $h$ has a pole at zero. If the <Taylor series> of $f$ is $f(w)=\sum_{n\geq0}a_nw^n$, then
$$
h(z)=\sum_{n\geq0}a_nz^{-n}.
$$
A pole has only finitely many negative powers, so $a_n=0$ for all sufficiently large $n$. Thus $f$ is a <polynomial>.
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