Prove that there doesn't exists a holomorphic function on $U^c$ satisfying this particular condition I am trying some questions in complex analysis and got struck upon this particular problem.

Question: Prove that there doesn't exists a holomorphic function f on $U^c$ such that $|f(z)| \to \infty$ as $|z|\to 1$ .

Edit: U is open unit disc with center at origin .
I tried this problem by assuming that there exists such a holomorphic function and trying to find a contradiction.
But unfortunately I am unable to find what result should I use to contradict.
I am not good at solving problems in complex analysis but trying to learn.
Your help would be really appreciated.
Thanks!!
 A: Conisder $g=1/f$ so $g(z) \to 0, |z| \to 1, |z|>1$. By the reflection principle $g$ extends to a holomorphic function defined on a neighborhood of the unit circle so to a domain $|z| > 1-\delta, \delta >0$ which is zero on the unit circle; by the identity principle $g$ is $0$ everywhere, hence $f$ is infinity everywhere, contradiction!
Edit later - as asked a concrete way of extension is to take $g(w)=g(\frac{1}{w}), 0< |w| <1$; then $g$ is clearly analytic in the punctured unit disc and $g(w) \to 0, |w| \to 1, |w| \ne 1$;
extending $g(w)=0$ when $|w|=1$, gives us a continuous function $g$ on a domain $D$ that can be split as $D_1 \cup C \cup D_2$, $g$ analytic on the domains $D_1, D_2$ and $C$ a smooth Jordan curve that is a common intersection of (in general part of) the boundaries of $D_1$ and $D_2$; the symmetry principle says that $g$ is analytic on $D$ and the proof can be done using Morera but also using Cauchy which I prefer (note that the reflection principle usually refers to the case when $C$ is part of a circle/line indeed like here but the proof below works for any nice enough - eg rectifiable - Jordan curve and any $g$ with the stated hypothesis)
Proof of the symmetry principle using Cauchy:
pick some $|w|=1$ and a small circle arc $C_w$ containing it; make small domains $U_1 \subset D_1, U_2 \subset D_2$ by joing the ends of $C_w$ with small arcs $C_{1,2}$ contained in $D_{1,2}$ with same ends as $C_w$ and let $\Gamma_{1,2}$ the two closed paths (traversed counterclockwise) that are the boundary of $U_{1,2}$ and $g_{1,2}(z)=\frac{1}{2\pi i}\int_{\Gamma_{1,2}}\frac{g(\zeta)d\zeta}{\zeta-z}$
Then by Cauchy $g_1(z)=g(z), z \in U_1, g_1(z)=0, z \in U_2$ and same for $g_2$ with roles reversed; adding them and noting that that $C_w$ is traversed in opposite directions so that part of the two integrals cancels out, we get that $g_1(z)+g_2(z)=\int_{\Gamma=C_1 \cup C_2}\frac{g(\zeta)d\zeta}{\zeta-z}$ is an analytic function in the domain that is bounded by $\Gamma$ and for which $w$ is now an interior point, while $g_1+g_2=g, |z| \ne 1$, hence by continuity $g_1+g_2=g, |z|=1$ so indeed $g$ is analytic at $w$ too!
