Show that there is no non-constant bounded analytic function on $\mathbb{C}\setminus \mathbb{Z}$.

In this homework problem i tried to proceed in the following way: If possible, let $f$ be a non-constant bounded analytic function on $\mathbb{C}\setminus \mathbb{Z}$. Define $F(z)=0$ if $z\in \mathbb{Z}$ and $f(z)$ otherwise. Then $F$ is analytic on $\mathbb{C}$, i.e., $F$ is entire and also by construction it is bounded. Then by Liouville's theorem $F$ is constant. Which in turn would mean that $f$ is constant. Which is a contradiction.

Am i correct? Can i define such a function $F$? I would be more than happy to get a feedback and if i am incorrect please hint me an idea to proceed. Thanks in advance.

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    $\begingroup$ You can't blindly set $F(z)=0$ at $z \in \mathbb{Z}$ and expect $F$ to be analytic, almost always this won't even produce a continuous function. $\endgroup$ – Omar Antolín-Camarena May 22 '14 at 21:54
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    $\begingroup$ Hint: what kind of singularity has $f$ at $0$? $\endgroup$ – Daniel Fischer May 22 '14 at 21:56
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    $\begingroup$ What kinds of singularities do you know? Which kinds can be excluded by the conditions on $f$? $\endgroup$ – Daniel Fischer May 22 '14 at 22:05
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    $\begingroup$ What do you know about the behaviour of a holomorphic function near an essential singularity? $\endgroup$ – Daniel Fischer May 22 '14 at 22:16
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    $\begingroup$ What do you know about the mapping behaviour? Does Casorati-Weierstraß ring a bell? $\endgroup$ – Daniel Fischer May 22 '14 at 22:30

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