Show that $f(z) = C \cdot \sin(\pi z)$. Suppose $f: \mathbb{C} \rightarrow \mathbb{C}$ is an entire function such that $|f(z)| \leq K e^{\pi |\text{Im}(z)|}$ for some $K > 0$, and that $f(n) = 0$ for all $n \in \mathbb{Z}$. Show that $f(z) = C \cdot \sin (\pi z)$.
I have seen very similar problems on here asking this question with (what I think is) a stronger assumption. That is, assuming $f(z+1) = -f(z)$ for all $z \in \mathbb{C}$. The argument itself is an application of Liouville's theorem. But with only assuming that $f(n) = 0$ for $n \in \mathbb{Z}$ I am not able to recreate the argument.
Bounding the function $g(z) = f(z)/\sin(\pi z)$ on a strip like $|\text{Im}(y)| \geq 1$ comes down to bounding the sine function from below on this region. However we cannot do the same for $|\text{Im}(y)| \leq 1$. How can I proceed?
Here is the link to the post I mentioned above: Show that $f(z) = c \sin (\pi z)$.
 A: The idea is that $|\sin(z)|$ is not only bounded below for $|\operatorname{Im}(z)| \ge c > 0$, but also on the lines $\operatorname{Re}(z) = (k+1/2)\pi$ for all integers $k$.
Consider the rectangles
$$
 R_n = \{ x+ iy \mid -n-\frac 12 \le x \le n + \frac 12, -n \le y \le n \}
$$
where $n$ is a positive integer, and note that
$$
 |\sin(x+iy)|^2 = \sin^2(x) + \sinh^2(y) 
$$
for $x, y \in \Bbb R$.
On those parts of the boundary of $R_n$ with $|y| \ge 1/\pi$ you can proceed as in the referenced answer:
$$
|\sin(\pi z)| \ge |\sinh( \pi y)| \ge \frac 14 e^{\pi |y|}
$$
and therefore
$$
 |g(z)| = \frac{|f(z)|}{|\sin(\pi z)|} \le 4 K \, .
$$
And on those parts of the boundary of $R_n$ with $|y| \le 1/\pi$ is $x=\pm(n + 1/2)$ so that
$$
|\sin(\pi z)| \ge |sin(\pi x)| = 1
$$
and therefore
$$
 |g(z)| = \frac{|f(z)|}{|\sin(\pi z)|} \le e K \, .
$$
Using the maximum modulus principle it follows that $|g(z)| \le 4K$ in every rectangle $R_n$, so that $g$ is bounded in $\Bbb C$, and consequently, constant.
