Show that $f(z) = c \sin (\pi z)$ 
Let $f: \mathbb{C} \to \mathbb{C} $ be an entire function  such that  $f(0)=0, f(z+1) = -f(z)$ and $|f(z)| \leq e^{\pi |\text{Im}(z)|}$ for all $z \in \mathbb{C}$. Show that $f(z) = c \sin (\pi z)$, for some $c \in \mathbb{C}$.

I'm trying to use Liouville's theorem.
If $z \notin \mathbb{Z}$, then:
$$\bigg|\frac{ f(z)}{(e^{i\pi z}- e^{-i\pi z})} \bigg| = 
\frac{|f(z)|}{|e^{i\pi z}- e^{-i\pi z}|} \leq
\frac{e^{\pi |\text{Im}(z)|}}{|e^{i\pi z}- e^{-i\pi z}|}  \leq
\frac{e^{\pi |\text{Im}(z)|}}{|e^{i\pi z}| - |e^{-i\pi z}|} $$
$$= \frac{e^{\pi |\text{Im}(z)|}}{e^{-\pi \text{Im}(z)} - e^{\pi \text{Im}(z)}} =  \frac{1}{e^{-\pi (\text{Im}(z) + |\text{Im}(z)|)} - e^{\pi (\text{Im}(z) - |\text{Im}(z)|)}}$$
I'm stuck here, I don't know how to prove that the last term is bounded.
And I don't know how to proceed when  $z \in \mathbb{Z}$.
Could somebody help me out?
New approaches are welcome.
 A: Note that $f$ and $g \colon z \mapsto \sin (\pi z)$ are both periodic with period $2$. Also, $g$ has simple zeros at the integers, and $f$ also vanishes at the integers. Thus $q = f/g$ has removable singularities at the integers, and considering them removed, $q$ is an entire function with period $2$. By continuity, $q$ is bounded on $K = \{ z = x + iy : \lvert x\rvert \leqslant 1,\, \lvert y\rvert \leqslant 1\}$. By periodicity, $q$ is bounded on $\{ z : \lvert \operatorname{Im} z\rvert \leqslant 1\}$.
It remains to see that $q$ is bounded on $\{ z : \lvert \operatorname{Im} z\rvert \geqslant 1\}$. Since $\sin \overline{w} = \overline{\sin w}$, we have $\lvert g(\overline{z})\rvert = \lvert g(z)\rvert$, hence it suffices to look at the lower half-plane. Now if $y \geqslant 1$ and $x\in \mathbb{R}$, then
$$\lvert e^{i(x-iy)} - e^{-i(x-iy)}\rvert = e^y\lvert e^{ix} - e^{-ix-2y}\rvert \geqslant e^y (1 - e^{-2y}) \geqslant (1 - e^{-2})e^y \geqslant \frac{1}{2}e^y,$$
whence $\lvert \sin (\pi z)\rvert \geqslant \frac{1}{4} e^{\pi\lvert \operatorname{Im} z\rvert}$ for $\lvert \operatorname{Im} z\rvert \geqslant 1/\pi$. And this yields $\lvert q(z)\rvert \leqslant 4$ for $\lvert \operatorname{Im} z\rvert \geqslant 1/\pi$. Together with the boundedness on the strip $\lvert \operatorname{Im} z\rvert \leqslant 1$, we see that $q$ is globally bounded, hence constant by Liouville's theorem.
