Show $\frac{f(z)}{\sin(\pi z)} = \sum_{-\infty}^{\infty} \frac{(-1)^nf(n)}{(z-n)}$ Let $f(z)$ be entire function satisfying $|f(x+iy)| \leq Ce^{a|y|}$ for $C > 0$ and $a \in (-\pi, \pi).$
Show $\frac{f(z)}{\sin(\pi z)} = \sum_{-\infty}^{\infty} \frac{(-1)^nf(n)}{(z-n)}$
All I have noticed in this problem is that the poles on the LHS matches the poles of the RHS. I thought about using residue theorem, but I don’t know how that will determine if two functions are equal. Most importantly, I don’t understand the bounded condition. Please help me, any hint will be nice. 

Put $g(z)=f(z)\sin(\pi z)^{-1}$. Since $f(z)$ is entire, $g(z)$ has simple poles at $a_n=n$ with residues $b_n=\text{Res}(g(z); n) = f(n)\text{Res}(\csc(\pi z); n) =\frac{(-1)^n}\pi f(n)$. On the contours suggested by Yiorgios S. Smyrlis, $|\sin(\pi z)|$ is bounded below by either $\cosh(\pi\text{Im}z)$ or $|\sinh(\pi(n\pi + \pi/2))|$ so $|g(z)|$ is bounded by $Ce^{a|\text{Im}z|}\cosh(\pi|\text{Im}z|)^{-1}$ on the vertical sides of $\gamma_n$  or $Ce^{a|\text{Im}z|}\sinh(\pi|\text{Im}z|)^{-1}$ on the horizontal sides. Applying L'Hopital's rule shows $Ce^{a|\text{Im}z|}\sinh(\pi|\text{Im}z|)^{-1}$ tends to infinity like $Ce^{a|\text{Im}z|}\cosh(\pi|\text{Im}z|)^{-1}$. By the hypothesis on $a$, this shows that for $z\in \gamma_n$, $|g(z)|\to 0$ as $|\text{Im}z|\to \infty$, while remaining bounded near the real line. In particular, $\left|\frac w{2\pi i}\int_{\gamma_n}\frac{g(z)}{z-w}\right|\to 0$ as $n\to \infty$. If $w$ is not a pole of $g$ and $\gamma_n$ encircles $w$, the Residue Thm. gives
$$\frac 1{2\pi i}\int_{\gamma_n}\frac{g(z)}{z-w}\text{d}z=g(w)+\sum_{a_k \text{ in }\gamma_n} \frac{b_k}{a_k-w}=g(w)+\frac 1\pi\sum_{|k|\le n}\frac{(-1)^kf(k)}{k-w}$$ 
Letting $n\to \infty$ yields
$$\frac{f(z)}{\sin(\pi z)}=\frac 1\pi \sum_{k=-\infty}^\infty\frac{(-1)^kf(k)}{w-k} $$
