Usbrth

Let $f(z)$ be entire function satisfying $|f(x+iy)| leq Ce^y$ for $C > 0$ and $a in (-pi, pi).$

Show $fracf(z)sin(pi z) = sum_-infty^infty frac(-1)^nf(n)(z-n)$

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 $b_n=textRes(g(z); n) = f(n)textRes(csc(pi z); n) =frac(-1)^npi f(n)$. On the contours suggested by Yiorgios S. Smyrlis, $|sin(pi z)|$ is bounded below by either $cosh(pitextImz)$ or $|sinh(pi(npi + pi/2))|$ so $|g(z)|$ is bounded by $Ce^cosh(pi|textImz|)^-1$ on the vertical sides of $gamma_n$ or $Ce^sinh(pi|textImz|)^-1$ on the horizontal sides. Applying L'Hopital's rule shows $Ce^sinh(pi|textImz|)^-1$ tends to infinity like $Ce^cosh(pi|textImz|)^-1$. By the hypothesis on $a$, this shows that for $zin gamma_n$, $|g(z)|to 0$ as $|textImz|to infty$, while remaining bounded near the real line. In particular, $left|frac w2pi iint_gamma_nfracg(z)z-wright|to 0$ as $nto infty$. If $w$ is not a pole of $g$ and $gamma_n$ encircles $w$, the Residue Thm. gives
$$frac 12pi iint_gamma_nfracg(z)z-wtextdz=g(w)+sum_a_k text in gamma_n fracb_ka_k-w=g(w)+frac 1pisum_frac(-1)^kf(k)k-w$$
Letting $nto infty$ yields
$$fracf(z)sin(pi z)=frac 1pi sum_k=-infty^inftyfrac(-1)^kf(k)w-k $$

Hint: use the given inequality to show that LHS remains bounded as $|z| to infty$ through non-integer values. Then verify that LHS $-$ RHS is an entire function and apply Loiville's Theorem.

Hint. Apply Mittag-Leffler's Theorem, to
$$
g(z)=fracf(z)sin pi z
$$
using the sequence of contours $gamma_n$, $ninmathbb N$, which are squares with vertices $(n+1/2)(pm 1pm i)$.