Suppose $f$ is entire and $|f(z)| \leq 1/|Re z|^2$ for all $z$. Show that $f $ is identically $0$.
Solution 1:
Fixed $w\in\mathbb{C}$, let $R>|w|$ and $g(z)=(z^2+R^2)^2f(z)$
If $z\in \partial B(0,R)$ then $|g(z)|\le 4R^2$ (Note that $z^2+R^2=2z \,\mathfrak{Re}(z)$)
Applying Maximum modulus principle we have that $$|g(z)|\le 4R^2 \;,\;\forall z\in B(0,R)$$
Therefore $|f(w)|\le\displaystyle \frac{4R^2}{|R^2+w^2|^2}$ taking limit $R\longrightarrow\infty$ , $f(w)=0.$