Suppose $f$ is entire and $|f(z)| \leq 1/|Re z|^2$ for all $z$. Show that $f $ is identically $0$.

Question

This is a problem from my complex analysis textbook. The hint is to consider $g(z)=(z-iR)^2(z+iR)^2 f(z)$ and to show that $|g(z)| \leq 8R^2$.

This is what i have tried: Consider $Re z \geq 0$, then $|(z-iR)^2 f(z)|\leq \left(\frac{|z-iR|}{Re z}\right)^2\leq 2$

But then I don't know what to do next. Any help I'd appreciate :)

Answer 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 $R\longrightarrow\infty$ , $f(w)=0.$