why $\frac{2}{|f(z)|}= 3R$?

Question

Prove that if $f$ is an entire function satisfying $|f(z) | \le \frac{1}{\Im z}$ for all $z$ , then $f \equiv 0$

My attempt : I got the answer here.

My doubt : My doubt is written in red colour given below

Let $w \in \mathbb{C}$ and let $R > |w|$. We first bound the analytic function $(z^2 - R^2)f(z)$ on the circle $|z| = R$.

If $|z| = R$ and $\operatorname{Re}z \ge 0$, then

$$|(z- R)f(z)| \le \frac{|z - R|}{\operatorname{Im}z} = \sec \theta$$

where $\theta \in [0, \dfrac{\pi}{4}]$ so $|(z - R)f(z)| \le \sqrt{2}$.

For $\operatorname{Re} z \le 0$ and $|z| = R$, we have

$$|(z + R)f(z)| \le \sqrt{2}$$

Hence,

$$\color{red}{|(z^2 - R^2)f(z)| \le 3R}$$

on $|z| = R$. By the maximum modulus principle the same bound holds in the interior of the disk $|z| \le R$ and so

$$|f(w)| \le \frac{3R}{|w^2 - R^2|}$$ Letting $R \to \infty$, we get the desired result.

My attempt : $|(z+R)|\le \frac{\sqrt 2}{|f(z)|}$ and $|(z-R)|\le \frac{\sqrt 2}{|f(z)|}$

take $g(z) = (z^2 -R^2 )f(z)$

Then $|g(z)|= |z+R||z-R|f(z)| \le \frac{\sqrt 2}{|f(z)|}\frac{\sqrt 2}{|f(z)|}|f(z)|=\frac{2}{|f(z)|}$

My confusion is that why $\frac{2}{|f(z)|}= 3R$ ?

Answer 1

$|f(z)| \leq \sqrt 2 \frac 1{|z+R|}$. So $|z^{2}-R^{2}||f(z)| \leq \sqrt 2 \frac {|z^{2}-R^{2}|} {|z+R|}=\sqrt 2 |z-R|\leq \sqrt 2 (2R)$. Now just note that $2 \sqrt 2 <3$.