$f$ entire, $f$ satisfies $|f(x+iy)|\leq\frac{1}{|y|}$ for all $x,y\in\mathbb{R}$. Prove that $f\equiv 0$. [duplicate]

First approach: The first thing that comes to mind is Jensen's Formula.

For any point $\omega \in \mathbb{R}$ and $r > 0$, we have:

$$\log |f(\omega)| \le \frac{1}{2\pi}\int_0^{2\pi} \log |f(\omega + re^{i\theta})|\,d\theta$$

Since, $\displaystyle |f(\omega + re^{i\theta})| \le \frac{1}{|r\sin \theta|}$, we have:

$$\log |f(\omega)| \le -\frac{1}{2\pi}\int_0^{2\pi} \log |r\sin \theta|\,d\theta = - \log r - \frac{1}{2\pi}\int_0^{2\pi} \log |\sin \theta|\,d\theta$$

Since, the integral on the RHS is a constant,

$$|f(\omega)| \le \frac{K}{r}$$

Hence, $f(\omega)$ must be identically zero on $\mathbb{R}$ and hence on $\mathbb{C}$.

Second approach: Let us first fix a $r > 0$ and $z = re^{i\theta}$, then

$$|(z^2 - r^2)f(z)| = 2r|\Im(z).f(z)| \le 2r$$

Then, $g(\omega) = (\omega^2 - r^2)f(\omega)$ is holomorphic in the unit disc $D_r = \{\omega : |\omega| \le r \}$

Hence, by Maximum Modulus Principle:

$$|g(\omega)| \le 2r \textrm{ in the disc } D_r$$

and hence, $$|f(\omega)| < \frac{2r}{|\omega^2 -r^2|} \textrm{ in the interior of the disc }D_r$$

Letting $r \to \infty$, it follows that $f$ must be identically $0$.

Here's another approach: Note that $z-\bar z=2iy.$ Thus

$$\tag 1 |(z-\bar z)f(z)| \le 2$$

for all $z.$ Let's write $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ Setting $z=re^{it},$ we get

$$\tag 2 (z-\bar z)f(z) = (re^{it} - re^{-it})(\sum_{n=0}^{\infty}a_nr^ne^{int}).$$

Play with $(2)$ a little to see it equals

$$\tag 3 \sum_{n=0}^{\infty}(a_nr^{n+1} - a_{n+2}r^{n+3})e^{i(n+1)t} -a_0re^{-it} - a_1r^2.$$

Let's now integrate the modulus squared of $(3)$ over $[0,2\pi]$using the orthogonality of the exponentials. We get, recalling $(1),$

$$\sum_{n=0}^{\infty}|a_nr^{n+1} - a_{n+2}r^{n+3}|^2 +|a_0r|^2 +|a_1r^2|^2 \le 4.$$

This inequality holds for all $r>0.$ Verify that this holds iff $a_n=0$ for all $n,$ i.e., iff $f\equiv 0.$