Does $\sin\left(\frac1z\right)$ violates Riemann's theorem for isolated singularity??? (Yes/No). I'm not getting plz help

Question

$f(z)=\sin\left(\frac1z\right)$ has an isolated singularity at $z=0$, further $|f(z)|\le1$ (i.e, $f$ is bounded) in any deleted neighbourhood of $z=0$. But $z=0$ is an essential singularity not removable.

Answer 1

It is not true that $\left\lvert\sin\left(\frac1z\right)\right\rvert\leqslant1$ for each $z\in\mathbb C\setminus\{0\}$, although it is true when $z$ is real. Actually, it turns out that $z\mapsto\left\lvert\sin\left(\frac1z\right)\right\rvert$ is unbounded on any neighborhood of $0$. In fact, if $\lambda\in\mathbb R\setminus\{0\}$, then$$\sin\left(\frac1{\lambda i}\right)=-\sin\left(\frac i\lambda\right)=-\frac{e^{-1/\lambda}-e^{1/\lambda}}{2i}$$and$$\lim_{\lambda\to0^+}-\frac{e^{-1/\lambda}-e^{1/\lambda}}2=-\infty.$$