Essential singularity $z=a$

Question

Suppose $f$ has an essential singularity at $z=a$, is it true that $f(z)$ is not equal to 3 for $z$ not equal to $a$, then $\frac{1}{f(z)-3}$ is bounded in some punctured disk $D'(a,s)$ for some $s>0$?

Answer 1

No.

By the Casorati–Weierstrass theorem, if $f$ has an essential singularity at $z=a$, then $f$ gets arbitrary close to every complex number in a neighborhood of $a$. So if $f(z) \neq 3$ for $z=a$, $f(z)$ still gets arbitrarily close to $3$ in any punctured neighborhood of $z =a$ so in particular there exists a sequence $(z_n)$ with limit $a$ such that $|f(z_n) - 3| \to 0$ as $n \to \infty$. Hence, $\lim_{n \to \infty} \frac{1}{|f(z_n) - 3|} = \infty$.