How to show that $z=0$ is an essential singularity of $\sin(\frac{1}{z})$

Question

My attempt: $0$ is clearly a singularity since $f$ is unbounded as $z\rightarrow 0$.

Using Taylor expansion we see that $\sin(\frac{1}{z})= -\frac{1}{z} + \frac{1}{3!z^3} - \frac{1}{5!z^5}+\dots=\frac{1}{z}[-1 + \frac{1}{3!z^2} - \frac{1}{5!z^4}+\dots] $ so that as $z\rightarrow 0$ we have $f(z)$ is unbounded and therefore $0$ is not a removable singularity.

My question: How do we know that this is not a pole and hence it is an essential singularity. If I check out $1/f$ nothing really simplifies and it's just $1$ over a messy expansion. I can see that the polynomial on the bottom will diverge faster than the linear $z$ will converge to $0$ on the top so that $1/f$ diverges so it can't be a pole, but this doesn't seem rigorous enough.

Are there any better ways to see this, or to think about this problem?

Answer 1

If it was a pole, then this would be valid: $\lim_{z\to 0}\sin\left(\frac{1}{z}\right)=\infty$.

Then, for every sequence $z_n\to0$ ($n\to\infty$) we would have $\lim_{n\to\infty}\sin\left(\frac{1}{z_n}\right)=\infty$.

However - compare that with $z_n=\frac{1}{\frac{\pi}{2}+n\pi}\to 0$ and $\sin\left(\frac{1}{z_n}\right)=(-1)^n$.

Answer 2

If $h$ is an analytic function on a punctured disc $D - \{c\}$, then $h$ has an absolutely convergent Laurent expansion

$$h(z) = \sum\limits_{n \in \mathbb Z} a_n (z-c)^n$$

for all $z \neq c$ sufficiently near $c$.

If $a_n = 0$ for all $n < 0$, the singularity at $h$ is removable.

If $a_n \neq 0$ for finitely many $n < 0$, the singularity is a pole.

If $a_n \neq 0$ for infinitely many $n < 0$, the singularity is essential.

The proofs of these statements are easy, but depend on what definitions you are using of pole and essential singularity.