Essential singularity question for $ f(z) = \frac{1}{\sin(z)}$

Question

I have a question regarding essential and isolated singular points.

So I am analyzing the behavior of various functions at $\infty$. I'm looking at this one right now:

$$ f(z) = \frac{1}{\sin(z)}$$

So the condition for an essential singular point in the book is:

$f(z)$ has an essential singularity iff $f(1/w)$ has an essential singularity at $w=0$

So I try to solve for $f(1/w)$, which yields:

$$ f(1/w) = \frac{1}{\sin\left(\frac{1}{w}\right)}$$

Wouldn't this be an essential singularity because it doesn't converge to $\infty$ or any particular value when you take the $\lim_{w \rightarrow 0} $? The back of the book said it is not an isolated point which I don't understand. Are they the same thing?

Answer 1

An essential singularity is one kind of isolated singularity. $f$ has an isolated singularity at $z_0$ if $f$ is not analytic at $z_0$, but $f$ is analytic in some punctured disk $0<|z-z_0| < r$ for some real number $r$. In your case, it is clear that $0$ is not an isolated singularity of $\sin(1/w)$ so it can't be an essential singularity.

Answer 2

$1/\sin{z}$ has poles at the points $\{n\pi:n \in \mathbb{Z}\}$. Therefore $1/\sin{(1/z)}$ has poles at the points $\{1/(n\pi):n \in \mathbb{Z}\}$. This set has $0$ as an accumulation point, so there is no open disk containing $0$ that does not contain a singularity of $1/\sin{(1/z)}$. Therefore $z=0$ is not an isolated singularity.

This is in contrast to the situation for $\sin{(1/z)}$, which is analytic on $\mathbb{C} \setminus \{0\}$, so $0$ is an essential singularity of it.

Answer 3

Let $g (w)=f (1/w) $. If $w_n=\frac {1}{n \pi}\quad ,(n \in \mathbb N) $, then $w_n $ is an isolated singularity of $g $. Therefore $w=0$ is an accumulation point of singularities of $g $.