Types of singularities, why is this an essential singularity

Question

I'm learning about different types of singularities:

1) removable singularities.
2) poles.
3) essential singularities.

I think I understand 1 and 2 but I don't really get the 3rd one, could someone please explain me?

1) For a removable singularity $|f|$ is bounded near $z_o$ so as an example we'd have a removable singularity at $z_0=0$ for $\frac{sin(z)}{z}$.
2) For a pole $|f|$ goes to infinity near $z_o$ , so as an example we'd have a pole at $z_0=0$ for $\frac{1}{z}$.
3) For an essential singularity $|f|$ doesn't go to infinity near $z_o$ nor is bounded here. In the book they give the example: we'd have an essential singularity at $z_0=0$ for $e^{\frac{1}{z}}$.

I don't get this last one, since in my head: (I know it's not mathematically correct what I'm gonna write but I do it just to explain my problem) $e^{\frac{1}{0}}=e^\infty=\infty$

Thanks in advance.

Answer 1

Well! As far as Complex exponential is concerned $e^\infty$ is not actually $\infty$ it is not defined to be precise.

While understanding types of singularities you should also learn them by their laurentz series expansions That way it would be easier and you will not get confused.

Hope it works

Answer 2

Note that a function $f(z)$ can be expanded as a Laurent series about a singularity $z_0$ as: $$f(z) = \sum_{n=0}^{\infty} a_n(z-z_0)^n + \sum_{n=1}^{\infty} \frac{b_n}{(z-z_0)^n}$$

In case, the second sum (the principal part) has infinitely many terms, then the singularity $z=z_0$ of $f(z)$ is called an essential singularity.

Hence, the given function: $$f(z) = e^{\frac1{z}} = \sum_{n=0}^{\infty} \frac{1}{n!z^n}$$ has $z=0$ as an essential singularity.