How to check the type of singularity

Question

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$$f(z) = \frac{e^{2z}-1}{z^2 }$$

Can you please explain how here or in other similar tasks I can find the type of singularities? I tryed to make Taylor series on this function and got this enter image description here And I think the answer is removable because there are no negative powers as in Lauren series similar to power series.

But I am also confused because If I do this request, I got this answer enter image description here "Simple pole"

Can you help me what is correct answer and how I can find and check solution?

Answer 1

For each $z\in\Bbb C$, $e^{2z}-1=2z+\frac{(2z)^2}{2!}+\frac{(2z)^3}{3!}+\cdots$, and therefore, if $z\ne0$,$$\frac{e^{2z}-1}{z^2}=2\color{red}{z^{-1}}+2+\frac43z+\cdots$$So, yes, $f$ has a simple pole at $0$.

Answer 2

Here's the way I see types of singularities:

1. A "pole" is basically where an otherwise asymptote would occur, but when you go beyond two dimensions, the vertical line that would be the asymptote is now literally a pole shooting up out of the complex plane. These occur when you have a $0$ denominator situation and the order of the pole is the multiplicity of the solutions that makes the denominator equal to zero. In your case it's a little harder to see since you have to consider the expansion of $e^{2z}$, which gives cancellation with the $-1$ in the numerator, leaving only non-constant terms in the numerator, so $z^2$ in the denominator cancels down to $z$, hence the simple (order 1) pole at $z=0$.
2. A "removable" singularity is one that appears to be a pole put cancels out completely in the end. For example, if we instead had ${{e^{2z}-1}\over{z}}$ then there wouldn't actually be a pole; think of it as a "$0\over0$" indeterminate form versus "$\#\over0$," which would give an infinite value instead of a finite one.
3. "Essential" singularities are the hard/"other" ones, usually when you have something like $e^{1\over z}$. This is an oversimplification, but if you consider the expansion of $e^{1\over z}$, there is no highest negative power, something to think about.

I hope this helps, Complex Analysis is awesome!