Check boundedness of $\frac{z^2}{(e^z-1)^2}$ near $0$

Question

Let $$f(z)=\frac{z^2}{(e^z-1)^2}$$ I have already checked the limit $\lim_{z\rightarrow 0}\left|f(z)\right| = 1$ using l'Hopital.

I'd like to check whether $f(z)$ is bounded on $V\setminus\{0\}$ for some neighborhood $V$ of $0$ to determine whether $f(z)$ has a removable or non-removable singularity at $0$.

Is there a tool or a different way to write the function such that we may determine the (non)-boundedness of $f$ easily?

Answer 1

Write down the Taylor series for small values of $\;|z|\;$ :

$$\frac{z^2}{(e^z-1)^2}=\frac{z^2}{\left(z+\frac{z^2}2+\frac{z^3}6+\ldots\right)^2}=\frac1{\left(1+\frac z2+\frac{z^2}6+\ldots\right)^2}=$$

$$=\frac1{1+z+\ldots}=1-z+z^2-z^3+\ldots$$

We got a Taylor series (and thus the function's analytic at zero) and thus the singularity at $\;z=0\;$ is a removable one and the function's bounded in some neighborhood of zero.

Of course, the way you did it is way faster and effective: as the limit exists finitely, the function is bounded... as simple as that

Answer 2

The nominator and the denominator in $f $ both have a zero of order $2$ at $0$. Hence there is an entire function $g $ such that $f=g $ on $\mathbb C \setminus \{0\}. $

Conclusion?

Answer 3

The correct expansion of the given $f(x)$ is $$f(x)=\frac{x^2}{(e^x-1)^2}=1-x+\frac{5x^2}{12}-\frac{x^3}{12}+\frac{x^4}{240}+...$$