What is behind $\frac{z^2-1+e^{-z}}{z^2(e^{-z}-1)}\sim\frac{1}{z^2}$?

Question

Determine the singularities of $f(z)=\frac{1}{e^{-z}-1}+\frac{1}{z^2},\:z_0=0$.

I know tha if I take $z=2\pi k\:\:k\in\mathbb{N}$ I have a pole. In order to determine the order an answer was presented to me:

$f(z)=\frac{1}{e^{-z}-1}+\frac{1}{z^2}=\frac{z^2-1+e^{-z}}{z^2(e^{-z}-1)}\sim\frac{1}{z^2}\:\text{as}\:z\to 0$

Question:

I am not seeing how this approximation was done $\frac{z^2-1+e^{-z}}{z^2(e^{-z}-1)}\sim\frac{1}{z^2}\:\text{as}\:z\to 0$. How was it done? What is the procedure behind it?

Thanks in advance!

Answer 1

Since $z^2-1+e^{-z}=-z+\frac32 z^2+o(z^2)$ while $e^{-z}-1=-z+\frac12 z^2+o(z^2)$, these functions' ratio is $$\frac{-z+\frac32 z^2+o(z^2)}{-z+\frac12 z^2+o(z^2)}=\frac{1-\frac32 z+o(z)}{1-\frac12 z-o(z)}=1+o(1).$$Hence $$\frac{z^2-1+e^{-z}}{e^{-z}-1}\sim1\implies\frac{z^2-1+e^{-z}}{z^2(e^{-z}-1)}\sim\frac{1}{z^2}.$$

Answer 2

Using Taylor expansion $$A=\frac{z^2-1+e^{-z}}{z^2(e^{-z}-1)}=\frac{z^2-1+\left(1-z+\frac{z^2}{2}-\frac{z^3}{6}+\frac{z^4}{24}+O\left(z^5\right) \right) } {z^2\left(\left(1-z+\frac{z^2}{2}-\frac{z^3}{6}+\frac{z^4}{24}+O\left(z^5\right) \right)-1\right)}$$ $$A=\frac{ -z+\frac{3 z^2}{2}-\frac{z^3}{6}+\frac{z^4}{24}+O\left(z^5\right)} {-z^3+\frac{z^4}{2}-\frac{z^5}{6}+\frac{z^6}{24}+O\left(z^7\right) }$$ Now, long division $$A=\frac{1}{z^2}-\frac{1}{z}-\frac{1}{2}-\frac{z}{12}+O\left(z^2\right)\sim \frac{1}{z^2}$$