I have the function $f(z) = \frac{1}{e^z - 1} - \frac{1}{z}$. I need to determine wither the singularities $z_0$ of the function are removable, a pole, or essential.
$f(z) = \frac{1}{e^z - 1} - \frac{1}{z} = \frac{1}{z + z^2/2! + z^3/3! + ...} - \frac{1}{z} = \frac{-z^2/2! - z^3/3! - z^4/4!}{z(z + z^2/2! + z^3/3! + ...)}$
I don't really know where to go from here. I know $\frac{1}{e^z - 1}$ has a pole at $z_0 = 2n\pi i$. I don't know how to apply that here.
$$\frac1{e^z-1} -\frac1{z} = \frac1{z} \frac1{1+s(z)} - \frac1{z} $$
where $s(z) =\frac{z}{2} + \cdots$ is small for small values of $|z|$. Thus the term involving $s(z)$ may be Taylor expanded as
$$\frac1{z} (1-s(z) + \cdots) - \frac1{z}$$
and I hope you can see that the singularity at $z=0$ is removable.