$$f(z)=\frac{z^3+1}{z^2(z+1)} $$
has singularities $z=0, z=-1$ right?
How can I determine the type of singularity of this points?
Classifications:
1.)Removable pole (Then $f(z_0)$ is bounded, $f(z)$ has a limit if $z \to \infty$
2.) Pole of order $m \implies |f(z)|\to \infty $ as $z \to z_0$
3.) Essential singularity : not bounded, does not go to infinity.
(Another way to descibre is to look at the coefficients of the Laurent Series.
Best Answer
A function $f$ which is holomorphic for all $z$ near $z_0 \in \mathbb{C}$ (with $z \neq z_0$) has a pole of order $m>0$ at $z=z_0$ if and only if
$$ \lim_{z \to z_0} (z-z_0)^m f(z) $$
is finite and nonzero. If $f$ has a singularity at $z=z_0$ and
$$ \lim_{z \to z_0} f(z) $$
is finite and nonzero then the singularity is removable, and vice versa.