Singular Points and their Classification Suppose I have these functions:


*

*$\frac{1}{z-z^3}$

*$\frac{1}{(z^2+4)^2}$

*$\frac{e^z}{1+z^2}$
I need to find the singular points of these functions, and classify them ($\infty$ included). So far, I understand that finding the singular points themselves is as simple as setting the denominator to be $0$. However, I do not comprehend at all how to classify them. 
Either way, I decided to try something to see where it'd lead me. For the first function, the singular points are at $z = -1, z = 0, z = 1$, because the function can be rewritten as
$$ \frac{1}{z(1-z^2)}$$
I was told that if the limit is infinite for the singular point in question, then that point is a pole. How do I determine which order of a pole it is? Also, I was told that I could also expand these in a Laurent series and find the principal part, but I fail to see how that would help me (and I believe it would take too long anyways).
Any suggestions/comments? Help would be much appreciated.
 A: There are three kinds of singularities:
 1. Removable Singularity 
 2. Poles
 3. Essential Singularity
The Limit Way of Classification: 
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.
If the singularity is not of the above kinds, then it is essential. 
Laurent Series Method
Personally, I think this is easier in most cases and doesn't take that long. 
Once you expand your function in its Laurent Series form, a simple analysis of the principal part - that is the part where the ($z$ - $z_0$) occurs in the denominator - will give away the answer. 
If there are no terms in the principal part, then your singularity is removable. 
Finitely many terms mean it is a pole. 
The greatest order m such that ($z - z_0)^m$ is in the denominator - is the order of the pole. If it is 1, then it's known as a simple pole. 
If there are infinite such terms, then the singularity is essential.
Hope this helps. 
Example for Laurent Series Usage :
$f(z)$ = $\frac{1}{z-z^3}$ = $\frac{1}{z(1-z^2)}$ = $\frac{1}{z}$$(1+z^2+z^4+...)$ 
(We've used $\frac{1}{z}$ = $1+z+z^2+..$ for $z^2$ )
= $\frac{1}{z}$ + $z+z^3+..$ 
Here, the principal value is just $\frac{1}{z}$. Therefore, this singularity is a pole of order 1. 
