# Laurent Series of a function in general

I need to find the laurent series of a function. the function in question has the interesting characteristic where the singularities of the function are:

z= 0, +i, -i

and my $z_0$ = 0

when my $z_0$ is not equal to zero hence in the other parts of the problem $z_0$ = z

I am trying to find laurent for one function but in multiple domains hence the different parts of the one problem, there are exactly 8 parts to my one problem each with a different domain, I have looked at other examples in the similar questions area BUT i have not found one similar enough to my question where as i say: $z_0$ = z which in my understanding kinda blows the equations that i want to use up...

mainly for my term $$\frac{-1}{z_0}$$ since my $z_0$ =z as stated above z=0 for one of my domains and this kills my entire Laurent series...

my original function that i am working with is called: $$f{(z)} = \frac{1}{z(z^2+1)}$$

anyone with any help please as i have said i have 8 parts to this one question that i WANT TO ANSWER.

Knowing the singularities is not enough. For example, the function $z \mapsto z^{-n}$, where $n$ is a positive whole number, all have singularities at $z=0$. However, they all have different Laurent Series. In fact, they are their own Laurent Series. In conclusion: you need to know the order of each pole.
Let's assume that $\operatorname{f}$ has simple poles at $z=0,\pm\operatorname{i}$. Then, we have: $$\operatorname{f}(z) = \frac{\operatorname{g}(z)}{z(z^2+1)}$$ where $\operatorname{g}$ is some function with the properties that $\operatorname{g}(0) \neq 0$ and $\operatorname{g}(\pm\operatorname{i}) \neq 0$. If $z=0,\pm\operatorname{i}$ are the only poles of $\operatorname{f}$ then $|\operatorname{g}(z)|<\infty$ for all $z$. One such example is your function: $\operatorname{g}(z) \equiv 0$ and $$\operatorname{f}(z) = \frac{1}{z(z^2+1)}$$
The find the Laurent Series about $z=0$, notice that $$\frac{1}{1+z^2} = 1-z^2+z^4-z^6\pm \cdots$$ for all $|z| < 1$. Hence: $$\frac{1}{z(z^2+1)} = \frac{1}{z}-z+z^3-z^5 \pm \cdots$$
For all $z \in \mathbb{C} \backslash \{0,\pm\operatorname{i}\}$, the Laurent Series will be the ordinary Taylor Series, since $\operatorname{f}(z)$ is well-defined at such points.
To find the Laurent Series around $z=\pm\operatorname{i}$, decompose $\operatorname{f}$ into partial fractions:
$$\frac{1}{z(z^2+1)} \equiv \frac{1}{z}-\frac{1}{2(z+\operatorname{i})}-\frac{1}{2(z-\operatorname{i})}$$