# Finding Taylor/Laurent series for complex functions? - confused about general method

I'm confused about the correct method for finding Taylor/Laurent series for complex functions. For example, I'm given this function $$f(z) = \frac{z^2-1}{(z+2)(z+3)}$$ and I need to find the Taylor/Laurent series centered at $0$ for all $z \in \mathbb{C}, |z| \neq 2, 3$. From my understanding, the technique is to split the problem into three regions: (i) $|z| < 2$, (ii) $2 < |z| < 3$ and (iii) $|z| > 3$, and then manipulate the denominator to obtain a geometric series. However, I'm having trouble understanding what series representation is valid in what region. For example, for (i) $|z| < 2$ I would do first: $$f(z) = \frac{z^2-1}{(z+2)(z+3)} = \bigg( \frac{1}{2+z} - \frac{1}{z+3} \bigg) z^2 - 1.$$ Then the first term I would write it as $$\frac{1}{2+z} = \frac{1}{2} \frac{1}{1 + \frac{z}{2}} = \frac{1}{2} \sum_{n=0}^{\infty} (-1)^n \big(\frac{z}{2}\big)^n.$$ Similarly for the second term $$- \frac{1}{z+3} = - \frac{1}{3} \frac{1}{1 + \frac{z}{3}} = - \frac{1}{3} \sum_{n=0}^{\infty} (-1)^n \big( \frac{z}{3} \big)^n.$$ So for the first region I would have $$f(z) = \bigg(\frac{1}{2} \sum_{n=0}^{\infty} (-1)^n \big(\frac{z}{2}\big)^n - \frac{1}{3} \sum_{n=0}^{\infty} (-1)^n \big( \frac{z}{3} \big)^n \bigg) z^2 - 1.$$ Now, for the second region, where $2 < |z| < 3$, I'm not sure what I would have to change in the above representation. How do I know when a certain series is valid?

In the first region you have, as observed, $|\frac z2|<1$ and $|\frac z3|<1$.

In the second region, $|\frac 2z|<1$ and $|\frac z3|<1$.

In the third region, $|\frac 2z|<1$ and $|\frac 3z|<1$.

For the geometric series you use that, for example, $$\frac1{z+2}=\frac12·\frac1{1+\frac z2} = \frac1z·\frac1{1+\frac 2z}.$$