Decide which term represents the negative exponent on Laurent series.

Question

This might be a stupid question, but here we go. I cannot understand the logic behind each part of a Laurent series, for $n\geq0$ and $n<0$. Here is an example:

Find the Laurent series in the given domain:
$f(z)=(z^2+1)^{-1}, A={\{z\in C/1<|z-2i|<3\}}$
The solution I found was:
$\frac{1}{z^2+1}=\frac{1}{(z+i)(z-i)}$, using partial fractions:
$\frac{1}{2i}[\frac{1}{z+i}-\frac{1}{z-i}]\Rightarrow$ so, for each of these terms, we have:
(1) $\frac{1}{z+i}=\frac{1}{z+i+2i-2i}=\sum_{n=0}^{\infty}{(-1)^n\frac{(z-2i)^n}{(3i)^{n+1}}}$
(2) $\frac{1}{z-i}=\frac{1}{z-i+2i-2i}=\sum_{n=0}^{\infty}{(-1)^n\frac{i^n}{(z-2i)^{n+1}}}$

However, I did not solve it like this initially, I had both (1) and (2) with positive exponents. I saw the answer then and noticed my mistake. The questions is: How do I decide which of the two represents the negative exponent series?

My initial approach was to replace $z=i$ into the norm $|z-2i|$ which gives me 1, but this value should be out of the series convergence ring. How should I proceed?

Answer 1

Based on the comment by @reuns, I guess I figured out how it works. For each term, there are 2 series, one of them converges and the other diverges. The same happens for the other term. This is connected to the interval of convergence for the problem.

An alternative example:

Let $A=\{z\in C/1<|z+i|<\sqrt{2}\}$, find the Laurent series of $f(z)=(z^2-z)^{-1}$

$\frac{-1}{z}+\frac{1}{z-1}=\frac{-1}{z+i-i}+\frac{1}{z-1+i-i}$, so, for each term:
$\frac{1}{-i}\frac{1}{(1+\frac{z+i}{-i})}=\sum_{n=0}^{\infty}(-1)^n\frac{(z+i)^n}{(-i)^{n+1}}$ OR $\frac{1}{(z+i)}\frac{1}{(1+\frac{-i}{z+i})}=\sum_{n=0}^{\infty}\frac{(-i)^n}{(z+i)^{n+1}}$, but only the second converges within the inequality.

$\frac{1}{(-1-i)}\frac{1}{(1+\frac{z+i}{-1-i})}=\sum_{n=0}^{\infty}\frac{(z+i)^n}{(-1-i)^{n+1}}$ OR $\frac{1}{(z+i)}\frac{1}{(1+\frac{-1-i}{z+i})}=\sum_{n=0}^{\infty}\frac{(-1-i)^n}{(z+i)^{n+1}}$, but only the first one converges.

Therefore $f(z)=\sum_{n=0}^{\infty}\frac{(-i)^n}{(z+i)^{n+1}}+\sum_{n=0}^{\infty}\frac{(z+i)^n}{(-1-i)^{n+1}}$