I want to calculate the Laurent-series of $f: \mathbb C\setminus \{\pm i \} \to \mathbb C, \, \, f(z) = \frac{1}{1+z^2}$ within the domain $D = \{ z \in \mathbb C : 0 \lt \lvert z-i \rvert \lt 2\}$.
My attempt is to use partial fractions and then try to use the geometric series: $$ \frac{1}{1+z^2} = \frac{1/2}{1+iz} + \frac{1/2}{1-iz}$$
Now consider $$\frac{1}{1+iz} = \frac{1}{1-(-iz)} = \sum_{n=0}^{\infty} (-iz)^n \,\,\,$$ for $\lvert -iz \rvert = \lvert z \rvert < 1$ . But for my set $\lvert z \rvert $ has $[0,3)$ as possible values.
On the other hand, $$\frac{1}{1+iz} = \frac1z \frac{1}{i+1/z} = \frac1z \frac{-i}{1-i/z} = \frac{-i}{z}\sum_{n=0}^{\infty}(\frac{i}{z})^n$$ for $\lvert i/z\rvert = \lvert \frac 1z \rvert \lt 1$ or $\lvert z \rvert \gt 1$.
What can I do to find a series representation that includes both cases ?
You have developed the partial fractions into a power series or Laurent series at $z_0 =0$ and $z_0 = \infty$, which unfortunately does not help to get the Laurent series at $z_0 = i$.
Instead you have proceed as follows: For $|z-i| < 2$ you have $$ \frac{1}{z+i} = \frac{1}{(z-i) + 2i} = \frac{1}{2i(\frac{z-i}{2i} + 1)} = -\frac i2 \frac{1}{1 - \frac i2(z-i)} \\ = -\frac i2 \sum_{n=0}^\infty \bigl(\frac i2\bigr)^n (z-i)^n = \sum_{n=0}^\infty -\bigl(\frac i2\bigr)^{n+1} (z-i)^n $$ and therefore for $0 < |z-i| < 2$ $$ f(z) = \frac{1}{1+z^2} = \frac{1}{(z-i)(z+i)} = \sum_{n=0}^\infty -\bigl(\frac i2\bigr)^{n+1} (z-i)^{n-1} = \sum_{n=-1}^\infty -\bigl(\frac i2\bigr)^{n+2} (z-i)^n $$ which is the Laurent series of $f$ at $z_0 = i$.