Question: So I want to calculate the Laurent series of this function 

$$ f: \mathbb{C} \to \mathbb{C}, \quad f(z) = \frac{1}{z^{2}+1}.$$

The Laurent series has to be in this form: 

$$\sum\_{n=- \infty }^{ \infty } a\_{n} (z-i)^n$$

for a circular disc
$$ 0<| z-i|<p,$$ 
where $p$ has to be found. 

With partial fraction expansion I am getting $$ f(z) =\frac{i}{2}\left( \frac{1}{z+i} - \frac{1}{z-i}\right).$$

For the first summand, $$\frac{1}{z+i} = \frac{1}{2i} \frac{1}{1+\frac{z-i}{2i}} = \frac{1}{2i} \sum\_{n= 0 }^{ \infty }\left(\frac{-(z-i)}{2i}\right)^n = \frac{1}{2i} \sum\_{n= 0 }^{ \infty } \left(\frac{i}{2}\right)^n (z-i)^n $$
for $$\left|\frac{-(z-i)}{2i}\right| < 1 \Longrightarrow \left| z-i \right| < 2.$$
Now I don't know how to continue with $$\frac{1}{z-i} .$$

Answer: 
In my opinion it is easier without partial fraction decomposition: let $z=w+i$ then for $0<|w|<2$
$$f(z) = \frac{1}{z^{2}+1}=\frac{1}{w(w+2i)}=\frac{1}{2iw(1-iw/2)}=-\frac{i}{2w}\sum\_{k=0}^{\infty}(iw/2)^k.$$
Hence the Laurent expansion of $f$ in $0<|z-i|<2$ is
$$f(z)=-\frac{i}{2(z-i)}+\sum\_{k=0}^{\infty}\frac{i^{k}(z-i)^{k}}{2^{k+2}}$$
**Note:** Despite the question at the end regarding *how to continue* note that already all calculations were done in order to solve the problem.

> 
> The function
>  \begin{align\*}
> f: \mathbb{C} \to \mathbb{C}, \quad f(z) &= \frac{1}{z^{2}+1}\\
> &=\frac{i}{2}\left( \frac{1}{z+i} - \frac{1}{z-i}\right)\tag{1}
> \end{align\*}
>  is to expand in a Laurent series at $z=i$.
> 
> 
> 

We observe in (1) that
\begin{align\*}
-\frac{i}{2}\cdot\frac{1}{z-i}\tag{2}
\end{align\*}
is the *principal part* of the Laurent series of $f$. On the other hand we know that
\begin{align\*}
\frac{i}{2}\cdot\frac{1}{z+i}=\frac{1}{4} \sum\_{n= 0 }^{ \infty } \left(\frac{i}{2}\right)^n (z-i)^n\tag{3}
\end{align\*}
is the power series representation of $f$ at $z=i$ with region of convergence $|z-i|<2$.

> 
> We conclude from (2) and (3) the Laurent series expansion of $f$ around $z=i$ is
>  \begin{align\*}
> f(z)&=\frac{1}{z^{2}+1}\\
> &\color{blue}{=-\frac{i}{2}\cdot\frac{1}{z-i}+\frac{1}{4}\sum\_{n= 0 }^{ \infty } \left(\frac{i}{2}\right)^n (z-i)^n}
> \end{align\*}
> 
> 
>