Why $\frac{1}{z^2+1}$ has simple poles at $\pm i$. $\frac{1}{(z+i)(z-i)}$ has simple poles at $\pm i$. I know that pole is simple if Laurent series has only $a_{-1} \ne 0$. So, i tried to derive the series:
\begin{align*}
\frac{1}{(z+i)(z-i)} &= \frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right).
\end{align*}
But here i have two terms $\frac{1}{z-i}-\frac{1}{z+i}$. So instead of single $a_{-1}(z-z_0)^{-1}$ i have two terms. Does this mean that the function has two poles? And what if i had three such terms?

$\frac{1}{z+i}$ is holomorphic in a neigborhood of $i$, hence
$\frac{1}{z+i}= \sum_{n=0}^{\infty}a_n(z-i)^n$. The Laurent expansion around reads now as follows:
$\frac{1}{z^2+1}= \frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)=\frac{1}{2i}\left(\frac{1}{z-i}-\sum_{n=0}^{\infty}a_n(z-i)^n \right)$.
