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?

This isn't really an answer, but I cannot comment yet. I don't understand why you need Laurent series for this - if there's a simple pole at $i$ then $(z-z_0)f(z)=(z-i)\frac{1}{(z+i)(z-i)}=\frac{1}{z+i}$ should be holomorphic, which it is, right?
