$\frac{1}{(z+i)(z-i)}$ 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?
On
When talking about a Laurent expansion, you should define which point the expansion is being carried around. In your example, if the expansion is around $i$, you conclude that $i$ is a pole with your criterion. If the expansion is around $-i$, same conclusion. If you perform the expansion around any other point other than $\pm i$, you deduce that this point is not a pole.
On
$\frac{1}{z+i}$ 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:
$\frac{1}{z^2+1}= \frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)=\frac{1}{2i}\left(\frac{1}{z-i}-\sum_{n=0}^{\infty}a_n(z-i)^n \right)$.
No, you don't have two terms. Near $i$ we have\begin{align}\frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)&=\frac1{2i}\frac1{z-i}-\frac1{2i}\frac1{i+z}\\&=\frac1{2i}\frac1{z-i}-\frac1{2i}\frac1{2i+(z-i)}\\&=\frac1{2i}\frac1{z-i}+\frac14\frac1{1+\frac{z-i}{2i}}\\&=\frac1{2i}\frac1{z-i}+\frac14\sum_{n=0}^\infty\frac{(-1)^n}{(2i)^n}(z-i)^n.\end{align}This series converges when $|z-i|<2$. So, $a_{-1}=\frac1{2i}$ and $a_n=0$ if $n<-1$. Therefore, your function has a simple pole at $i$. At $-i$ it's similar.