In a problem, I have been asked to find the poles of the following function:
$$f(z) = \frac{1}{z^4-1}$$
Of course, $f(z) = \frac{1}{(z+1)(z-1)(z+i)(z-i)}$. So the singularities are $z=\pm1$ and $z=\pm i$.
In order to determine if they are poles (and not removable or essential singularities), we need to find the Laurent series of $f(z)$ and show that the principal part is finite. So how would we go about finding Laurent series, say centered around $z=1$?
Also, is there a way to determine that $z=\pm1$ and $z=\pm i$ are poles indeed, without finding the Laurent series?
To find the laurent series of a polynomial it is usually helpful to compute the partial fraction decomposition and rewriting the denominator as a limit of a geometric series usually leads to the desired laurent series. For example \begin{align}\frac{1}{z^4-1} = \frac{1/4}{z-1} + \frac{-1/4}{z+1} + \frac{-1/2}{z^2-1}\end{align} Consider $\frac{1}{z-1}$: \begin{align}\frac{1}{z-1} = \frac{1}{z}\frac{1}{1-1/z} = \frac{1}{z}(\sum_{k=0}^\infty 1/z^k) = \sum_{k=0}^\infty 1/z^{k+1} \dots\end{align}