Poles and the Laurent series for $\frac{1}{z^4-1}$

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?

• Are you aware that $f$ has a pole at $z_0$ exactly if $\lim_{z \to z_0} |f(z)| = \infty$? it is also equivalent to $1/f$ having a removable singularity with value zero. Commented Jul 23, 2023 at 18:30
• No, did not know that. I know what $|z|$ would mean, but what does $|f(z)|$ mean? What would it be for the function in my case? Commented Jul 23, 2023 at 18:36
• $f(z)$ is a complex, thus $|f(z)|$ is its module, exactly as $|z|$ is the module of $z$. Another way to see that these are poles is : if a singularity is removable, the function is bounded in its neighborhood, and if it's essential, the image of a neighborhood is dense in $\mathbb{C}$.
– hdci
Commented Jul 23, 2023 at 18:46
• Try writing $f(z) = {A \over z-1} + {B \over z+1} + {C \over z-i} + {D \over z+i}$. Alternatively, if you can write $f(z) = g(z) {1 \over z-p}$ where $g$ is analytic at $z=p$ then the Laurent expansion of $f$ around $p$ will look like $f(z) = {g(p) \over z-p} + g'(p) + \cdots$. Commented Jul 23, 2023 at 19:01
• Let me see if I got that right. You know the meaning of $|z|$ for any complex number, but you don't know the meaning of $|f(z)|$, in spite of the fact that $f(z)$ is some complex number. Did I get that right? Commented Jul 23, 2023 at 19:10

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}
• The question is about characterizing the singularities at the points $\pm 1$ and $\pm i$ via Laurent series. So you would need the Laurent series centered at those points, and not the Laurient series centered at $z=0$ (as you did). Commented Jul 24, 2023 at 3:55