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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?

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  • $\begingroup$ 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. $\endgroup$
    – Martin R
    Commented Jul 23, 2023 at 18:30
  • $\begingroup$ 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? $\endgroup$
    – FrontEnd-Python
    Commented Jul 23, 2023 at 18:36
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    $\begingroup$ $f(z)$ 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 $\mathbb{C}$. $\endgroup$
    – hdci
    Commented Jul 23, 2023 at 18:46
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    $\begingroup$ 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 $f(z) = {g(p) \over z-p} + g'(p) + \cdots$. $\endgroup$
    – copper.hat
    Commented Jul 23, 2023 at 19:01
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    $\begingroup$ 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? $\endgroup$
    – José Carlos Santos
    Commented Jul 23, 2023 at 19:10

1 Answer 1

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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}

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  • $\begingroup$ Great first answer! Hoping to see more input from you in the future :) $\endgroup$
    – Kamal Saleh
    Commented Jul 23, 2023 at 21:39
  • $\begingroup$ @KamalSaleh Thanks! $\endgroup$
    – VincentV
    Commented Jul 23, 2023 at 21:58
  • $\begingroup$ 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). $\endgroup$
    – Martin R
    Commented Jul 24, 2023 at 3:55

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