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Locate and classify the singularity of $\frac{e^{z-1}-1}{z^4-1}$.

I may be a bit confused with the idea of the classification of the singularities, but if I have that:

$\frac{e^{z-1}-1}{z^4-1}=\frac{e^{z-1}-1}{(z+1)(z-1)(z+i)(z-i)}$ I am not allowed to say that the poles $\pm 1, \pm i$ are simple because I did not expand the numerator yet. Correct?

I know that $e^{z-1}-1 = \sum_{n \geq 0} \frac{(z-1)^n}{n!}-1 =(z-1) \sum_{n \geq 2}\frac{(z-1)^n}{n!}$. Now I see that:

$\frac{e^{z-1}-1}{z^4-1} = \frac{(z-1) \sum_{n \geq 2}\frac{(z-1)^n}{n!}}{(z+1)(z-1)(z+i)(z-i)} = \frac{\sum_{n \geq 2}\frac{(z-1)^n}{n!}}{(z+1)(z+i)(z-i)}$.

Now, I can conclude that $-1, \pm i$ are simple poles and $1$ is actually a removable singularity. Is that correct? Is there a "general procedure" to start with such tasks?

Correction: There is a mistake in the expansion. See the comment below.

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    $\begingroup$ You misexpanded $e^{z-1}-1$. It should be $\sum_{n\geq1}\frac{(z-1)^n}{n!}=(z-1)\sum_{n\geq1}\frac{(z-1)^{n-1}}{n!}$. $\endgroup$
    – Moko19
    Jun 28 at 17:04

2 Answers 2

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Just note that $$\lim_{z\to 1}(z-1)\frac{e^{z-1}-1}{z^4-1}=0$$ so its removable. For the others just note that $-1,i,-i$ are simple zeros of $z^4-1$, and since there are no problems in that points in the numerator, they are simple poles of the function

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  • $\begingroup$ What do you mean by "no problems in that points in the numerator"? Why are we event concerned about the numerator and not just the denominator? $\endgroup$
    – R.S.
    Jun 28 at 17:01
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    $\begingroup$ I mean that they are no poles or zeros of the numerator. For example $1$ is a zero of the denominator so it's a candidate for pole, but also is a zero of the numerator so it could be a removable singularitie, in this case that's true, But since the numerator is not zero in $-1,i,-i$ if a point is a zero of order $m$ of $g$ then is a pole of order $m$ of $1/g$. That's also true for $f/g$ if $f$ does not have a pole or singularitie in those points. $\endgroup$ Jun 28 at 17:05
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Your general methodology is correct (other than the small miscalculation mentioned in my comment).

In general, to classify singularities of a function:

  1. Look for any essential singularities. In general, these will either be branch points, such as if you calculate $\ln(z-a)$ or points where you do something like $e^{(\frac1{z-a})}$

  2. Write the function in the form $\frac{f(z)}{g(z)}$, where $f(z)$ and $g(z)$ are continuous everywhere (except at the essential singularities that you found)

  3. Find the roots of $g(z)$, $r_i$, and note their multipliicties $m_i$.

  4. Find the zeroes of $f(z)$, $p_j$, and note their multiplicities $n_j$.

  5. If $r_i\not\in\{p_j\}$, then it is a pole of order $m_i$, If $r_i=p_j$, then if $m_i>n_j$, it is a pole of order $m_i-n_j$, and if $m_i\leq n_j$, it is a removable singularity

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  • $\begingroup$ In the fifth point it should be $m_i \leq n_j$, right? $\endgroup$
    – R.S.
    Jun 28 at 18:51
  • $\begingroup$ @R.S. Yes, fixed $\endgroup$
    – Moko19
    Jun 29 at 10:35

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