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

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.

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

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

• 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?
– R.S.
Jun 28 at 17:01
• 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. Jun 28 at 17:05

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

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