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I'm learning about different types of singularities:

1) removable singularities.
2) poles.
3) essential singularities.

I think I understand 1 and 2 but I don't really get the 3rd one, could someone please explain me?

1) For a removable singularity $|f|$ is bounded near $z_o$ so as an example we'd have a removable singularity at $z_0=0$ for $\frac{sin(z)}{z}$.
2) For a pole $|f|$ goes to infinity near $z_o$ , so as an example we'd have a pole at $z_0=0$ for $\frac{1}{z}$.
3) For an essential singularity $|f|$ doesn't go to infinity near $z_o$ nor is bounded here. In the book they give the example: we'd have an essential singularity at $z_0=0$ for $e^{\frac{1}{z}}$.

I don't get this last one, since in my head: (I know it's not mathematically correct what I'm gonna write but I do it just to explain my problem) $e^{\frac{1}{0}}=e^\infty=\infty$

Thanks in advance.

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    $\begingroup$ If you approach $0$ through not positive real numbers, e.g. through negative real numbers, or through purely imaginary numbers, it looks different. $\endgroup$ Commented Dec 27, 2017 at 11:03
  • $\begingroup$ See math.stackexchange.com/questions/1284316/… $\endgroup$ Commented Dec 27, 2017 at 13:51

3 Answers 3

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Note that a function $f(z)$ can be expanded as a Laurent series about a singularity $z_0$ as: $$f(z) = \sum_{n=0}^{\infty} a_n(z-z_0)^n + \sum_{n=1}^{\infty} \frac{b_n}{(z-z_0)^n}$$

In case, the second sum (the principal part) has infinitely many terms, then the singularity $z=z_0$ of $f(z)$ is called an essential singularity.

Hence, the given function: $$f(z) = e^{\frac1{z}} = \sum_{n=0}^{\infty} \frac{1}{n!z^n}$$ has $z=0$ as an essential singularity.

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  • $\begingroup$ But the overal expansion for $e^z$ still has infinitely many terms right? $\endgroup$
    – Amaluena
    Commented Dec 27, 2017 at 12:09
  • $\begingroup$ @Amaluena Yes, you can see that the index of $n$ goes from $0$ to infinity. $\endgroup$
    – user371838
    Commented Dec 27, 2017 at 12:10
  • $\begingroup$ It helped but I'm hoping there will be a more intuiative answer, since this is a rule I had not heard of before $\endgroup$
    – Amaluena
    Commented Dec 27, 2017 at 12:19
  • $\begingroup$ @Amaluena Note that the expansion of f(z) has a z^n term in the denominator. It keeps on increasing as n increases. There is no way that we can multiply by some power of z to remove everything. Hence, z=0 is an essential singularity. $\endgroup$
    – user371838
    Commented Dec 27, 2017 at 12:21
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Well! As far as Complex exponential is concerned $e^\infty$ is not actually $\infty$ it is not defined to be precise.

While understanding types of singularities you should also learn them by their laurentz series expansions That way it would be easier and you will not get confused.

Hope it works

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The issue with your simply treating $\lim_{x \to 0} e^{\frac{1}{x}}$ as $e^{\frac{1}{0}} = e^\infty$ is that while $-0 = 0$, $e^{-\infty} \neq e^{\infty}$.

What this means is that as you approach 0 from the right, you get increasinly large function values, tending towards infinity as you get closer to 0. But if you come from the left, the exponent becomes more and more negative, and the exponential tends towards 0, not infinity.

This can be easily seen when looking at a plot of the function, such as the one found on Wikipedia:

Plot of exp(1/z)

In this image, dark colors indicate values close to 0, bright values indicate values close to $\infty$ (in terms of their absolute value).

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