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How can I determine the type of a singularity of a function $$f(z)={e^{1/z} \over z-1}+{\pi z \over 2\sin(\pi z)}$$

at $z_0=0$?

I don't see an easy way to represent it using Laurent series, neither I don't see how I can find the limit of the function with $z\to 0$.

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2 Answers 2

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Look at $e^{1/z}$ around $z = 0$. This function approaches every point in the complex plane (take the limit along the real axis, positive or negative or along the imaginary axis). Dividing by and adding a meromorphic function does not change this, so there is no singularity that can be characterized in the normal way with Laurent series, residues, etc. This is called an essential singularity.

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  • $\begingroup$ That's clear, I have the same result, that made me more confident I was thinking in a correct way. But that means that I can't get use of the Residue theory for integration to find a path integral where the path is a circle with a middle point $z_0=0$? $\endgroup$ Jul 19, 2014 at 11:38
  • $\begingroup$ Yes, that is correct. The theory of residues is only defined for meromorphic functions - that is, functions on the complex plane that have poles rather than essential singularities. $\endgroup$
    – Dorebell
    Jul 19, 2014 at 11:40
  • $\begingroup$ So but it's still possible to do using Cauchy's integral formulae, isn't it? $\endgroup$ Jul 19, 2014 at 11:41
  • $\begingroup$ No, Cauchy's integral formula is also for meromorphic functions. Basically, any normal complex analysis is completely useless for these sorts of functions near essential singularities. Now, if you work in a part of the complex plane that avoids the origin entirely, you can use all of these things. $\endgroup$
    – Dorebell
    Jul 19, 2014 at 11:42
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    $\begingroup$ @DmitryKazakov This is becoming another question on itself. Please ask it separately. $\endgroup$
    – Git Gud
    Jul 19, 2014 at 11:46
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It is essentially because the part $e^{\frac{1}{z}}=\sum_{n=0}^{\infty}\frac{\frac{1}{z^n}}{n!}=\sum_{n=0}^{\infty}\frac{1}{z^n\cdot n!}$ attains infinity times often addends in the form $\frac{a_k}{z^k}$ (an the other parts are holomorphic or liftable in $0$, in a neighborhood of $0$) so the principal part has infinity many addeds. Thus $z_0=0$ as an essentially singularity.

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