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Why it is true , that the logarithmic residue of function at isolated singularity point always integer number?

BeesaFangirl DOTO
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    More needs to be asked of the function for the 'logarithmic residue' to be defined at an isolated essential singularity. For a pole $z_0$ of order $m$ you can write the function as $f(z)=(z-z_0)^{-m}h(z)$ for $h$ analytic with $h(z_0)\neq0$. Then $\frac{f'(z)}{f(z)}=\frac{-m(z-z_0)^{-m-1}h(z)+(z-z_0)^{-m}h'(z)}{(z-z_0)^{-m}h(z)}=\frac{-m}{z-z_0}+\frac{h'(z)}{h(z)}$. Integrating, and taking into account that $h(z_0)\neq0$ and therefore $\frac{h'(z)}{h(z)}$ is analytic near $z_0$, you get that the integral is $-m$. – egorovik Dec 13 '19 at 20:01
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    The residue is an integer because it's a winding number. (If the point is an isolated singularity of $f'/f$.) – Daniel Fischer Dec 13 '19 at 20:41
  • Here is a nice (analytical) argument why the winding number is always an integer. – Martin R Dec 13 '19 at 21:57
  • @egorovik , And what about essential singularities? – BeesaFangirl DOTO Dec 14 '19 at 07:30
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    @BeesaFangirlDOTO The 'logarithmic residue' may not be defined at all, or be infinite. – egorovik Dec 14 '19 at 12:15

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