I understand how residues work for single-layered functions, but one question type from my assignment has me puzzled.



has a simple pole at $z=z_0$ and its residue is

$$\text{Res}_{z\to{z_0}} f(z)=\sqrt{z_0}$$

But I'm asked to "find the residues for all layers of a function". What does the second layer "mean" for the residue $\sqrt{z_0}$?

Thanks in advance

  • $\begingroup$ Could you rephrase your question? What does the second layer "mean" for the residue $\sqrt{z_0}$? does not sound clear to me. $\endgroup$ – Adayah Jul 27 '18 at 21:26
  • $\begingroup$ @Adayah Well that's the problem, I'm not even sure what to ask here. "If I'm asked to find the residues for all layers of a the said $f(z)$, is it enough if I write $\sqrt{z_0}$?" - Does this make the question any clearer? $\endgroup$ – Andrii Kozytskyi Jul 27 '18 at 21:33
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    $\begingroup$ It does and the answer is no. Each layer is a separate function and needs its own residue computation. So the solution to your problem will be a pair of two numbers, one of which you correctly identified to be $\sqrt{z_0}$ (for one of the layers). $\endgroup$ – Adayah Jul 27 '18 at 21:35
  • $\begingroup$ @Adayah May I ask you to show me how exactly it is done on this example? I don't have any idea on how the function should be separated, and I can't seem to find it anywhere in my literature or the web $\endgroup$ – Andrii Kozytskyi Jul 27 '18 at 22:05

Suppose $z_0 \neq 0$ and let $s : U \to \mathbb{C}$ be a fixed branch of $\sqrt{z}$ in a small neighborhood $U$ of $z_0$. Then $f$ has two branches:

$$\frac{s(z)}{z-z_0} \quad \mathrm{and} \quad \frac{-s(z)}{z-z_0}.$$

The first one has the residue at $z=z_0$ equal to $s(z_0)$, the second one has it equal to $-s(z_0)$, that is, the residues are both square roots of $z_0$.

  • $\begingroup$ Ahh I am retarded. Should've probably listened to my own question - if the function is branching it "means" you should look at what happens near around the pole on each layer. Thanks a whole lot! $\endgroup$ – Andrii Kozytskyi Jul 27 '18 at 22:42

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