1
$\begingroup$

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

Say,

$$f(z)={\sqrt{z}\over{(z-z_0)}}$$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $z_0$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.