# Residue on all layers of a complex function

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 $\sqrt{z_0}$?

• Could you rephrase your question? What does the second layer "mean" for the residue $\sqrt{z_0}$? does not sound clear to me. – Adayah Jul 27 '18 at 21:26
• @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? – Andrii Kozytskyi Jul 27 '18 at 21:33
• 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). – Adayah Jul 27 '18 at 21:35
• @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 – 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 $z_0$.