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I'm wondering what would one do when one wishes to find the residue of a function $$\text{Res}_{z\to z_0} f(z)$$ where $f(z)$ has multiple branch points, for instance $f(z)$ may be a function such as

$$f(z) = \frac{1}{\sqrt{z}+\ln z}\quad \text{or} \quad f(z)=\frac{1}{\ln(\sqrt{z}+1)}$$

I'd clearly have to do a Laurent series expansion, but I'm wondering will the branch point(s) affect the validity of my expansion?

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You have to do the expansion using the particular branch you're interested in. It's quite possible that a point may be a singularity for one branch but not for another, e.g. $f(z) = \dfrac{1}{\sqrt{z} - 1}$ has a pole at $z=1$ for a branch where $\sqrt{1} = 1$, but not for a branch where $\sqrt{1}=-1$.

I hope you're not trying to find a residue at a branch point: that doesn't exist. You only have a residue when you have an isolated singularity, and a branch point is not an isolated singularity. Thus your second example has no isolated singularities: to make the denominator $0$ you'd need $z=0$, but that's a branch point of the square root.

Your first example $1/(\sqrt{z} + \ln(z))$ will have poles where $\sqrt{z} + \ln(z) = 0$, namely $4 W(\pm 1/2)^2$ where $W$ is a branch of the Lambert W function. One such point (using the "main" branch $W_0$ and $+1/2$) is approximately $0.4948664144$, and requires using the principal branches of $\sqrt{z}$ and $\ln(z)$. If that point is $p_0$, we have $$\sqrt{z} + \ln(z) = \left(\dfrac{1}{2 \sqrt{p_0}} + \dfrac{1}{p_0}\right) (z - p_0) + O\left((z-p_0)^2\right)$$ so the residue at $p_0$ is $$\dfrac{1}{\dfrac{1}{2 \sqrt{p_0}} + \dfrac{1}{p_0}}$$ using the same branch of $\sqrt{p_0}$ that you are using for $\sqrt{p_0} + \ln(p_0) = 0$.

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  • $\begingroup$ Why don't residues exist at non-isolated singularities? Any references? $\endgroup$ – Eulerian Apr 29 '18 at 15:22
  • $\begingroup$ The definition of residue requires an isolated singularity. Basically, the reason for this is that the integral $\oint_C f(z)\; dz$ around a loop $C$ enclosing the point $p$ is independent of $C$ in a deleted neighbourhood $U$ of $p$ if $f$ is analytic in $U$. However, there is no reason for path-independence if $f$ is not analytic in $U$ (i.e. if there are other singularities there). $\endgroup$ – Robert Israel Apr 30 '18 at 1:19

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