# Residue of function with different branch points

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?

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 $\sqrt{p_0} + \ln(p_0) = 0$.
• 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). – Robert Israel Apr 30 '18 at 1:19