A pole of multiplicity $m$ can be identified as the root of the denominator of a function like so:
$$f(z) = \frac{g(z)}{(z-z_0)^m}$$
This also makes it easy to see that the $m$-th "order" residue calculation cancels out this part of the denominator.
How does one define/find the multiplicity of a pole (is it?) coming from a square root? For example:
$$f(z) = \frac{1}{\sqrt{1+z^2}}$$
The alternative method of finding the multiplicity of a pole, the Laurent series, is useless here, as there is no $\frac{1}{z}$ terms. Or does this really mean $z=\pm \mathrm{i}$ is not a pole of the function at all? One could of course multiply by $\frac{\sqrt{1+z^2}}{\sqrt{1+z^2}}$, which would make the pole simple, but the actual function such that the residue (well, tor really the residue, but the almost complete small circle integral around the point) is zero there (as the denominator is also present in the numerator). Is this a general property?
I'm working on calculating a somewhat complicated contour integral containing two different square roots in the denominator, and would like to keep everything as simple as possible.
Best Answer
This isn't a pole, it is an essential singularity (an algebraic one). Such a singularity introduces a cut line into the complex plane, and finding contours that avoid such to apply Cauchy's residue theorem becomes challenging. Your best bet is to avoid the singularities altogether.