If I wanted to integrate
$$\oint \frac{1}{\sqrt{z^{2}-1}}$$
Say around a circular contour radius $2$ centre $0$, how would I do that? Does the function have poles at $\pm 1$ or are they just "branch points" without residue? Would the definition of $\sqrt{}$ make this integral ambiguous somehow?
$1$ and $-1$ are branch points of the function, so the residue theorem cannot be applied directly.
Assuming that we take the branch cut $[-1,1]$, or at least that the branch cut stays within the disk of radius $2$ centered at the origin, so that we don't run into difficulties with the branch cut, we can evaluate the integral by using Cauchy's integral theorem to shift the contour to a larger circle, and the standard estimate (ML lemma). The value of course depends on the chosen branch of the square root, but changing the branch only changes the value by a factor of $-1$.
Choosing the branch with $\sqrt{z^2-1}$ real and positive for $z$ real and $> 1$, we can write
$$\sqrt{z^2-1} = z\cdot \sqrt{1-\frac{1}{z^2}},$$
where the principal branch of $\sqrt{1-w}$ on the unit disk is used. Then we have
$$\begin{align} \int_{\lvert z\rvert = 2} \frac{dz}{\sqrt{z^2-1}} &= \int_{\lvert z\rvert = R} \frac{dz}{\sqrt{z^2-1}}\\ &= \int_{\lvert z\rvert = R} \frac{dz}{z\sqrt{1-\frac{1}{z^2}}}\\ &= \int_{\lvert z\rvert = R} \frac{1}{z}\left(\sum_{k=0}^\infty (-1)^k\binom{-\frac{1}{2}}{k}z^{-2k}\right)\,dz\\ &= \int_{\lvert z\rvert = R} \frac{1}{z}\left(1+ \frac{1}{2z^2} + O\left(z^{-4}\right)\right)\,dz\\ &= 2\pi i + \int_{\lvert z\rvert = R} O(R^{-3})\,dz\\ &= 2\pi i + O(R^{-2}). \end{align}$$
Since by Cauchy's theorem the integral doesn't depend on $R \geqslant 2$, we have the result
$$\int_{\lvert z\rvert = 2} \frac{dz}{\sqrt{z^2-1}} = 2\pi i$$
for the chosen branch.
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\bbox[5px,#ffd]{\oint_{\verts{z}\ =\ 2} {\dd z \over \root{z^{2} - 1}}} = \int_{0}^{2\pi} {2\ic\expo{\ic\theta}\,\dd\theta \over \root{4\expo{2\ic\theta} - 1}} \\[5mm] = &\ \ic\int_{0}^{2\pi}\pars{1 - {\expo{-2\ic\theta} \over 4}}^{-1/2}\,\dd\theta \\[3mm]&=\ic\sum_{n = 0}^{\infty}{-1/2 \choose n}{\pars{-1}^{n} \over 4^{n}}\ \overbrace{\int_{0}^{2\pi}\expo{-2\ic n\theta}\,\dd\theta} ^{\ds{=\ 2\pi\,\delta_{n0}}}\ =\ \color{#66f}{\large 2\pi\,\ic} \end{align}
Let's say you're using a branch of this function that is analytic outside the interval $[-1.1]$, and positive on $(1,\infty)$. The integral will be the same over the circle of any radius $R > 1$. So take $R \to \infty$, using the fact that as $|z| \to \infty$, $$ \dfrac{1}{\sqrt{z^2-1}} =\dfrac{1}{z} + O\left(\dfrac{1}{|z|^3}\right)$$
