Integrating $1/\sqrt{z^{2}-1}$ on some contour 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?
 A: $1$ 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.
A: $\newcommand{\+}{^{\dagger}}
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\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}
A: 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 $|z| \to \infty$, 
$$ \dfrac{1}{\sqrt{z^2-1}} =\dfrac{1}{z} + O\left(\dfrac{1}{|z|^3}\right)$$
