TITLE: Integrating $1/\sqrt{z^{2}-1}$ on some contour
QUESTION [6 upvotes]: 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?

REPLY [7 votes]: $\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}