$\int \frac{dz}{z\sqrt{(1-{1}/{z^2})}}$ over $|z|=2$ I need help in calculating the integral of $$\int \frac{dz}{z\sqrt{\left(1-\dfrac{1}{z^2}\right)}}$$ over the circle $|z|=2$. (We're talking about the main branch of the square root).
I'm trying to remember what methods we used to calculate this sort of integral in my CA textbook. I remember the main idea was using the remainder theorem, but I don't remember the specifics of calculating remainders....
Thanks you for your help!
 A: Let $w = z^{-1}$. Then $dz/z = -dw/w$ and
$$ \int\limits_{|z|=2} \frac{dz}{z \sqrt{1-z^{-2}}}
= \int\limits_{|w|=1/2} \frac{dw}{w \sqrt{1-w^{2}}}
= 2\pi i \mathrm{Res}_{w=0} \frac{1}{w \sqrt{1-w^{2}}}
= 2\pi i. $$
(Here, the minus sign of $-dw/w$ is compensated by the inversion of the path of integration.) Of course, direct calculation is also possible. Let $z = 2 e^{i\theta}$ for $-\pi < \theta < \pi$. Then
\begin{align*}
\int\limits_{|z|=2} \frac{dz}{z \sqrt{1-z^{-2}}}
&= i \int_{-\pi}^{\pi} \frac{d\theta}{\sqrt{1 - \frac{1}{4}e^{-2i\theta}}} \\
&= i \sum_{n=0}^{\infty} \binom{-1/2}{n} \left(-\frac{1}{4} \right)^{n} \int_{-\pi}^{\pi} e^{-2in\theta} \, d\theta\\
&= 2\pi i.
\end{align*}

Added. Uunder the standard branch cut, the function $$ z \mapsto \frac{1}{\sqrt{1 - z^{-2}}} $$ fails to be holomorphic along $[-1, 1]$. That is, it is holomorphic only on $\Bbb{C} \setminus [-1, 1]$. So in principle we cannot apply the residue theorem to calculate the integral.
To resolve this problem, I inverted the integral. To be precise, let us consider the Riemann sphere $\hat{\Bbb{C}} = \Bbb{C} \cup \{\infty\}$ and call $\infty$ the north pole and $0$ the south pole. Then the path of integration $|z| = 2$ is a circle winding this sphere. But in view of the Riemann sphere, there is little distinction between the 'inside the circle' and 'outside the circle'. Rather, a circle divides $\hat{\Bbb{C}}$ into two regions where one contains $\infty$ and the other not.

Thus instead of considering the poles inside the circle, we may consider the poles outside the circle. In actual calculation, this can be achieved by introducing the inversion $w = z^{-1}$ and writing everything in terms of $w$.
In summary, I did not eliminated the branch cut (where the holomorphy breaks down so that the residue theorem is inapplicable), but rather circumvented it by applying the residue theorem to the outside of the circle $|z| = 2$.

Added 2. Felix Marin, since both me and Random Variable did not make any mistake, it says that Mathematica is doing something wrong. Indeed, Mathematica 8 yields 

A: $$\frac{1}{z \sqrt{1-\frac{1}{z^{2}}}} =  \frac{1}{z} \Big( 1 - \frac{1}{2z^{2}} + O(z^{-4}) \Big) \text{for} \ |z| >1 \implies \int_{|z|=2} \frac{1}{z \sqrt{1-\frac{1}{z^{2}}}} \ dz = 2 \pi i (1) = 2 \pi i $$
