Contour integral of $\sqrt{z^2-1}$ on $|z| = 2$ I've been wrestling with this problem (available here ):
Evaluate the integral of $f(z) = \sqrt{z^2-1}$ around the circle $\{z: |z| = 2\}$, where the branch of the square root function is chosen so that $\sqrt{2^2-1} > 0$.
I came up with the answer
$$\int\limits_{|z| = 2} \sqrt{z^2-1} dz = \pi i$$
using the following argument, which seems a little like voodoo to me.
You notice that there are two points, at $z \pm 1$, where you might have to pick a branch of $log()$ to define the square root. Then, I convinced myself that as long as you exclude from your domain the real interval $-1 \le x \le 1$, the function $f(z)$ can be continuously defined (maybe the two branches somehow cancel?), and so is analytic on its domain.
Therefore you can shrink the contour of integration continuously using the Cauchy Closed Curve Integral Theorem, so that you wind up reducing it to the real integral
$$2\int_{-1}^1 \sqrt{x^2-1} dx = \pi i$$
I'm not really sure the answer is correct, and I'm even less certain about the way I got it. I had a hard time convincing myself that $\sqrt{z^2-1}$ was not multi-valued on the domain. To try to get at that I tried $\sqrt{z^2-1} = \sqrt{z+1} \sqrt{z-1}$ and reasoned that the product makes the discontinuity disappear, using
$$e^{\frac{\pi}{2}i}\cdot e^{\frac{\pi}{2}i} = -1 = e^{-\frac{\pi}{2}i}\cdot e^{-\frac{\pi}{2}i} $$
for the different branches of $log()$ you can choose.
I'd appreciate any input that would help me get a a better way to look at this.
Thanks!

EDIT
An attempt to use the Residue at infinity:
By substituting $z = 1/w$ you get
\begin{eqnarray}
\int\limits_{|z| = 2} \sqrt{z^2-1} dz &=& \int\limits_{|w| = 1/2} \frac{1}{w^2}\sqrt{\frac{1}{w^2}-1} dw \\
&=& \int\limits_{|w| = 1/2} \frac{1}{w^2}\sqrt{\frac{1-w^2}{w^2}} dw \\
&=& \int\limits_{|w| = 1/2} \frac{1}{w^3}\sqrt{1-w^2} dw
\end{eqnarray}
The integrand is analytic in $|w| \le 1/2$ except at $w = 0$, so we can apply the Residue Theorem. Near $w = 0$, $\sqrt{1-w^2} = 1 - \frac{1}{2}w^2 + \dots$, so the residue of $\frac{1}{w^3}\sqrt{1-w^2} $ at $w = 0$ is $-\frac{1}{2}$. Then the integral is
$$\int\limits_{|z| = 2} \sqrt{z^2-1} dz = 2\pi i \cdot (-1/2) = -\pi i$$
 A: You are correct that by making the branch cut along $[−1,1]$ you get a well-defined analytic function. Each square root picks up a factor of $−1$ when you go around its branch point, so the product picks up $(−1)(−1)=1$ and is thus well-defined.
The nicest way to compute the integral is to apply the Residue Theorem, computing the residue at infinity. Yes, although they don't tell you this in a beginning complex variables class, you're actually computing residues of the meromorphic $1$-form $f(z)\,dz$, so when you substitute $z=1/w$, this becomes $-f(1/w)\,dw/w^2$; but, as you noted, the orientation of the circle as the boundary of a neighborhood of $\infty$ reverses, so we get $-2\pi i \text{ res}_\infty(f(z)\,dz)=2\pi i\text{ res}_0(f(1/w)\,dw/w^2)$. 
Conceptually, the Riemann surface of $\sqrt{z^2-1}$ is made by removing the interval $[-1,1]\subset\mathbb R\subset\mathbb C$ and attaching two sheets, as usual with the square root function. If we compactify, we get two points at infinity, one on each branch; the compact Riemann surface is analytically equivalent to $\mathbb CP^1 \cong S^2$.
A: An idea:You could also try the following
$$\int\limits_\gamma \sqrt{z^2-1}\;dz=i\int\limits_\gamma \sqrt{1-z^2}\;dz$$
Now substitution
$$z=\sin t\;,\;u'=\cos t\;,\;\;0\le t\le 2\pi\;,\;\;\text{so that we "shrink" the circle as you did ,}$$
$$i\int\limits_0^{2\pi}\cos^2t\,dt=\left.i\frac{t+\cos t\sin t}{2}\right|_0^{2\pi}=\pi i$$
