Let \(C\) denote the positively oriented circle \(\abs{z}=2\), \(z\in\CC\). Evaluate the integral
\[ \int_C\sqrt{z^2-1}\,dz \]
where the branch of the square root is chosen so that \(\sqrt{2^2-1}>0\).
Solution
1. On \(\abs{z}>1\), the required branch is\[ \sqrt{z^2-1}=z\sqrt{1-z^{-2}}, \]where the second square root is the principal branch.
For \(\abs{z}>1\), one has \(\abs{z^{-2}}<1\), so \(1-z^{-2}\) lies in the open disk of radius \(1\) centered at \(1\). The principal square root is analytic on this disk, so the displayed function is analytic on \(\abs{z}>1\), and its square is \(z^2(1-z^{-2})=z^2-1\). At \(z=2\) it gives \[ 2\sqrt{1-\frac14}=\sqrt3>0, \] so it is the branch required in the problem.
2. For \(\abs{z}>1\), the branch of step 1 has Laurent expansion\[ \sqrt{z^2-1} =z-\frac1{2z}-\frac1{8z^3}-\cdots, \]whose coefficient of \(z^{-1}\) is \(-\tfrac12\).
For \(\abs{w}<1\), the binomial series gives \[ \sqrt{1-w} =1-\frac12w-\frac18w^2-\cdots. \] Put \(w=z^{-2}\) and multiply by \(z\).
3. One has\[ \int_C\sqrt{z^2-1}\,dz=\boxed{-\pi i}. \]
The Laurent series of step 2 converges uniformly on the circle \(\abs{z}=2\), so termwise integration around the positively oriented circle gives \[ \int_C\sqrt{z^2-1}\,dz =2\pi i\left(-\frac12\right) =-\pi i. \]
4. Q.E.D.
Step 3 gives the requested value.