On a complex function $f(z)$ which is $\sqrt{z^2-1}\in\Bbb R $ when $x＞1$ Let $f(z)$ be a complex function whose value is $\sqrt{z^2-1}\in\Bbb R$ when $z\in \Bbb R$ is $＞１$, and let's suppose $f(z)$ is holomorphic if  $1＜｜z｜＜\infty$.
I want to integrate $f(z)$ over $C$, a circle of radius $R$ whose center is the origin $0\in\Bbb C$. However, to integrate it, I firstly should figure out what is the explicit form $f(z)$. If I could do that, I may be able to integrate it by using residue theorem: so 

what is the analytic expression of $f(z)$ when $z\in \Bbb C\setminus \Bbb R$ or better, how can I analytically continue $f(z)$ (hopefully in a unique way) for non real values of the variable $z$?

Thank you for your help.
 A: There is indeed an analytic continuation of $\sqrt{z^2-1}$ to the set $E = \{ z\in\Bbb C\colon 1 < |z|\}$.
It might seem that there is not, since $\sqrt{z^2-1} = \sqrt{z-1}\sqrt{z+1}$ has singularities at both $z=1$ and $z=-1$; however, one can connect the two singularities with a branch cut, and $\sqrt{z^2-1}$ is continuous in the complement of that branch cut, hence certainly in $E$ (the exterior of the unit disk).
In any case, write $\sqrt{z^2-1} = z g(z)$ where $g(z) = \sqrt{1-1/z^2}$; it suffices to find an analytic continuation of $g(z)$ to $E$. But $g(z) = h(1/z)$ where $h(z) = \sqrt{1-z^2}$, and $h(z)$ is definitely analytic in the unit disk $D = \{ z\in\Bbb C\colon |z|<1\}$ and thus in $D\setminus\{0\}$; hence $g(z)$ is analytic in $1/(D\setminus\{0\}) = E$. (Indeed $h(z)$ is even analytic at $\infty$ when viewed as a function on $E\cup\{\infty\}$ on the Riemann sphere, and thus $\sqrt{z^2-1}$ has a simple pole at $\infty$, which—if one is careful—would even allow for computing the integral of $\sqrt{z^2-1}$ around a circle in $E$ that goes around the origin.)
