Showing that $\sqrt{z^2-1}$ has analytic square root in $|z|>1$ and finding an integral I want to show that $\sqrt{z^2-1}$ has analytic square root in $|z|>1$ and to find the integral 
$$\int_{C^+(0,R)}\frac {\sqrt{z^2-1}} {z^2}.$$ 
Can I apply Residue theorem?
 A: Define
$$
f(w)=\sqrt{3}\exp\left(\int_2^w\frac12\left(\frac1{z-1}+\frac1{z+1}\right)\,\mathrm{d}z\right)
$$
Where the path of integration does not cross $[-1,1]$. This definition is unambiguous. Suppose we have two paths, $\gamma_1$ and $\gamma_2$ from $2$ to $w$ leading to $f_1$ and $f_2$ respectively.
$$
f_1/f_2=\exp\left(\oint_{\gamma_1-\gamma_2}\frac12\left(\frac1{z-1}+\frac1{z+1}\right)\,\mathrm{d}z\right)
$$
Since the paths don't cross $[-1,1]$, $\gamma_1-\gamma_2$ circles the poles at $-1$ and $1$ the same number of times. Since the residue at each pole is $\frac12$, the integral along $\gamma_1-\gamma_2$ is an integral multiple of $2\pi i$. Therefore, $f_1/f_2=1$.
Looking at the logarithmic derivative of $f$ shows that $f(w)=\sqrt{w^2-1}$.
Note that although we usually think of $\sqrt{x^2-1}$ as an even function, here, $f$ is an odd function.
A: Take the branch of $f(w)=\sqrt w$ that is undefined on the negative reals. Where is $f(z^2-1)$ undefined? You should get $[-1,1]$.
Since the integrand is not meromorphic in the region you are considering, you cannot apply the residue theorem. 
