Holomorphicity of $(z^2-1)^{1/2}$ In one workshop question I am asked to find a branch of $f(z) = (z^2-1)^{\frac{1}{2}}$ such that it's holomorphic on the exterior of the unit disk $|z|>1$. Hint for the question is to separate $z^2-1$ into $z^2(1-z^{-2})$ and take the logarithm only for the second term when computing the power, as $(z^2)^{\frac{1}{2}} = z$. Thus we can find a branch.
I am really reluctant to the solution. In theory $f(z) = \exp(\frac{1}{2}\log_\theta(z^2-1))$, for some $\theta$ which defines the branch cut. Thus, I need to find a $\theta$ such that the image of $z^2-1$ doesn't cross the cut, which is impossible.
Any help please :)?
 A: First note that $f(z) = \sqrt{z^2-1}$ has branch points at $z=\pm 1$.  
We cut the plane with branch cuts from $(-1,0)$ to $(\infty, 0)$ and from $(1,0)$ to $(\infty,0)$ such that 
$$0\le \arg(z-1)<2\pi$$
and 
$$0\le \arg(z+1)<2\pi$$

Next, we note that on the real axis for $\text{Re}(z)>1$, coming from the upper-half plane, 
$$\begin{align}
\arg(\sqrt{z^2-1})&=\frac12 \arg(z-1)+\frac12 \arg(z+1)\\\\
&=\frac12 (0)+\frac12 (0)\\\\
&=0
\end{align}$$
while on the real axis for $\text{Re}(z)>1$, coming from the lower-half plane, 
$$\begin{align}
\arg(\sqrt{z^2-1})&=\frac12 \arg(z-1)+\frac12 \arg(z+1)\\\\
&=\frac12 (2\pi)+\frac12 (2\pi)\\\\
&=2\pi
\end{align}$$
And since $e^{i2\pi }=1$, we find that $f(z)$ is continuous across that part of the real axis for which $\text{Re}(z)>1$.

Similarly, we find that $f(z)$ is continuous across that part of the real axis for which $\text{Re}(z)<-1$.

Finally, let's examine the behavior of $f$ along the real axis between $z=-1$ and $z=1$.  Coming from the upper-half plane, we find that
$$\begin{align}
\arg(\sqrt{z^2-1})&=\frac12 \arg(z-1)+\frac12 \arg(z+1)\\\\
&=\frac12 (\pi)+\frac12 (0)\\\\
&=\pi/2
\end{align}$$
while coming from the lower-half plane, we find that
$$\begin{align}
\arg(\sqrt{z^2-1})&=\frac12 \arg(z-1)+\frac12 \arg(z+1)\\\\
&=\frac12 (\pi)+\frac12 (2\pi)\\\\
&=3\pi/2
\end{align}$$
Note that $e^{i\pi/2}=-e^{i3\pi/2}$ and the function $f(z)$ is, therefore, discontinuous across the line segment from $z=-1$ to $z=1$.


Putting all of this together, we conclude that the two individual chosen branch cuts have a "cancellation" effect for $\text{Re}(z)>1$.  The net result is a branch cut from $z=-1$ to $z=1$ and $f(z)$ is analytic outside the disk $|z|=1$.


NOTE:
A word of caution is in order here.  This "cancellation" effect does not happen in general.  The line segment between the two branch points of the function $g(z)=(z^2-1)^{1/3}$, for example, is not a suitable branch cut for $g(z)$.
A: Actually, you just need to know that any closed loop crosses the cut an even number of times; since each crossing represents a sign change, you get consistency along paths. 
In this case, the log of $z^2$ will have branch cuts on, say, the positive and negative real axes, and the only homotopically nontrivial loops are those that go around the disk some integral number of times, hence hit these branch cuts an even number of times. 
