Do both of these functions define the same branch of $\sqrt{z^2 - 1}$? Let $\sqrt{\cdot}$ be the function defined by 
\begin{align}
\sqrt{z} &= r^\frac{1}{2} e^{i \frac{\theta}{2}}, \\ 
r &= |z|, \\ 
\theta &= \arg(z), \\ 
-\pi &< \theta \le \pi
\end{align}
Let $$f(z) = z\sqrt{1 - \frac{1}{z}}\sqrt{1 + \frac{1}{z}}$$
and let 
$$g(z) = \sqrt{z +1}\sqrt{z - 1}.$$
The functions $f$ and $g$ are both branches of the multivalued function implicitly defined by $w^2 = z^2 - 1$. Indeed $f^2(z) = g^2(z) = z^2 - 1$.
Furthermore $f(x) = g(x)$ for $x > 1$, therefore $f = g$ on $\mathbb C \setminus [-1, 1]$ 
by the Identity theorem.
Is there a more direct way to see this from how $\sqrt{\cdot}$ depends on the argument of a number?
It's tempting to factor $\sqrt{z}$ from both factors of $g$ however for $\alpha, \beta \in \mathbb C$, in general $\sqrt{\alpha\beta} \neq \sqrt{\alpha}\sqrt{\beta}$. 
Thanks.
 A: [Arguments in this answer will always be defined to be in $(-\pi,\pi]$, as in your definition of $\sqrt{z}$.]
The way you have defined square roots, $\sqrt{\alpha\beta}=\sqrt{\alpha}\sqrt{\beta}$ is valid as long as the arguments of $\alpha$ and $\beta$ add to give something inside the interval $(-\pi,\pi]$.
Now suppose $z$ is not real.  Then the arguments of $z$ and $1/z$ have opposite sign, and hence so do the arguments of $z$ and $1+1/z$ (the sign of the argument is the sign of the imaginary part).  Thus the sum of the arguments of $z$ and $1+1/z$ must be in $(-\pi,\pi]$.  That is, $\sqrt{z+1}=\sqrt{z}\sqrt{1+1/z}$.
Now let us consider $z$ and $1-1/z$.  The arguments of these have the same sign, so we must give a more careful argument.  First, note that the arguments of $z$ and $-1/z$ sum to $\pm\pi$.  Now when you add $1$ to $-1/z$, that will shrink the absolute value of its argument (draw a picture if this isn't obvious).  Thus the absolute value of the sum of the arguments of $z$ and $1-1/z$ is less than $\pi$.  This again means $\sqrt{z-1}=\sqrt{z}\sqrt{1-1/z}$.
Thus if $z$ is not real, $\sqrt{1+z}\sqrt{1-z}=\sqrt{z}\sqrt{1+1/z}\sqrt{z}\sqrt{1-1/z}=z\sqrt{1+1/z}\sqrt{1-1/z}$.  We can also easily check that this identity holds for $z>1$ and $z<-1$, so it is valid on all of $\mathbb{C}\setminus[-1,1]$.
