$\sqrt{z+1}\sqrt{z-1} = -\sqrt{z^2 - 1}$ when $\Re(z) < -1$ Question:

Show $\sqrt{z+1}\sqrt{z-1} = -\sqrt{z^2 - 1}$ when $\Re(z) < -1$
  Every square root is assumed to be the principal value.  

This is my attempt:
(note that $\mathrm{Log}$ denotes the principal Logarithm)
$$\sqrt{z+1}\sqrt{z-1} := \exp(\frac{1}{2}\mathrm{Log}(z+1))\exp(\frac{1}{2}\mathrm{Log}(z-1)) \\ = \exp(\frac{1}{2}(\mathrm{Log}(z+1) + \mathrm{Log}(z-1))) \\ = \exp(\frac{1}{2}\mathrm{Log}(z^2 - 1)) $$  
Now I'm stuck (and not even sure if that last equality is justified).  
A previous question made me found that $\sqrt{z+1}$ is analytic in $\mathbb{C} \backslash \{z = x+iy | x \leq -1\}$.
 A: Be $\,a:=|a|e^{i\alpha}\,$ and $\,b:=|b|e^{i\beta}\,$ with $\,-\pi<\alpha,\beta\leq \pi\,$ .
For $\,\Re(a)\geq 0\,$ and $\,\Re(b)\geq 0\,$ we have $\sqrt{a}\sqrt{b}=\sqrt{|ab|}e^{\frac{\alpha+\beta}{2}}=\sqrt{|ab|e^{\alpha+\beta}}=\sqrt{ab}$ 
with $\,-\pi< \frac{\alpha+\beta}{2}\leq\pi\,$ .
Be $\,\Re(z)\geq 0\,$ and $\,\Im(z)\ne 0\,$. 
Then it’s $\,\sqrt{-z}= -i\cdot \text{sgn}(\Im(z))\cdot\sqrt{z} \,$ , because by definition it’s always $\,\Re(\sqrt{w})\geq 0\,$ 
for any $\,w\in\mathbb{C}\,$ . This is equivalent to $\,\displaystyle \cos(\frac{\alpha-\pi}{2})\cdot\text{sgn}(\sin \alpha)\ge 0\,$ for $\,-\pi<\alpha\leq \pi\,$ .  
For $\,\Re(a)\geq 0\,$ and $\,\Re(b)\geq 0\,$ follows 
$\,\sqrt{-a}\sqrt{-b}=[-i\cdot\text{sgn}(\Im(a))\cdot\sqrt{a}][-i\cdot\text{sgn}(\Im(b))\cdot\sqrt{b}]= -\text{sgn}(\Im(a))\text{sgn}(\Im(b)) \sqrt{ab}$ .       
With $\,a:=-z-1\,$ and $\,b:=-z+1\,$ and $\,\Re(z)<-1\,$ we get 
$\sqrt{z+1} \sqrt{z-1}=-\text{sgn}^2(\Im(-z)) \sqrt{(-z-1)(-z+1)}=-\sqrt{(-z)^2-1}=-\sqrt{z^2-1}\,$ .
A: Note that, in general:
$$\operatorname{Arg}z_1+\operatorname{Arg}z_2\ne\operatorname{Arg}(z_1z_2)$$
Being $\operatorname{Arg}z\in(-\pi,\pi]$,  $\operatorname{Arg}z_1+\operatorname{Arg}z_2\in(-2\pi, 2\pi]$ and so:
$
\begin{array}{l|l}
\operatorname{Arg}z_1+\operatorname{Arg}z_2&\operatorname{Arg}(z_1z_2)\\
\hline
(-2\pi,-\pi]&\operatorname{Arg}z_1+\operatorname{Arg}z_2+2\pi\\
(-\pi,\pi]&\operatorname{Arg}z_1+\operatorname{Arg}z_2\\
(\pi,2\pi]&\operatorname{Arg}z_1+\operatorname{Arg}z_2-2\pi\\
\end{array}\tag{1}
$
In your case, being $\Re{(z)}<-1$, when $\arg(z)\in(\frac{\pi}{2},\pi]$ it results that $\operatorname{Arg}(z+1),\,\operatorname{Arg}(z-1)\in(\frac{\pi}{2},\pi]$ and $\operatorname{Arg}(z+1)+\operatorname{Arg}(z-1)\in(\pi,2\pi]$, therefore, from $(1)$, $$\operatorname{Arg}(z^2+1)=\operatorname{Arg}(z+1)+\operatorname{Arg}(z-1)-2\pi,~\text{ if }\arg(z)\in(\frac{\pi}{2},\pi]\tag{2a}$$.
At the same time when $\arg(z)\in(-\pi,-\frac{\pi}{2})$ it results that $\operatorname{Arg}(z+1),\,\operatorname{Arg}(z-1)\in(-\pi,-\frac{\pi}{2})$ and $\operatorname{Arg}(z+1)+\operatorname{Arg}(z-1)\in(-2\pi,-\pi)$, therefore, from $(1)$, $$\operatorname{Arg}(z^2+1)=\operatorname{Arg}(z+1)+\operatorname{Arg}(z-1)+2\pi,~\text{ if }\arg(z)\in(-\pi,-\frac{\pi}{2})\tag{2b}$$.
So your computation must proceed this way: if $\Re(z)<1$, then
$\sqrt{z-1}\sqrt{z+1}=\sqrt[+]{|z-1|}\sqrt[+]{|z+1|}e^{i\operatorname{Arg}(\sqrt{z-1})+i\operatorname{Arg}(\sqrt{z+1})}\stackrel{(2)}=\begin{cases}\stackrel{\arg(z)\in(\frac{\pi}{2},\pi]}=&\sqrt[+]{|z^2-1|}e^{i\operatorname{Arg}(\sqrt{z^2-1})-i\pi}=-\sqrt[+]{|z^2-1|}e^{\operatorname{Arg}(\sqrt{z^2-1})}=-\sqrt{z^2-1}\\\stackrel{\arg(z)\in(-\pi,-\frac{\pi}{2})}=&\sqrt[+]{|z^2-1|}e^{i\operatorname{Arg}(\sqrt{z^2-1})+i\pi}=-\sqrt[+]{|z^2-1|}e^{\operatorname{Arg}(\sqrt{z^2-1})}=-\sqrt{z^2-1}\end{cases}$
You can also use $\operatorname{Log}$ instead of $\operatorname{Arg}$ but it is a bit more involved.
First note that 
$$\operatorname{Log}z=\log|z|+i\operatorname{Arg}z$$
and then:
$\sqrt{z-1}\sqrt{z+1}=e^{\frac{1}{2}\operatorname{Log}(z+1)}e^{\frac{1}{2}\operatorname{Log}(z-1)}=e^{\frac{1}{2}\log(|z-1|)\log(|z+1|)+\frac{1}{2}i\operatorname{Arg}(z+1)\operatorname{Arg}(z-1)}\stackrel{(2\text{a})\text{ and }(2\text{b})}=\sqrt[+]{|z^2-1|}e^{\frac{1}{2}i\operatorname{Arg}(z^2+1)\pm i\pi}=-\sqrt{z^2-1}$ 
A: Since $\left|\,\arg\left(\sqrt{z^2-1}\right)\,\right|\lt\frac\pi2$, when
$$
\left|\,\arg\left(\sqrt{z-1}\right)+\arg\left(\sqrt{z+1}\right)\,\right|\lt\frac\pi2\tag{1}
$$
we have $\sqrt{z-1}\sqrt{z+1}=\sqrt{z^2-1}$; otherwise, $\sqrt{z-1}\sqrt{z+1}=-\sqrt{z^2-1}$ because the argument of the left side needs to be adjusted by $\pi$ to put it back into $\left[-\frac\pi2,\frac\pi2\right]$.
$(1)$ is equivalent to
$$
\left|\,\arg(z-1)+\arg(z+1)\,\right|\lt\pi\tag{2}
$$
The situation for $\operatorname{Im}(z)\lt0$ is the same as for $\operatorname{Im}(z)\gt0$, just with the args negated; that is, $\sqrt{\bar{z}}=\overline{\sqrt{z}}$ and $\arg\left(\bar{z}\right)=-\arg(z)$.
So assume $\operatorname{Im}(z)\gt0$. Then $(2)$ becomes
$$
\arg(z-1)+\arg(z+1)\lt\pi\tag{3}
$$
$\operatorname{Im}(z)\gt0$ is pictured below:

Note that $(3)$ becomes
$$
\arg(z+1)\lt\pi-\arg(z-1)\tag{4}
$$
which is just saying that $\operatorname{Re}(z)\gt0$.
Thus,
$$
\operatorname{Re}(z)\gt0\implies\sqrt{z-1}\sqrt{z+1}=\sqrt{z^2-1}
$$
and
$$
\operatorname{Re}(z)\lt0\implies\sqrt{z-1}\sqrt{z+1}=-\sqrt{z^2-1}
$$
