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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 $\mathbb{C} \backslash \{z = x+iy | x \leq -1\}$.

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    $\begingroup$ Since "principal value" is not a standard concept for square roots in $\mathbb{C}$, could you state how, exactly, the author defined it? $\endgroup$ – Daniel Fischer Aug 22 '17 at 14:34
  • $\begingroup$ Sorry for the late response: it is defined as $$PV\sqrt{w} = \begin{cases} |w|^{\frac{1}{2}} e^{i \mathrm{Arg}(w)/2}, \quad w \neq 0, \\ 0 \qquad \qquad \quad \ \ \quad w = 0.\end{cases}$$ where $\mathrm{Arg}$ is the principal Argument value on $(-\pi,\pi]$. $\endgroup$ – Twenty-six colours Aug 23 '17 at 9:57
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    $\begingroup$ You've got three answers. Is there something left which you like to know ? :-) $\endgroup$ – user90369 Sep 5 '17 at 13:55
  • $\begingroup$ No, I accidentally didn't assign the bounty reputation before it expired, so I started a new one (chose the "additional bounty" one) to assign it but I have to wait 24 hours. $\endgroup$ – Twenty-six colours Sep 5 '17 at 13:57
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    $\begingroup$ Thanks a lot for your bounty! :-) $\endgroup$ – user90369 Sep 7 '17 at 14:24
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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

$\sqrt{z+1} \sqrt{z-1}=-\text{sgn}^2(\Im(-z)) \sqrt{(-z-1)(-z+1)}=-\sqrt{(-z)^2-1}=-\sqrt{z^2-1}\,$ .

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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:

$\sqrt{z-1}\sqrt{z+1}=e^{\frac{1}{2}\operatorname{Log}(z+1)}e^{\frac{1}{2}\operatorname{Log}(z-1)}=e^{\frac{1}{2}\log(|z-1|)\log(|z+1|)+\frac{1}{2}i\operatorname{Arg}(z+1)\operatorname{Arg}(z-1)}\stackrel{(2\text{a})\text{ and }(2\text{b})}=\sqrt[+]{|z^2-1|}e^{\frac{1}{2}i\operatorname{Arg}(z^2+1)\pm i\pi}=-\sqrt{z^2-1}$

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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:

enter image description here

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 $$ \operatorname{Re}(z)\lt0\implies\sqrt{z-1}\sqrt{z+1}=-\sqrt{z^2-1} $$

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