Here is a question that is driving me insane:
Show that $p.v \sqrt{z-1}\times p.v\sqrt{z+1}=-p.v.\sqrt{z^2 -1}$ for $Re(z)<-1.$(p.v. stands for the principal singular valued logarithmic function.
Show that $p.v. \sqrt{z^2-1}$ is analytic for $Re(z)<-1$.
Show that $(p.v.\sqrt{z+1})(p.v.\sqrt{z-1})$ is analytic on the complement of $z:-1\leq Re(z)\leq 1,Im(z)=0$
How that negative appears, I do not know. Working backwards I see that $-1=exp(Log(-1))$. That is all I can see. Thank you to all for your help !
EDIT: I have added to two following questions which I still can not do. This is really bothering - I seem to be missing something very important.
The branch cut of the principal value square root is the real interval $(-\infty,0]$. Note that $z^2-1 \in (-\infty,0]$ for $z$ on the imaginary axis or on the real interval $[-1,1]$, while $z-1 \in (-\infty,0]$ for $z \in (\infty,1]$ and $z+1 \in (-\infty,0]$ for $z \in (-\infty,-1]$. In each of the regions ${\text Re}(z) < 0$, ${\text Im}(z) > 0$ and ${\text Re}(z) < 0$, ${\text Im}(z) > 0$ all three square roots are analytic, and either $\sqrt{z-1} \sqrt{z+1} = +\sqrt{z^2-1}$ or $\sqrt{z-1} \sqrt{z+1} = -\sqrt{z^2-1}$. So which is it? E.g. you might try it for $z = -1+\epsilon i$ and $z=-1-\epsilon i$ where $\epsilon > 0$ is small.