Principal Logarithmic QuestionHere is a question that is driving me insane:Show that p . v z − 1 × p . v z + 1 = − p . v . z 2 − 1 for R e ( z ) &lt; − 1. (p.v. stands for the principal singular valued logarithmic function.Show that p . v . z 2 − 1 is analytic for R e ( z ) &lt; − 1Show that ( p . v . z + 1 ) ( p . v . z − 1 ) is analytic on the complement of z : − 1 ≤ R e ( z ) ≤ 1 , I m ( z ) = 0How that negative appears, I do not know. Working backwards I see that − 1 = e x p ( L o g ( − 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.

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Principal Logarithmic QuestionHere is a question that is driving me insane:Show that   p  .  v      z    −    1    ×  p  .  v      z    +    1    =  −  p  .  v  .            z      2        −    1   for    R  e  (  z  )  &amp;lt;  −  1. (p.v. stands for the principal singular valued logarithmic function.Show that   p  .  v  .            z      2        −    1   is analytic for   R  e  (  z  )  &amp;lt;  −  1Show that   (  p  .  v  .      z    +    1    )  (  p  .  v  .      z    −    1    ) is analytic on the complement of   z  :  −  1  ≤  R  e  (  z  )  ≤  1  ,  I  m  (  z  )  =  0How that negative appears, I do not know. Working backwards I see that   −  1  =  e  x  p  (  L  o  g  (  −  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.

Maximo Sweeney · Accepted Answer

The branch cut of the principal value square root is the real interval    (  −  ∞  ,  0  ]. Note that       z    2    −  1  ∈  (  −  ∞  ,  0  ] for   z on the imaginary axis or on the real interval   [  −  1  ,  1  ], while   z  −  1  ∈  (  −  ∞  ,  0  ] for   z  ∈  (  ∞  ,  1  ] and   z  +  1  ∈  (  −  ∞  ,  0  ] for   z  ∈  (  −  ∞  ,  −  1  ]] . In each of the regions       R    e    (  z  )  &amp;lt;  0,       I    m    (  z  )  &amp;gt;  0 and       R    e    (  z  )  &amp;lt;  0,       I    m    (  z  )  &amp;gt;  0 all three square roots are analytic, and either       z    −    1        z    +    1    =  +            z      2        −    1   or       z    −    1        z    +    1    =  −            z      2        −    1  . So which is it? E.g. you might try it for   z  =  −  1  +  ϵ  i and   z  =  −  1  −  ϵ  i where   ϵ  &amp;gt;  0 is small.

Principal Logarithmic Question Here is a question that is driving me insane: Show that p .

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