Defining a branch of $(1-\zeta^2)^{-1/2}$ In this question I brought up a passage from Stein/Shakarchi's Complex Analysis page 232: 

...We consider for $z\in \mathbb{H}$, $$f(z)=\int_0^z
 \frac{d\zeta}{(1-\zeta^2)^{1/2}},$$ where the integral is taken from
  $0$ to $z$ along any path in the closed upper half-plane. We choose
  the branch for $(1-\zeta^2)^{1/2}$ that makes it holomorphic in the
  upper half-plane and positive when $-1<\zeta<1$. As a result,
  $$(1-\zeta^2)^{-1/2}=i(\zeta^2-1)^{-1/2}\quad \text{when }\zeta>1.$$

One thing I'm still not quite clear on: why is there a factor of $i$ between $(1-\zeta^2)^{-1/2}$ and $(\zeta^2-1)^{-1/2}$? If we look at the argument of $(1-\zeta)(1+\zeta)$, it seems like it should change by $\pi$ when we go by $\zeta=1$, and change again by $\pi$ as we go by $\zeta=-1$. Therefore it is changing by $2\pi$ total...halve that and you get $\pi$, apply the exponential and you get a factor of $-1$. So why is the factor $i$?
Also, explicitly what branches of ${\sqrt{1-\zeta}}$ and $\sqrt{1+\zeta}$ are we choosing to make it real and positive on $(-1,1)$?
 A: $\newcommand{\+}{^{\dagger}}%
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I found some way which makes clear the integration in cases like this one: Split the integral into two pieces and translate the integrand to start both from zero. This is a 'brute force trick' but, in my particular case, I found this one quite clear and not confusing.
Besides the fact that the integral is performed in the complex plane ( we can always fix that ):
\begin{align}
&\int_{0}^{z}{\dd \zeta \over \pars{1 - \zeta^{2}}^{1/2}}
=
\half\bracks{\int_{0}^{z}{\dd \zeta \over \pars{1 - \zeta^{2}}^{1/2}} + \int_{0}^{z}{\dd \zeta \over \pars{1 - \zeta^{2}}^{1/2}}}
\\[3mm]&=
\half\braces{\bracks{%
\int_{0}^{-1}{\dd \zeta \over \pars{1 - \zeta^{2}}^{1/2}}
+
\int_{-1}^{z}{\dd \zeta \over \pars{1 - \zeta^{2}}^{1/2}}} + \bracks{%
\int_{0}^{1}{\dd \zeta \over \pars{1 - \zeta^{2}}^{1/2}}
+
\int_{1}^{z}{\dd \zeta \over \pars{1 - \zeta^{2}}^{1/2}}}}
\\[3mm]&=
\half\int_{-1}^{z}{\dd \zeta \over \bracks{\pars{1 - \zeta}\pars{1 + \zeta}}^{1/2}}
+
\half\int_{1}^{z}{\dd \zeta \over \bracks{\pars{1 - \zeta}\pars{1 + \zeta}}^{1/2}}
+
\half\int_{-1}^{1}{\dd \zeta \over \pars{1 - \zeta^{2}}^{1/2}}
\\[3mm]&=
\half\int_{0}^{z + 1}{\xi^{-1/2}\dd \zeta \over \pars{2 - \zeta}^{1/2}}
+
\half\int_{0}^{z - 1}{\pars{-\xi}^{-1/2}\dd \zeta \over \pars{2 + \zeta}^{1/2}}
+ \half\,\pi\,,\qquad z \not= \pm 1
\end{align}
Indeed, the 'fine' procedure will involve 'path parametrization' but I hope this illustrates the general idea.
A: The defining property of $(1-\zeta)^{-1/2}$ is that its square is $1/(1-\zeta^{2})$. Both of the forms you gave check out that way because $i^{2}=-1$. This is why using the traditional view of branch cuts can be confusing. If you consider $f(z)=\sqrt{z}$ to be the branch cut where $z= x+i0$ for $-\infty \le x \le 0$, then $\sqrt{z}$ and $i\sqrt{-z}$, while both square roots of $z$, have different branch cuts when considered as functions of $z$. Using logarithms to define the powers makes the branch cuts more obvious.
