Defining a branch of $(1-\zeta^2)^{-1/2}$

Question

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 $(-1,1)$?

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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.