Choose the branch for $(1-\zeta^2)^{1/2}$ that makes it holomorphic in the upper half-plane and positive when $-1<\zeta<1$

Question

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, $$(1-\zeta^2)^{-1/2}=i(\zeta^2-1)^{-1/2}\quad \text{when }\zeta>1.$$

I'm a little confused on the branch they're choosing. Do they mean that the we cut along $\mathbb{R}\setminus [-1,1]$? But then $(\zeta^2-1)^{-1/2}$ is undefined for $\zeta>1$...

Answer 1

A good way to think about powers is through the logarithm $$ (1-\zeta^{2})^{1/2} = e^{\frac{1}{2}\log(1-\zeta^{2})}. $$ Wherever the logarithm is holomorphic, so is $(1-\zeta^{2})^{-1/2}$. Wherever the logarithm is real, the square root is real and positive. Logarithm of $f(z)=(1-z^{2})$ is done using an integral of $f'(z)/f(z)$. In this case, $$ \log(1-\zeta^{2})=\int_{0}^{\zeta}\frac{-2z}{1-z^{2}}\,dz. $$ The only trick in designing the logarithm is to restrict the domain so that you cannot circle a point where there is a non-zero residue. So the definition of the above works just fine, so long as you choose any path from 0 to $\zeta$ along a path which does cross the real axis on $(-\infty,-1]$ or on $[1,\infty)$. That's how branch cuts are handled. Branch cuts can be along just about any curve you like, chosen to make sure you cannot circle the places where the integrand has non-zero residues. Then $$ (1-\zeta^{2})^{1/2} = \exp\left\{-\frac{1}{2}\int_{0}^{\zeta}\frac{2z}{1-z^{2}}dz\right\}=\exp\left\{\frac{1}{2}\int_{0}^{\zeta}\frac{1}{z-1}+\frac{1}{z+1}\,dz\right\}. $$ You can see that this function is positive and real on the real axis $(-1,1)$ because the integral reduces to one on the real line where the integrand is real. Furthermore, function is discontinuous across the real axis on $(-\infty,1]$ by a multiplicative constant equal to $\exp\{\frac{1}{2}(-2\pi i)\}=-1$ because of the residue of $\frac{1}{1+z}$ at $z=-1$.

Answer 2

Another (probably not as good) way to look at it is to think $$ \sqrt{1-\zeta^2} = i \sqrt{\zeta -1} \sqrt{\zeta + 1} $$ Choose the (possibly different) branches for each of the square roots on the right side to give you what you need. It might look at first like you will have problems on $z \in [-1,1]$, but the discontinuities cancel each other.

For example, use the principle branch of $\sqrt{w}$ on $\zeta - 1$, but for $\zeta + 1$ use the branch of $\sqrt{w}$ where $-2\pi < \arg(w) < 0$.

EDIT:
Use $-2 \pi \le \arg(\zeta+1) < 0$ and $-\pi < \arg(\zeta -1) \le \pi$.
Then $-\pi \le \arg(\sqrt{\zeta+1}) < 0$ and $-\pi/2 < \arg(\sqrt{\zeta -1}) \le \pi/2$.

So on $-1 < z < 1$ you get $$ \arg(i \sqrt{\zeta -1} \sqrt{\zeta + 1} ) = \pi/2 + (-\pi) + \pi/2 = 0 $$ On $1 < z$ you get $$ \arg(i \sqrt{\zeta -1} \sqrt{\zeta + 1} ) = \pi/2 + (-\pi) + 0 = -\pi/2 $$ so $$ \arg(\frac{1}{i \sqrt{\zeta -1} \sqrt{\zeta + 1}} ) = \pi/2 = \arg(\frac{i}{\sqrt{t^2-1}}) $$ where $t = |\zeta| > 1$ and $\sqrt{t^2-1}$ uses the regular square root on $\mathbb{R}^+$.