Branch of $\sqrt{1-z^2}$ Show that a branch of $\sqrt{1-z^2}$ can be defined in any region $\Omega$ where the points $1,-1$ are in the same component of its complement.
This is a question in Ahlfors' Complex Analysis (P.148 Q5) that I came across while trying to self-study the book. I tried to tackle the problem by considering $\Omega=\mathbb{C} \backslash [-1,1]$ first, and tried the approach as in Section 4.4 Corollary 2, namely find a branch of the corresponding log first. For this $\Omega$, the image of $1-z^2$ is $\mathbb{C} \backslash [0,1]$, so a branch of $log(1-z^2)$ cannot be defined; evidently one needs to construct the branch of $\sqrt{1-z^2}$ directly. Here is where I ran out of ideas... Any help is appreciated!
 A: A branch of $\sqrt{1-z^2}$  is a lift of $f(z)=1-z^2$ with respect to the covering map $p:\mathbb{C}^\times\to\mathbb{C}^\times,\,p(z)=z^2 $. Fix a basepoint $z_0\in\Omega$, and let $\tilde{w_0}\in \mathbb{C}^\times$ be a point such that $p(\tilde{w_0})=f(z_0)$. By the theory of covering spaces, $f$ has a lift with respect to $p$ if and only if $f_\ast(\pi_1(\Omega,z_0))\subset p_\ast(\pi _1(\mathbb{C}^\times,\tilde{w_0}))$. Now we may identify $\pi_1(\mathbb{C}^\times,{f(z_0)})$ with $\mathbb{Z}$ via the integration $$\gamma\mapsto \frac{1}{2\pi i}\int_\gamma \frac{1}{z}dz,$$ and under this identification $\operatorname{im}p_\ast$ corresponds to $2\mathbb{Z}$. THerefore, we have $\operatorname{im}f_\ast\subset\operatorname{im}p_\ast$ if and only if $$\frac{1}{2\pi i} \int_{f\ast\gamma}\frac{1}{z}dz\in 2\mathbb{Z}$$ for (every piecewise $C^1$) curve. On the other hand, $$\frac{1}{2\pi i} \int_{f\ast\gamma}\frac{1}{z}dz=\frac{1}{2\pi i}\int_{\gamma}\frac{2z}{z^2-1}dz=\frac{1}{2\pi i}\int_\gamma\left(\frac{1}{1-z}+\frac{1}{1+z}\right)dz$$ is the sum of the winding number of $\gamma$ around $1$ and $-1$. By  our hypothesis on $\Omega$, these winding numbers are equal, and we are done.
A: Let $\gamma$ be a closed curve in $\Omega$. Then
$$
\int_\gamma \bigl( \frac{1}{z-1} -  \frac{1}{z+1} \bigr) \, dz
= 2 \pi i \bigl( n(\gamma, 1) - n(\gamma , -1) \bigr)
$$
where $n$ denotes the winding number. $1$ and $-1$ are in 
the same component of $\mathbb C \setminus \Omega$ and
therefore also in the same component of $\mathbb C \setminus |\gamma|$.
Since the winding number is constant on each component
of $\mathbb C \setminus |\gamma|$, the integral is zero.
This holds for all closed curves in $\Omega$.
It follows that the integrand has a holomorphic antiderivative 
$$
  F : \Omega \to \mathbb C, \quad F'(z) = \frac{1}{z-1} -  \frac{1}{z+1} \, .
$$
Now define
$$
 f : \Omega \to \mathbb C, \quad f(z) = (z+1)e^{F(z)/2} \, ,
$$
this is (almost) the holomorphic branch of $\sqrt{1-z^2}$.
The function
$$
g : \Omega \to \mathbb C, \quad g(z) = \frac{f(z)^2}{z^2 -1}
= \frac{z+1}{z -1} e^{F(z)}
$$
satisfies
$$
\frac{g'(z)}{g(z)} = \frac{1}{z+1} -  \frac{1}{z-1} + F'(z) = 0
$$
It follows that $g$ is constant and therefore
$$
 f(z)^2 = C (1-z^2)
$$
for some constant $C$. Then $f(z)/\sqrt{C}$ is a holomorphic branch of $\sqrt{1-z^2}$ in $\Omega$.

The idea behind those calculations was to show that
$
F(z) = \log \frac{z-1}{z+1}
$ has a holomorphic branch in $\Omega$, then
$$
e^{F(z)} = \frac{z-1}{z+1} = \frac{z^2-1}{(z+1)^2}
$$
which leads to the "candidate" $f(z) = (z+1)e^{F(z)/2}$.
