How to show that a complex function have a branch in a domain I've given as homework to show that the function $$f(z)=\sqrt{\frac{z+1}{z-1}} $$ has a branch on $G = \mathbb C \backslash [-1,1] $.
I'm having a hard time in finding the way to approach this kind of questions, so what I'm actually looking is on a way to approach that kind of questions and not the solution for this specific question.
Thanks!
 A: Note that $f$ is the composition of two functions: a Möbius transformation and the square root. So, the domain of the Möbius transformation is $\mathbb{C}\backslash\{1\}$ and its image is all the plane (Why?). On the other hand, the square root is only defined in a branch of the logarithm. 
Where does the Möbius transformation maps the interval $[-1,1]$? Since it maps $-1$ to $0$, $0$ to $-1$ and $1$ to infinity, it maps the interval to the nonpositive axis! (Why?) So the principal branch of the logarithm is defined in the image of $\mathbb{C}\backslash[-1,1]$ under the Möbius transformation, which is $\mathbb{C}\backslash(-\infty,0]$.
A: It could be useful, even if not proper/formal, to integrate the function over a small loop around the suspicious point (withtin the domain of analiticity of the function in order to take advantage of Cauchy theorems): this should result in an expression involving quantities path-related (eg. the radius of a small circle: try with $\sqrt{z}$ around the origin) or phase-dependent. 
Another improper but useful way is to study the behaviour of the function around $z_0e^{\pm i\epsilon}$ and $z_0e^{i(2\pi\pm\epsilon)}$, $\epsilon>0$: this should result in different values for $f(z)$ even if evaluated in two "different (for a phase factor) but identified" points.  
A: Lemma. If $a,b$ belong to the same connected component of $\mathbb C\smallsetminus\mathbb \Omega$, where $\Omega\subset\mathbb C$ open, then 
$$
f(z)=\frac{z-a}{z-b},
$$
possesses an analytic logarithm in $\Omega$ (i.e., there exists a $g\in\mathcal H(\Omega)$, 
such that $\exp(g(z))=f(z)$, for all $z\in\Omega$).
Applying this lemma, we obtain an analytic logarithm $g$ for $\frac{z+1}{z-1}$ in $\mathbb C\smallsetminus [-1,1]$, and we can define 
$$
\sqrt{\frac{z+1}{z-1}}=\mathrm{e}^{\frac{1}{2}g(z)}.
$$
Proof of the Lemma. Fix $z_0\in\Omega$. A nowhere vanishing analytic function $f\in\mathcal H(\Omega)$ possesses an analytic logarithm if and only if the function 
$$
g(z)=\int_{z_0}^z\frac{f'(\zeta)}{f(\zeta)}\,d\zeta,
$$
is univalent (i.e., has only one value independent of the path in $\Omega$ 
connecting $z_0$ and $z$). Equivalently
$$
\int_\gamma\frac{f'(\zeta)}{f(\zeta)}\,d\zeta=0,
$$
for every closed path in $\Omega$. But
$$
\frac{f'(\zeta)}{f(\zeta)}=\cdots=\frac{1}{z-a}-\frac{1}{z-b},
$$
and hence
$$
\int_\gamma\frac{f'(\zeta)}{f(\zeta)}\,d\zeta=2\pi i\big(\mathrm{Ind}_\gamma(a)-\mathrm{Ind}_\gamma(b)\big).
$$
Here $\mathrm{Ind}_\gamma(w)$ is the index of $w$ with respect to the closed path $\gamma$.
But, as $a$ and $b$ belong to the same connected component of $\mathbb C\smallsetminus\Omega$, they have the same index:  $\mathrm{Ind}_\gamma(a)=\mathrm{Ind}_\gamma(b)$. Thus
$$
\int_\gamma\frac{f'(\zeta)}{f(\zeta)}\,d\zeta=0.
$$
