TITLE: How to properly understand branches of complex functions
QUESTION [9 upvotes]: $\DeclareMathOperator{\Log}{Log}$
I have several problems to understand the concept of branches and how to find analytic branches.
From what I learned, for example for the complex logarithm, it is a multi valued function, and if we want it to be analytic we have to cut some part of the domain (because otherwise we get different limits in the same point).
I understand then why $\Log(z)$ is analytic in the branch $\mathbb{C} \setminus (-\infty ,0]$ , since we can never complete a full circle around $0$. Here it is a simple case so it is easy to see that we always need to throw a ray from the origin.
My confusion starts when the function is not that simple. Let's take the function $Log(z^2-1)$ . I can understand why on the domain $\{ |z| < 1\}$ an analytic branch would be $\mathbb{C} \setminus [0,\infty )$, since this function takes the unit circle to itself and moves it left by $1$. So, the ray $[0,\infty )$ doesn't intersect with it. 
But what if the domain is $\{ |z| > 1\}$ ? How do I work with it since there is not such a pretty way? I thought of maybe using the main branch of the logarithm, and seeing where $z^2-1 \in (-\infty ,0]$, but is it what needs to be done.
Moreover, what about functions like $\sqrt{z^2-1}$  ? How do I start to look for an analytic branch there? It seems logical that the points $1$ and $-1$ play a part here but I am not sure how.
Another thing is, how do I solve integral with such functions? For example $$\int_{|z| = 2} \sqrt{z^2-1}$$ When the branch is defined in the following way: $$\sqrt{z^2-1} = z\sqrt{1-\frac {1}{z^2}} = z\exp[\frac{1}{2}Log(1-\frac {1}{z^2})]$$
How does the definition of the branch even play a part here?
Another example could be the integral: 
$$\int_{|z|=2} \frac{1}{\sqrt{z^4+4z+1}}$$ when $\sqrt{25} = 5$
Help would be tremendously appreciated. I someone could walk me thorugh an entire example, I would be really glad.

REPLY [2 votes]: $\log(z^2-1)$ is a standard analytic function on $|z-2| < 1$ (or $\Bbb{C}- [1,\infty)-(-\infty,-1]$, or ...) then what you need is to see all its possible analytic continuations. In general those can be very complicated (see the branch point $z=1$ of $\log \log z$ which isn't in  $\log(2i \pi +\log z)$ )
Sometimes "all the analytic continuation" can be described in a very simple way. For this what you need is the monodromy group : that is how $f(z)$ transforms under analytic continuation along curves in the homotopy group. For example $ f(z) = z^{1/3}$, its homotopy group is the closed curves around $z=0$, along a counterclockwise rotation $C_0$ around $z=0$, $f(z)$ transforms into $f_{C_0}(z) = e^{2i \pi /3} f(z)$ so $f_{C_0^n}(z)$ transforms into $f_{C_0^{n+1}}(z)=e^{2i \pi /3} f_{C_0^n}(z)$, thus the monodromy group is  of order $3$. 
For $\log z$ the monodromy group is generated by $C_0$ which sends $f(z)$ to $f(z)+2i\pi$ and is $\cong \Bbb{Z}$. 
For $f(z) = (z-1)^{1/2} (z+1)^{1/2}$ the monodromy group is generated by $C_1,C_{-1}$, $f_{C_1}(z) = -f(z), f_{C_{-1}}(z) = -f(z), f_{C_1 C_{-1}}=f_{C_{-1} C_1} = f$.