Prove that there exist a branch The question is-
Prove that there is a branch for the function $\sqrt{z-a}\sqrt{z-b}$ where $a,b$ are complex numbers, in $C\backslash [a,b]$.
Generally,I have no idea how to prove that there is a branch for a function...
Any idea?
 A: What follows is mainly rigorous but happens to be a little intuitive in its presentation. Some supplementary work is necessary on your side...
You know that the definition of $\sqrt{z}=z^{1/2}$ (like the "log" function) needs the introduction of a so-called "cut", which is a half line issued from $0$ that is "forbidden" in $\mathbb{C}$. 
Rather often, this branch cut is taken along the negative real axis, in order to be in harmony with the real fonction $\sqrt{x}$ (it is called the principal determination of the complex square root). In the case of functiion $\sqrt{z-a}$, $a \in \mathbb{C}$, the branch cut is a half line issued from $a$. 
A cousin view is to consider that the "cut" can be crossed but according to the way it is crossed, we have to "pay a toll" $-\pi$ (resp. $+\pi$) for the "right" of crossing it, in the direct (resp. clockwise) direction.
Explanation in the case of the principal determination of  $\sqrt{z}$ for two very near points situated on each side of the cut (the border): 
If $z_1:=re^{i (\pi-\varepsilon)}, \ \sqrt{z_1}=\sqrt{r}e^{i \pi/2} e^{-i\varepsilon/2}$ whereas 
if $z_2:=re^{-i (\pi-\varepsilon)}, \ \sqrt{z_2}=\sqrt{r}e^{-i \pi/2}e^{+i \varepsilon/2}$.
But $z_1$ is arbitrarily close to $z_2$ (with $\varepsilon \to 0$); it is why, in order that the square roots of $z_1$ and $z_2$ are in agreement, we consider that crossing the cut in the direction $z_2 \to z_1$ makes a $2*\pi/2=\pi$ jump for the argument, and crossing the cut in the inverse direction contributes to a jump of $-\pi.$
Now, if we have at the same time $\sqrt{z-a}$ and $\sqrt{z-b}$, is it possible to have a thorough cut permitting to give a non-ambiguous signification to $\sqrt{(z-a)(z-b)}$. Let us have a look at Fig. 1 below. It represents the four candidate cases.
The top case is with the two rays going to the left (case of two principal determinations), the second case with the ray issued from $E$ going to the left, as before, and the cut issued from $F$ going to the right, is favorable. 
I will give a graphical illustration for the second case. See Fig. 2, which has been build with the following Mathematica instruction:

Plot3D[Arg[Sqrt[(x + I y) - 2]] + Arg[Sqrt[-(x + I y) - 2]], 
  {x, -6, 6}, {y, -6, 6}, ViewPoint -> {-1, 2, 1}, PlotPoints -> 200, Mesh -> 
      False]

which represents the argument of $\sqrt{(z-2)(2+z)}$. We see that the interval $[-2,2]$ is "harmoniously" crossed, but that, either on $(-\infty,-2]$ or $[2,+\infty)$, there is a "cliff" with heigth $\pi$ (remember : jump of $+\pi$ or $-\pi$ jump.) 
In the same vein, Fig. 3 illustrates the fourth case (with $A$ and $B$) somewhat "dual" to the previous one : in this case, only segment $[-2,2]$ is forbidden (cliff with height $\pi$). Elsewhere, the conflict between "paying $\pi$ and paying $-\pi$ is annihilated: it is like if you have two infinitely close toll boothes: one where you pay $\pi$ euros and the next one where you are given $\pi$ euros... (it never happens that way in true life).
Fig. 1:

Fig. 2:

Fig. 3:

