Why we need branch cut?[Continuity and analytic on Branch cuts] Why we need branch cut in Complex analysis? It just my guess though, complex function itself is multi-valued function, so To define the one-valued function we take the branch cut, right? 
Also what about the continuity and analytic on the branch cut for the function $f(z)$ for $z \in \mathbb{C}$? Is $f(z)$ always both discontinuous and non-analytic on branch cut and the branch point? (The below e.g. is reason that why I did think that.)
e.g.) $f(z) = z^{1\over2}$ for $D = \{z \vert -{3\pi\over2} \leq arg(z) \lt {\pi\over 2} \}$ (It is not analytic on the $\{z \vert arg(z) = -{3\pi\over2} \}$ )
Plus Does entire function not have a branch cut?
 A: I would like to give here an intuitive understanding of the way branch cuts can be mastered in practical situations.
First of all, a recall about a fundamental function needing a branch cut, "the" $\log$ function. I have placed quotes around "the" because a same name "log" is given in complex function theory to an infinite number of functions ; a specific $\log$ function is completely characterized by the way one places a cut (any halfline issued from the origin or another point ; if one takes the negative real axis as on the upper left part of Fig. 1, we define in this way the « principal branch » of $\log$ function).
Two different intuitive ways to consider a cut associated to a "log" function :

*

*either you consider you can't cross it.


*or you consider that, when crossing it, you pay a precise toll as when you enter a toll highway. The toll amount is $\color{red}{+}2i\pi$ ou $\color{red}{-}2i\pi$ (using an imaginary currency (!)) according to the direction the branch cut is crossed (minus/plus sign according to the fact you cross it in a clockwise/anticlockwise manner resp.).

Fig. 1.
Here is a practical situation where one can use this "toll image".
Consider function $$f(z)= \dfrac{1}{\sqrt{z(z-1)}} $$ $0$ and $1$ as critical values for $f$. How can be "tracked" the  argument around these critical values ? Let us consider figure 2.

Fig. 2. Surface $f(x,y)=arg(\dfrac{1}{\sqrt{z(z-1)}})$ where $z:=x+iy$. Please note the $2 \pi$ high "cliffs" materializing branch cuts. Why is there a "pass" through $[0,1]$ ?
In order to define $n$-th roots in complex function theory, one needs to refer to logarithm and exponential functions, here through the following expression :
$$\dfrac{1}{\sqrt{z(z-1)}}=\exp(-\tfrac12(\log(z)+\log(z-1)))$$
As the $\exp$ function is defined everywhere, let us concentrate on $\log$ components.
We have to consider 2 cuts :

*

*a cut issued from point $0$ in order that $\log(z)$ is defined.


*a cut issued from point $1$ in order that $\log(z-1)$ is defined.
On the bottom left and right figures, 2 particular solutions are displayed (with a slight preference for the right one). Both give the same result : the "total crossing fee" amounts to $0$ (cancellation of $2i\pi$ by $-2i\pi$).
Remark :
We could have used this formula :
$$\log(ab) = \log(a) + \log(b)$$
which can be applied (with caution!) to complex $\log$ function.
Appendix: Matlab script for fig. 2:

[X,Y]=meshgrid(-1:0.01:2);Z=X+i*Y;
Z1=angle(1./sqrt(Z.*(1-Z)));
surf(X,Y,Z1,'edgecolor','none');view([15,36]);


A: Yes, you are right. A function such as $z^{\frac 1 2}$ is multi-valued and is properly only defined on a Riemann surface. But we can think of the Riemann surface as being made of multiple copies of the complex plane (two copies for $z^{\frac 1 2}$) joined together in a certain way along one or more lines which we call branch cuts.
A branch cut will start and end at a branch point. $z^{\frac 1 2}$ has two branch points - one at $z=0$ and one at $z=\infty$. If we follow the value of $z^{\frac 1 2}$ around any loop that contains $z=0$ in its interior, we will find that $z^{\frac 1 2}$ does not return to its original value when we complete one circuit of the loop. Instead it changes sign. We have to go around the loop twice before $z^{\frac 1 2}$ returns to its original value.
A branch cut can be any line or curve that joins two branch points - its exact form is arbitrary. So we can take the negative real axis as a branch cut for $z^{\frac 1 2}$, or the positive real axis, or any other curve that joins $z=0$ and $z=\infty$. A branch cut is simply a reminder that if we cross the cut we need to move to another copy of the complex plane.
Once we define a branch cut (or cuts) then each copy of the complex plane is called a branch - so $z^{\frac 1 2}$ has two branches. The multi-valued function becomes single valued on each branch, and as long as we avoid branch points and branch cuts, it is analytic on each branch. 
Defining branch cuts becomes important if we are using contour integration and the Cauchy residue theorem to evaluate an integral. The location of branch cuts and the contour of integration are chosen so that the contour does not cross a branch cut.
A: You are right: "To define the one-valued function". Branch cut has to start at branch point and then goes arbitrary path to infinity. In this way Riemann surface is cut into Riemann sheets. On a given sheet function is on the cut discontinuous and non-analytic. Of course, function remains analytic if you go from one sheet to the next one. It has to be non-analytic at branch point because the Taylor expansion (as criteria for analyticity) cannot take "multiple-valued" character of the function into account, Taylor expansion is always single-valued. The same argument applies for cut. Imagine the function on one single Rimmann sheet would be analytic at cut: how it would be continued to a different sheet? There would be only one sheet and no cut.
By definition entire function have no cut neither branch point.
