complex-analysiscomplex-geometryriemann-surfaces
Recently I am studying branch points and branch cuts in complex analysis.
If there is a function like $f(z)=\sqrt{(z-1)(z-2)}$, then the algebraic branch points are $1$ and $2$ and the branch cut is the interval $[1 , 2]$.
One can ask a natural question: Let $f(z)=\sqrt{(z-a) (z-b) (z-c)}$ where $a, b, c$ are three different non zero complex numbers. How to find the branch points and branch cut?
To construct that branch of log $z$, you just define the half-parabola to be the branch cut. This would mean log $z$ = Log|$z$| + $i\theta$ where $\theta$ equals the value of arg $z$ between $\frac{\pi}{2}$ and $2\pi$ for $z$ in the second, third, or fourth quadrant, the value of arg $z$ between $0$ and $\frac{\pi}{2}$ for $z$ in the first quadrant above the half parabola, and the value of arg $z$ between $2\pi$ and $\frac{5\pi}{2}$ for z in the first quadrant below the half parabola.
Explicitly, you can solve for $\theta$ in terms of $r$ where $z=re^{i\theta}$. By the equation of the half parabola, $\frac{y}{x}=\frac{1}{y}$. Then, $\theta=\arctan(\frac{y}{x})=\arctan(\frac{1}{y})$.
$r=\sqrt{x^2+y^2}=\sqrt{x^2+x}$ , then $x^2+x-r^2=0$. By the quadratic formula, $$x=\frac{-1\pm\sqrt{1+4r^2}}{2}$$ You choose the positive square root because you want the upper half parabola so, $$y=\sqrt{\frac{\sqrt{1+4r^2}-1}{2}}$$ Therefore, the entire branch would be defined as $$\log z = \mbox{Log }|z| + i\theta , \mbox{where } \theta = \arctan \bigg(\sqrt{\frac{2}{\sqrt{1+4r^2}-1}}\bigg)$$
I will demonstrate with the example of the complex logarithm.
Recall that a complex number $z=x+iy$ can be put into "polar form" $z=Re^{i \theta}$, where $R$ is the distance from $z$ to the origin and $\theta$ is the angle to $z$ around the complex plane measured positively from the $x$-axis. The complex logarithm has this formula for $z=Re^{i \theta}$:
$$\log z = ln |R| + i \theta.$$
You (hopefully) know that if $z=Re^{i \theta}$ then we can also represent $z$ as $z=Re^{i (\theta + 2\pi)}$ and generally as $z = Re^{i (\theta + 2 \pi k)}$. Consider what happens when you plug these different representations into the formula for $\log z$. If you use $Re^{i \theta}$ you get $\log z = ln |R| + i \theta$ but if you use $Re^{i (\theta + 8\pi)}$ you get $\log z = ln |R| + i (\theta + 8 \pi)$.
If you plot these points you run into an apparent problem -- they don't hit the same place! The consequence of this is that the complex logarithm is not a function (because functions take a single input to a single output -- this function takes a single input to multiple outputs).
Here is the "Riemann surface" for the complex logarithm:
$\qquad\qquad\qquad$ 
-- don't worry too much about how it was generated but realize what it's telling you. If you pick the point $z = Re^{i \theta}$ on the complex plane and map it through the complex logarithm, we've seen you get these infinite number of different values; the picture just represents this.
What a branch cut does is restrict the outputs of the logarithm to one particular loop around this corkscrew surface (which one it is depends on the range of values you choose for your cut). Moreover, when we define branch cuts, there's a lot of points that we make "undefined" -- the reason for this is that while you can pick a particular loop to "go around" in the Riemann surface, you can never have continuity around the place you define the cut because the limits from different sides will be at different heights and so cannot be equal (and therefore it will not be continuous, which is generally a bad thing in calculus).
Let me give you something specific. For example, if you choose the branch cut to be the negative real axis, then you are allowing $-\pi < \theta < \pi$ -- that choice will roughly correspond to the "level" in the Riemann surface that has the red on it. Note that $z = Re^{-i \pi} = Re^{i \pi}$.
With this, you can see what I said earlier about continuity -- if you approach $Re^{i \pi}$ from values of $\theta$ greater than $\pi$, you would be on the yellow part of the surface but if you approach $Re^{i \pi}$ from values of $\theta$ less than $-\pi$, you are closer to the purple part of the image. We have approached $z = Re^{-i \pi} = Re^{i \pi}$ two different ways and got two different answers, in contradiction to continuity which we generally like to have in calculus.
edit: I'd like to add this domain coloring image of $\log z$ that I found in this Mathematica stack exchange post. The colors represent the argument of the output of the logarithm -- where it switches colors at angle $\pi$ is precisely the same jump discontinuity I'm talking about above:
$\qquad\qquad\qquad$ 
Best Answer
Let $s(z):=\sqrt{z}$. Then if $g(z):=(z-a)(z-b)(z-c)$ you have that $g$ is a polynomial, so it is an entire function, however you must choose a branch of the function $s$, as it is not continuous in the whole $\mathbb{C}$.
Choosing a branch of $s$ is the same thing than choosing a branch of the complex logarithm, as we generally defines $\sqrt{z}:=\exp\left(\frac1{2}\log z\right)$ for some branch of the complex logarithm. Setting some branch of the logarithm we will have a domain $D$ where our logarithm is analytic, therefore the domain of $s\circ g$ will be $\{z\in \mathbb{C}:g(z)\in D\}=g^{-1}(D)$.