Finding branch cuts Let $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$. I would like to find branch cuts so that the complex function
$$f(z)=\sqrt{z(z+1)(z-\omega)}$$
can be defined continuously off the branch cuts. I searched through various textbooks and websites, and couldn't find any worked examples explaining in detail how the branch cuts are found. I tried imitating some examples, and I think that the branch cut can be taken to be

However, I do not understand rigorously why this can be taken to be a branch cut, and would want a detailed explanation. Any help is appreciated.
 A: This is about the Monodromy Theorem, although in explicit examples we can do things directly without invoking a theorem.
That is, for an algebraic "multi-valued function" with branch points $z_1,\ldots,z_n$, the $\pi_1$ homotopy group of $\mathbb C-\{z_1,\ldots,z_n\}$ acts by permutations on the "sheets". That it only depends on homotopy of paths is the Monodromy Theorem. To make single-valued versions of "the function", we need to kill off enough of $\pi_1$ so that the image acts trivially on some "sheet".
In the present case, going around a single branch point flips the sign of the square root. Bad. Going around two is ok! With three branch points, it would suffice to connect two of them by a "cut", and make another cut from the third going out to infinity. Or, "cuts" from all three going out to infinity. There is not generally a unique choice of "cutting" to kill off suitable homotopy.
A: You want to prevent any curve from winding around any of the three branching points; when they do wind around, your function changes values, i.e., it's a multi-function. (a) Standard example is that of the square root function $\sqrt z $: working in polars, so that $ z= e^{i\theta}$, which is then its square root is  $z=e^{i\theta/2} $. Then, if you wind around it starting at $0$, you end up with $$ z'= e^{i(\theta +2\pi)/2}=e^{i\theta/2}e^{i\pi}=-e^{ i\theta/2} $$. Notice if you do another circuit about $0$, you return to your original value, i.e., $$e^{i( \theta +4 \pi)/2}= e^{i(\theta/2} e^{i2 \pi}=e^{ i\theta/2} $$
This is because your functions jump from one branch of the function, where they have a value, to another branch, where they have a different value. This is because your functions (square root) here, are multi-valued, and must be restricted to a region of the plane --a branch-- where they are single-valued.
What is a branch?
It's a subset of the plane where your multi-function (Square root, logz, etc), restricts to a single-valued function.
Notice that the order of your branch point may be infinite, as in the case of $Logz$, where you ca
This is related to the fact that functions such as the square root, logs, others are multi-valued. Your curves accomplish this,i.e., they prevent any so they together are a Branch Cut; a cut that prevents any curve from winding around any of the trouble (branch) points. Hope I was clear. Please ask otherwise.
