# 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.

• Since $0$, $-1$ and $\omega$ are the only branching points, shouldn't the cuts be three non intersecting rays beginning on each of those points and going to infinity? Feb 9, 2022 at 23:58
• it doesn't have to be an unbounded branch cut. some branch cuts are like from [i, -i] Feb 10, 2022 at 0:07

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.

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.

• Nice explanation. But in this case, it is permissible for a curve to wind around two of the branch points. Under $z \mapsto z(z+1)$, a curve that winds around $0$ and $-1$ once maps to a curve that winds around $0$ and $-1$ twice. (If you have trouble visualizing this, compare the images of a circle of radius $\frac{1}{2}$ and radius $2$ centered at $z = 0$.) Thus the value of $\sqrt{z(z+1)}$ is the same at the start of the loop and at the end of the loop. A standard branch cut for OP's function $f(z) = \sqrt{z(z+1)(z-\omega)}$ is a line from $0$ to $-1$ and a line from $\omega$ to $\infty$.
– Ant
Mar 4, 2022 at 1:22