Suppose I have some function like $f(z)=(z^2-1)^{1/3}$ and I know that it has branch cuts at $\pm1$. Suppose, I'm integrating, and I care about the value of the function on the contour, that I'm setting up in a specific way.
It goes from $-1$ to $1$, then makes a loop about $1$ clockwise, goes back to $-1$, then loops about $-1$ in clockwise, then comes back to $1$ and makes a loop counter-clockwise and goes back to $-1$ and makes a loop counterclockwise, and joins the starting point.
Every time, the function loops about a branch point, the overall phase changes. For example, after the first loop about $+1$, the phase changes by $\exp (2\pi i/3)$.
My question is, I've been told that, when the function makes a clockwise loop and then a counter-clockwise loop about a branch point, the overall phase remains the same. It is as if, the phase change due to the clockwise loop is cancelled by the phase change in the counterclockwise loop.
I'm not sure why this happens. Is it because, in clockwise, I'm letting $\theta \rightarrow \theta+2\pi$ and in anticlockwise, I'm letting $\theta\rightarrow \theta-2\pi$ ? If so, why am I doing this ? If not, can someone point out what is the obvious thing that I'm missing here ?
If $w= f(z) = (z^2-1)^{1/3}$ then $w^3 = z^2-1$. For any fixed $z\ne \pm 1$ there are three distinct roots $w$ which when plotted in the $w$ plane are vertices of an equilateral triangle. Image that you are a dancer walking in the $z$ plane. The three roots $w$ vary as you walk (the triangle rotates and expands.) You initially pick one root to be your dance partner and watch how your partner moves as you move on your path. You link hands with that partner and are not allowed to suddenly switch to another partner. (Called continuous path-continuation of an initial solution. ) This works fine as long as travel in a simply-connected region and you avoid the singular points $z=\pm 1$ where the three roots collapse into a triple root.
What happens at the singular points?
When you travel on a small circle about the point $z=1$ by setting $z= 1+ r e^{i t}$ you see that $w^3= (1+ r e^{it})^2 -1 = 2r e^{it} + o(r)$ and thus $w\approx (2r)^{1/3} e^{it/3}$. The phase change in $w$ is $(1/3)$ your phase change. That means that your dance partner is spinning around $w=0$ more slowly than you spin around $z=1$. Note that if you change your $\theta$ a small amount and them reverse that change, so will your partner.
As you make one complete rotation, the triplet of three roots will rotate through a third of a revolution, and you will see a different root in phase with yourself. However that new root is not the partner you have agreed to link with. (A similar analysis can be performed near $z=-1$.)
To make a 3D plot of how phase $\phi$ of $w$ varies as we wander in the $z$ plane, note $3 \phi$ is the phase of $w^3= z^2-1 = (x^2-y^2 -1 + 2 I x y$. Then $tan 3\phi= \frac{2 x y}{ x^2- y^2-1}$ and therefore $$(\sin 3 \phi) (x^2- y^2-1)- 2 \cos(3\phi) x y =0$$ is their relationship. The graph of this implicit surface is plotted below.
The phase spirals in corkscrews around the two singular points $z=\pm 1$ and the vertical scale represents one full complete change in $\phi$ of $2\pi$. Note that it takes several circuits around a singular point before you make one complete full change in the phase. The fact that the graph can be split locally into three congruent vertical pieces stacked over one another represents the fact that there are three related roots to the cubic over each point in the $z$ plane.