I am taking a complex analysis course and the professor confused me on the following matter:
On one hand, given a complex function like $f(z)=z^{1/2}$, when we complete a closed circuit around the origin there is a discontinuity problem - for example, if we start at $z=1$ with $\theta=0$ and end at $z=1$ with $\theta=2\pi$, the function $f(z)$ has a discontinuity along the positive x-axis. on the other hand. if we do not include the origin, the problem is gone and the function remains continuous and single-valued.
I understand that the problem stems from the fact that if we include the origin, while we return to the same point we started with, the argument can complete a full cycle, which causes a phase difference that in turn affects the value of the function when we approach the point $z=1$ from the first or fourth quadrants. But why does including the origin cause the problem, contrary to when we traverse a closed path that does not include the origin? why does including the origin cause this discontinuity, and why does leaving the origin out solve it?
Someone asked my professor this question, but he kept repeating that the reason for the discontinuity was that we included the origin in the circuit, and gave no more explanations...
Thank you in advance!
To illustrate Karl's and Paul Sinclair's comments geometrically:
(There's more explanation in this related question.)