[complex analysis] are branch points always isolated?

[nonequilibrium]

You can choose to limit yourself to continuous or analytical functions

 

[Bacle]

Well, if you define Logz, isn't every point in the branch cut a branch point?

If you draw a circle that winds around any point in the negative real axis ---

or same thing for any point in any branch cut you use to define log, you do not

end up where you started after going 2Pi around.

 

[nonequilibrium]

I understand what you say, but my professor in Complex Analysis told me that the branch cut points don't count as branch points, because by moving the branch cut they can be made analytical. So for the logarithm, we say only zero is a branch point.

 

[nonequilibrium]

I thought about it some more, and I think I might have an example of a function defined on the unit disk where the branch points are dense in any environment around zero:

Define
f: \mathbb C \backslash \overline D(0,1) \to \mathbb C: z \mapsto \prod_{n=2}^{\infty} \sqrt[n]{z-n}

I'm not sure if it converges, but say it does, then it has a branch point for every integer n > 1 (?). Then define:

g: D(0,1) \to \mathbb C: t \mapsto f \left( \frac{1}{t} \right)