Branches of $\arg(z)$ and $\log(z)$

Question

I do not understand this excerpt from my textbook:

It says that any open disk G that excludes the origin, G has a branch. But in the problem, it says there is no such branch? I do not understand the wording.

Answer 1

An "open disk that excludes the origin" is something like $$ \{z : |z-2| < 1 \} $$ -- and open disk centered at $2$, with radius $1$. The point $z = 0$ is not in this set. If I'd written $$ \{z : |z-2| < 3 \} $$ it would be an open disk that DOES include the origin.

The set given in Exercise IV.10.1 is not a open disk -- it's annulus (one that happens to "enclose" the origin).

I think perhaps your confusion might be that you're reading "open disk, with the origin removed" for "open disk that excludes the origin," but I may be misunderstanding.

Regardless, the claim is that on certain open disks --- those in which the origin is not one of the points --- the branch is defined. The set in the exercise is not an open disk, so there's no contradiction.