Why is $z=1$ not a branch point of the function $w=f(z)=z^{1/2}$?

Question

Consider the function $w=f(z)=z^{1/2}$ and the point $z=1$ on the $z$-plane. Next consider a closed circular loop of radius $2$ about the point $z=1$ so that $w=1$. As we go around $z=1$ along this loop $f(z)$ doesn't return to $w=1$ but $w=-1$. Why is $z=1$ not a branch point?

Answer 1

Distilling reuns's comment: the definition of a branch point is one for which traversing an arbitrarily small circle around that point results in a multivalued function. One single circle is not enough. Indeed, if your argument showed that $1$ was a branch point of $\sqrt z$, then it would also show that every single complex number was a branch point.

Answer 2

In addition to Greg's answer :

Note $\log \log z$ has a branch point at $z=1$ which disappears after one rotation.

So one possible general definition of a branch point is : a given branch of $f(z)$ has a branch point at $z=a$ iff $F(s) = f(a+e^{-s})$ is analytic (or meromorphic) for $\Re(s) > r, \Im(s) \in (c,d)$ and its analytic (or meromorphic) continuation exists and is analytic (or meromorphic) for $\Re(s) > R, \Im(s) \in (a,a+2i\pi+\epsilon)\supset (c,d)$ and $F$ is not $2i\pi$-periodic.

The (or meromorphic) is needed to allow things like $\frac{1}{\sin \frac{1}{z}} + \log z$