Is $z=\infty$ a branch point of $f(z)=(z^{2}+1)^{1/2}$?

Question

Define $f(z)=(z^{2}+1)^{1/2}$. My argument is as follows: if we want to find out if $z=\infty$ is a branch point of this function, we define $\zeta=\frac{1}{z}$. From this, $f(\zeta(z))=(\frac{1}{\zeta^2}+1)^{1/2}=(\frac{1}{\zeta^2})^{1/2}(1+\zeta^2)^{1/2}$. Near $\zeta=0$, $f(\zeta)\rightarrow\infty$, so it has a branch point. This means $f(z)$ has a branch point at $z=\infty$. However, this was marked wrong, and I don't know why. Where does my argument fail, and why?

Answer 1

No, it's not a branch point, because the local solutions are $\pm\xi^{−1}A(\xi)$ where $A(\xi)$ is analytic at $\xi=0$. Since the exponent $−1$ is an integer, one sees that $\xi=0$ is not a branch point.