Branch Points of $\sqrt{z - 1/z}$

Question

I'm working through Gamelin's Complex Analysis and am confused by the following

Example: Consider $\sqrt{z - 1/z}$. We rewrite this as $\sqrt{z-1}\sqrt{z+1}/\sqrt{z}$. The function has three finite branch points, at $0$ and $\pm 1$. We must also consider $\infty$ as a branch point, since there is a phase change corresponding to a phase factor $-1$ as $z$ traverses a very large circle centered at $0$.

Normally I find Gamelin very clear, but here he seem's a bit vague. Can anybody explain why $0$, and $\infty$ are both branch points of this function? Is there an standard way of finding branch points and some intuition for them?

Answer 1

I'm assuming, given how your question is targeted especially at $z=0$, that you understand what a branch point is.

Consider the fact that $\sqrt z$ has a branch point at $0$. It means that its value is going to change if you turn once around $0$ along a small circle. It follows that any function that depends on $\sqrt z$ is going to change as well (unless there is something in the function that cancels the branching due to $\sqrt z$, which is not the case here).

As for $\infty$, consider the fact that turning around $\infty$ simply means turning around a "large" circle. Turning around a large circle has the same effect on the function as turning around $1$, and $-1$, and $0$ (in the opposite sense, since you need to take into account the orientation of the Riemann sphere). This does change the function.

Answer 2

Near zero, $f(z)=(\sqrt{z})^{-1}g(z)$ where $g(z)=\sqrt{z-1}\sqrt{z+1}$. If we choose the branch of $g$ near zero with $g(0)=i$ then $g(z)$ is holomorphic for $|z|<1$. Also $g(0)\ne 0$. But $\sqrt{z}$ has a branch point at $0$. Moving round a small circle centred at $0$ reverses the sign of $\sqrt z$ but $g(z)$ returns to the same value. So, moving round a small circle centred at $0$ reverses the sign of $f(z)$; $f(z)$ has a branch point (of order $2$) at $0$.

At $\infty$ it's convenient instead to consider the behaviour of $F(w)=f(1/w)$ near $0$ instead. Then $$F(w)=\sqrt{\frac{1-w^2}w}=\pm i f(w).$$ For the same reason as $f$, $F$ has a branch point of order $2$ at zero, so that $f$ has a branch point of order $2$ at infinity.