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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?

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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.

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  • $\begingroup$ So if we center a circle at $\infty$ on the Riemann sphere, and go around the circle we are in effect going around $-1,0,1$ and since it is a square root only two will cancel out, so we would need to go around the circle twice to return to the same branch? Is that the correct way to think about it? $\endgroup$ – Emilio Minichiello Jun 16 '18 at 18:13
  • $\begingroup$ @EmilioMinichiello Exactly. $\endgroup$ – Arnaud Mortier Jun 16 '18 at 18:15
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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.

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