# Branch cuts to infinity

The function $$f(z)=\sqrt{z^2-1}$$ does not have a branch point at infinity. However, people often take branch cuts going from $z=\pm 1$ to $\infty$ along the real positive and negative axis respectively. Why is this acceptable?

My guess would be that technically this is a branch point going from $z=+1$ to $z=-1$ since in the complex plane the point $z=\infty$ is taken to be one point (I think anyway, although I have heard mention of cases when it is not??!).

The key requirement in choosing a branch cut --- which I will denote by $BC$ --- is that it contains all branch points, and for any point $z_0$ in the complement of $BC$ and any appropriate value of $w_0=f(z_0)$, this one value can be extended to an analytic function $w=f(z)$ on the complement of $BC$.
In the case of $f(z)=\sqrt{z^2-1}$, your question discusses two choices of branch cut, and each one satisfies the key requirement:
• For the branch cut $BC = \bigm( (-\infty,-1] \cup [1,\infty) \bigr) \times 0$, the complement $\mathbb{C}-BC$ is simply connected. In the simply connected case the key requirement always holds (assuming, of course, that $BC$ contains all branch points).
• For the branch cut $BC = [-1,+1] \times 0$, the complement is not simply connected, and yet the key requirement still holds. The underlying geometric reason for this is, as you say, that $f(z)$ is not branched at $\infty$.