A branch cut problem In Ahlfors' Complex Analysis text, chapter 3, section 4 the transformation $z=\zeta+\frac{1}{\zeta}$ is discussed. The author notes that for every $z$, there exists 2 solutions for $\zeta$ and they are inverses of each other. In order to get a unique $\zeta$ he suggests the restriction $|\zeta|<1$, and he says that in that case the interval $(-2,2)$ must be removed from the $z$ plane.
My question is: why must that interval be removed? In previous chapters it was shown that in order to get an analytic branch for $\sqrt{w}$ one might use a plane with a ray removed (say the negative real axis) - not an interval.
The inverse function is
$$\zeta=\frac{z \pm \sqrt{z^2-4}}{2}$$
and requiring that the argument of the square root is not negative doesn't get me to the interval $(-2,2)$. 
 A: This function has two branch points at $\pm2$ as your inversion formula shows. By contrast, $\sqrt z$ alone has only one finite branch point at $0$; however, formally, it also has one at infinity. (Set $w=1/z$, or look at the Riemann sphere to make sense of this.)
Branch cuts always join two branch points, hence the line segment he chose. (An alternative would be two line segments from $2\to\infty$ and $-2\to-\infty$, which is secretly one line going via infinity.)
I can explain more if you like, but this is the reason.

Take $f(z)$ to be mostly two-valued, like $f=\sqrt{g(z)}$ for polynomial $z$. Now suppose you try to define $f$ by starting at some $z_0$ with one of the branches and smoothly (analytically) extending it. When does this go wrong? When you come back to a point $z_1$ you've seen before but with a different value.
Now imagine drawing the two curves you followed from $z_0$ to $z_1$; they form a loop. Start tightening it by bringing them closer together. Since it must be impossible for the two paths to be smoothly deformed onto each other (since they have different values at the $z_1$ end), there must be some point $z_\star$ (along the line you almost formed) such that you can't give a single consistent value to $f$ in any neighbourhood of $z_\star$. $z_\star$ is a branch point.
For example, with $\sqrt{z}$ this happens at the origin. Intuitively, if you go around it in an arbitrarily tiny circle, you get an inconsistent result.
It also happens at infinity; taking a small circle in the variable $1/z$ or a large circle in $z$ makes it clear that you have the same problem out here.
By contrast, $\sqrt{z^2-4}$ has two branch points at $\pm 2$. At infinity, for example, because $$z=Re^{i\theta} \implies (z^2-4)^{1/2}\sim (R^2e^{2i\theta})^{1/2}=Re^{i\theta}$$ there is no inconsistency on going round by a circle, as $\theta\sim\theta+2\pi$. (The $z^2$ goes round twice as fast to compensate for the way that $\sqrt z$ lags at half speed.)
Anyway, the point is that if you want to make it impossible to follow a pair of contradictory paths like those described above, then it seems you can't include any region containing a branch point, or indeed any region enclosing a branch point since the discontinuity will be felt further way. The exception is that when you enclose two branch points, it might be that going around both of them doesn't lead to a contradiction. In fact, we saw this for $\sqrt{z^2-4}$ above already. (It's an equivalent calculation in this case to seeing that infinity isn't a branch point.)
Thus is turns out that excluding the region $(-2,2)$, which prevents us from making any path which encircles just one of the branch points, is sufficient to allow a smooth definition everywhere. Hooray!
