In studying Ahlfors' Complex Analysis, a map $T$ is constructed from $w$-plane to $z$-plane by $z=w+\dfrac1w$. And setting $w=\rho e^{i\theta}$, fixing $\rho$ gives circle which correspond to an ellipse in $z$-plane, while fixing $\theta$ gives ray which corresponds to hyperbola. (see https://i.sstatic.net/d8REI.png)
The purpose of this section is to find maps from these regions onto $D^2$. So we should consider its inverse map $G$ with domain as open connected subset of $D^2$ instead (given by one of the branch $w=z-\sqrt{z^2-1}$).
However, unlike what is claimed in the book, as the boundary circle of $\overline{D^2}$ is absent, the real axis is actually missing from the image set no matter what domain is chosen for $G$. Hence the region between branches of hyperbola is not being mapped onto.
Of course, the problem can be solved by relaxing the domain of $G$ to be $D^2\bigcup\{e^{i\theta}:\theta\in[0,\pi]\}$. However, I think the domain should be made open connected region.
Also, to actually construct the domain, we specify one hyperbola corresponding to ray of argument $\theta_0$ (in fact $-\theta_0$, $\pi+\theta_0$ are possible choice). Let $Q$ be the region between two branches, it correspond to two separate sector (see https://i.sstatic.net/oFL4T.png, should be https://i.sstatic.net/AvjrT.png). The disconnectedness may lead to difficulties in constructing further map to upper half plane.
So how to solve the above questions? Thanks for any help.
P.S. The question can be treated as the following as well: how to continue with this trick to take the corresponding circular region to $D^2$?
For if it is answered, then the above subtle question about connectedness of image set need not be concerned anymore.
P.S.2 Actually region $Q$ (with infinity) projected back on Riemann's sphere just looks like bracelet, which is not simply connected. So intuitively our elementary way (LFT, power map, exponential map restricted to be made injective) of analytic mapping retaining bijectivity is not applicable here since by open mapping theorem, it should be homeomorphism which contradicts the number of connected components they have.
(Plus @Maxim 's comment, $Q$ without infinity is simply connected)
So we should just take say $f(z)=z^m$ ($m$ large) as the answer?
Best Answer
fwiw, showing what I got from my comment.
Further solving $(u,v)$ in terms of$(x,y)$
$$ u (1 + 1/(u^2 + v^2)) = x, v (1 - 1/(u^2 + v^2)) = y $$
results in sixth order $(x,y)$ polynomials.