Determine if analytic mappings from $\mathbb{C}\setminus \{0\}$ and $\mathbb{C}\setminus [0,\infty)$ to open unit disk exist and if so, find them…

analytic-functionscomplex-analysisopen-map

This is a question from a previous complex analysis qualifying exam that I'm working through to study for my own upcoming exam. I'm not sure how to approach it. I've included my thoughts on it, but I'm looking for both the solution to this specific problem and any useful suggestions for this general type of problem.

Problem:

For the following domains $\Omega$, determine if $\Omega$ can be mapped analytically onto the unit disk $|z|<1$. Justify your answer. If your answer is yes, find such a map.

(a) $\Omega = \mathbb{C}\setminus \{0\}$.

(b) $\Omega = \mathbb{C}\setminus [0,\infty)$.

My thoughts:

First, I note that this question is asking for analytic maps, so not necessarily conformal maps. I know the general idea of how to find a map that shifts, scales, rotates, and inverts the domain.

For (a), I considered something like $f(z) = \frac{1}{z}$, which maps the inside of the unit disk to the outside and vice versa. In this line of thought, I considered something like
$$
f(z) = \begin{cases} z, \quad |z|<1 \\ \frac{1}{z}, \quad |z|>1 \end{cases}
$$

But in this case, I'm not sure what to do with $|z|$ = 1.

So then I considered something like
$$
f(z) = \frac{z}{|z|+1}
$$

Perhaps this works? (Edit: Nope. As noted in the comments, this isn't analytic).

My sense is that there may be some theorem that I don't yet know about the existence of analytic maps that differentiates between an isolated point being excluded from the complex plane, such as in part (a), and a range of non-isolated points, such as the whole positive real line in part (b).

I also believe that, by the Open Mapping theorem, the map would need to be open since it is mapping to an open domain, the open unit disk.

Best Answer

(a) This is not possible. Such a function is necessarily bounded, and hence $0$ is a removable singularity and $f$ is entire. By Liouville's theorem, $f$ must be constant.

(b) It is simply connected, hence by Riemann mapping theorem, there is a bihomolorphic map from it to the unit disc. For example, we may use $z\mapsto\sqrt z$ to map the domain to the upper half plane, and then map the upper half plane to the unit disc by a Mobius transform.