I struggle solving the following exercise:
Are the sets $U := \{z\in \mathbb C: |z| > 1\}$ and $V := \mathbb C\setminus \{0\}$ conformally equivalent?
Now I am pretty sure the answer is "no". My try:
Assume there existed a biholomorphism $g:V \to U$.
$U$ is conformally equivalent to $D' :=D\setminus \{0\}:= \{z\in \mathbb C : \lvert z|< 1, z \neq 0\}$ (for example via the map $z \mapsto 1/z$), so it suffices to show that $g: V \to D'$ can't be a conformal mapping.
To this end, distinguish between the following cases:
Case 1: $0$ is a removable singularity of $g$. It follows easily by Liouvilles theorem that then $g$ has to be constant, a contradiction.
Case 2: $0$ is an essential singularity of $g$. As an easy application of the Casorati-Weierstraß theorem one sees that $g$ can't be injective near $0$, a contradiction.
Case 3: Here's where I am stuck. I don't know how to get to a contradiction here and hope someone of you can give me a hint (or correct my proof to this point if it has flaws).
Hint: $1/z$ is a bounded analytic function on $U$. Is there a non-constant bounded analytic function on $\mathbb C \backslash \{0\}$?