I would like to look for a conformal map which maps $U_1=\mathbb{H}-\{yi:y\in[1,2]\}$ to $U_2=D(0,1)\setminus\{0\}$, where $\mathbb{H}$ means the upper half plane.
I am not sure where to start with, since translation would destroy the nice half-plane property.
Thank you!
There does not exist any conformal map between those two sets.
In general, if $S$ is a Riemann surface that is homeomorphic to the open annulus $A = \{z \mid 1 < |z| < 2\}$, then $S$ is conformally equivalent to one of the following Riemann surfaces:
1) The punctured plane $\mathbb C - \{0\}$
2) The punctured disc $D(0,1)-\{0\} = \{z \mid 0 < |z| < 1\}$
3) An annulus $A_r = \{z \mid 1 < |z| < r\}$ where $r > 1$
Furthermore, the one of these to which $S$ is conformally equivalent is unique, meaning that two Riemann surfaces in two different ones of these three categories are conformally inequivalent, and two Riemann surfaces $A_r$, $A_s$ in category 3) such that $r \ne s$ are conformally inequivalent.
It turns out that $U_1$ falls into category 3), it is conformally equivalent to $A_r$ for some $r > 1$ (this is a good exercise).
It follows that $U_1$ is not conformally equivalent to $U_2$ which is in category 2).