Are $\{ z: 1<|z|<2\}$ and $\{z: |z|>1\}$ conformally equivalent?

I don't know whether a ring, i.e. the domain $\{z: 1<|z|<2\}$ and the exterior of the closed unit disk $\{z: |z|>1\}$ are conformally equivalent?

I have tried to look on some topological properties, which must be preserved by conformal mapping. They are not simply connected so I can't use Riemann's Theorem. I would like to find a specific conformal mapping between them. I believe that it is possible to map the boundary of $D_2(0)$ to infinity somehow.

Thanks for any help

Your question is equivalent to asking if $A = \{z: 1<|z|<2\}$ and $B = \{w: 0<|w|<1\}$ are conformally equivalent, and they are not:
The conformal mapping $f : B \to A$ would have a removable singularity at $w=0$, and this easily leads to a contradiction (consider both cases $f(0) \in \partial A$ and $f(0) \in A$).
• Thank you! I didn't know that it would hold for ratio=$\infty$ as well. Jun 11, 2018 at 9:29
• I can see that the $f(0) \in A$ case is bad because then $f$ is no longer injective, but I don't see with certainty why $f(0) \in \partial A$ is problematic. Is it because the rest of $\partial A$ besides the point to which $f(0)$ maps to is not in the image of $f$? Aug 10, 2020 at 2:58
• @JoshuaSiktar: If $f(0) \in \partial A$ then either $|f|$ or $|1/f|$ has a maximum at $z=0$, and the maximum modulus principle would imply that $f$ is constant. Aug 10, 2020 at 3:28