Does there exist an analytic function whose domain is the complement of the closed unit disk and whose range is the open unit disk?
I am studying John B. Conway's book Functions of One Complex Variable for my own amusement. One of the exercises following Chapter III, Section 3, asks whether the open unit disk can be mapped conformally onto the punctured open unit disk. I solved that one (it's true), and then asked myself whether the converse were true as well. I immediately realized that that question is equivalent to whether the complement of the closed unit disk can be mapped conformally onto the open unit disk. Then I realized that I didn't even know the answer to the simpler question asked above.
Best Answer
Take $g(z)$ any Blaschke product of degree two that is not $cz^2$; then $g$ maps the punctured unit disc holomorphically onto the unit disc since $g(0)=g(w)$ for some $w \ne 0$. Then $f(z)=g(1/z)$ will map the exterior of the closed unit disc holomorphically onto the open unit disc.
For example $g(z)=\frac{(1-2z)^2}{(2-z)^2}$ is one such Blaschke product, giving
$f(z)=\frac{(z-2)^2}{(2z-1)^2}$
It is considerably more difficult (though doable) to find a conformal such $f$ (in other words with the additional condition that $f'(z) \ne 0, |z|>1$)