Is there a direct proof, using the Riemann mapping theorem for the Jordan domain, than every doubly connected domain in the complex plane can be mapped conformaly onto a round annulus.



Mapping the doubly connected domain to an annulus
Yes, here is a sketch. Let $A$ be your doubly connected domain, wlog $0$ and $\infty$ are in different components of the complement. Consider the preimage of
$A$ under $\exp(z)$. This is an unbounded simply connected domain $D$ with two boundary points at infinity. This domain will be periodic: $z\mapsto z+2\pi i$
will map it onto itself. Now using the Riemann mapping theorem, map $D$ by some function $g$ onto the vertical
strip $0<\Re z<1$, so that the infinite points on the boundary of $D$ correspond to the infinite points on $\partial D$. By the uniqueness in the Riemann theorem, your map $g$ will satisfy $g(z+2\pi i)=g(z)+c$ with some constant $c\neq 0$. Multiply $g$ on $k=2\pi i/c$ to obtain a new function $g\_1=kg$ which satisfies such an equation
with $c=2\pi i$. Function $g\_1$ will map $D$ on the strip $0<\Re z<k$.
Now the function $\exp\circ g\_1\circ\log$ will map $A$ onto the round ring
$1<|z|<\exp(k)$. 


