Let $(A,g)$ be a compact surface with boundary, diffeomorphic to the standard annulus $\{z\in\mathbb{C}:1\le|z|\le 2\}$, equipped with a smooth metric $g$. 



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> Does there always exist a conformal (bijective) map
>  $\phi:A\to\{z\in\mathbb{C}:1\le|z|\le R\}$, for some $1<R<\infty$?
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A possible idea is to use the well-known fact that $A\setminus\partial A$, as a Riemann surface (whose conformal structure is induced by $g$), is biholomorphic to either $\{1<|z|<R\}$ (for some $1<R<\infty$), $\{0<|z|<1\}$ or $\mathbb{C}\setminus\{0\}$. Is there a way to rule out the latter two cases and to prove that the biholomorphism extends nicely to $\partial A$?



Uniformization for annuli with boundary
An alternate proof that interior of A is not biholomorphic with punctured disc or punctured plane is to notice that that there is a harmonic function on the interior of A which is 0 and 1 on the different boundary components respectively . Such functions do not exist on the punctured disc or the punctured plane. In fact this harmonic function has no critical points and the corresponding holomorphic 1 form has its periods an imaginary multiple of integers .Hence composing the integral of a multiple of the holomorphic one form with the exponential function gives the uniformizing map.


