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I´ve been stuck in this problem for a long time, i wish you could help me to solve it:

Let U an open connected subset of the complex plane where $\mathbb{C}-U$ has a connected component with at least two points, show that U is analytically isomorphic to an open subset of $\mathbb{D}$.

My attempt: If U is simply connected is trivial. The condition that one connected component hast at least two points is equivalent to say that one of this points is on $\partial U$ and the Dirichlet´s problem has solution. Any ideas?

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    Let $\gamma $ denote a connected component of $\mathbb{C}- U$ which contains at least two points. Do you agree about the part that if $\gamma $ is unbounded, $V=\mathbb{C}-\gamma $ is a simply connected open set and $U\subset V$ ? – 2017-01-17
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    Yes, I agree with that. – 2017-01-17
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    OK. I will post my answer. – 2017-01-17

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Let $\gamma $ denote a connected component of $\mathbb{C}- U$ which contains at least two points. If $\gamma $ is unbounded, $V=\mathbb{C}-\gamma$ is a simply connected region which is not the whole palne $\mathbb{C}$. Hence we can map $V$ onto $\mathbb{D}$ by a conformal map $f$, i.e. $f(V)=\mathbb{D}$. Since $U\subset V$, we see $f(U)\subset \mathbb{D}$. Therefore $U$ is analytically isomorphic to an open subset $f(U)$ of $\mathbb{D}$.

If $\gamma $ is a bounded set, $\mathbb{C}-\gamma$ is doubly connected in $\mathbb{C}$. The above argument does not work. So we consider the extended complex plane $\hat{\mathbb{C}}=\mathbb{C}\cup \{\infty\}$. Then $V=\hat{\mathbb{C}}-\gamma$ is simply connected in $\hat{\mathbb{C}}$.
Let $a,b\in\gamma $ and define $\varphi (z)=\frac{1}{z-a}$. Since $W=\varphi (V)$ is a simply connected open set which is not the whole palne $\mathbb{C}$, we can map $W$ onto $\mathbb{D}$ by a conformal map $g$. Note that $\varphi (U)\subset W$. Therefore $U$ is analytically isomorphic to an open subset $(g\circ\varphi )(U)$ of $\mathbb{D}$.