Non constant analytic function from $\{z\in\mathbb{C}:z\neq 0\}$ to $\{z\in\mathbb{C}:|z|>1\}.$
Note that $\mathbb{C}\setminus\overline {\mathbb{D}}$ is conformally equivalent to $\mathbb{D}\setminus\{0\}$, via the map $z \to \dfrac{1}{z}$. So essentially you have to construct a map from $\mathbb{C}\setminus\{0\}$ to $\mathbb{D}\setminus\{0\}$. Suppose the map is $f$. We already have a map, $\mathbb{C}\to\mathbb{C}\setminus\{0\}$ given by $z \to e^z$. Compose this with $f$ to obtain a bounded entire function, which means $f$ is constant.
If so then $1/f$ is bounded. Hence $1/f$ has a removable singularity at the origin, giving a bounded entire function.