One of the problems assigned to me for homework was to construct an analytic isomorphism from $\mathbb H$ to $\mathbb D\backslash[1/2, 1)$. Here is my attempt at a solution, but I am worried that it is not analytic at a certain point. To be more specific, I don't think my solution is analytic when $z = \sqrt{2}i$.
First, consider the following Analytic Isomorphism: $$f(z) = -iz$$ It is clear that $f \in \text{Iso}(\mathbb H, \mathbb H_R)$. Next, consider the following isomorphism: $$g(z) = z^2$$ It is clear that $g \in \text{Iso}(\mathbb H_R, \mathbb C\backslash(-\infty, 0])$. Therefore, $g \circ f \in \text{Iso}(\mathbb H, \mathbb C\backslash(-\infty, 0])$: $$(g \circ f)(z) = -z^2$$ Next, consider the following isomorphism: $$h(z) = 1-z$$ It is clear that $h \in \text{Iso}(\mathbb C\backslash(-\infty, 0], \mathbb C\backslash[1,\infty))$. Therefore, $h \circ g \circ f \in \text{Iso}(\mathbb H, \mathbb C\backslash[1,\infty))$: $$(h \circ g \circ f)(z) = 1+z^2$$ Finally, consider the following isomorphism: $$j(z) = \frac{z}{|z|+1}$$ It is clear that $j \in \text{Iso}(\mathbb C\backslash[1,\infty), \mathbb D\backslash[1/2, 1))$. Therefore, $j \circ h \circ g \circ f \in \text{Iso}(\mathbb H, \mathbb D\backslash[1/2, 1))$: $$\frac{1+z^2}{|1+z^2|+1} \in \text{Iso}(\mathbb H, \mathbb D\backslash[1/2, 1))$$
Hints: $f(z) =(z-1)/(z+1)$ maps the right half plane to $\mathbb D$ in a nice way. The map $f(\sqrt {z})$ then maps $\mathbb H$ to the upper half of $\mathbb D.$ Square that map to yield a map onto $\mathbb D \setminus [0,1).$ Now compose with a biholomorphic self map of $\mathbb D$ that sends $\mathbb D \setminus [0,1)$ to $\mathbb D \setminus [1/2,1).$