Bijective Conformal Mapping onto the Open Unit Disc $\mathbb{D}$

Question

What is the explicit bijective conformal mapping $f(z):G_n\to\mathbb{D}$, $z\in\mathbb{C}$ for the following domain transformations:

$G_1=\{x+iy~|~x>1/2,y>0\}$ is the open region of the first quadrant.

$G_2=\{x+iy~|~x<0,y>0\}$ is the open second quadrant.

$G=_3\{x+iy~|~x<0,y<0\}$ is the open third quadrant.

$G_4=\{x+iy~|~x>0,y<0\}$ is the open fourth quadrant.

Where it is noted that $\mathbb{D}$ is the open unit disc defined by $\mathbb{D}=\{x+iy~|\sqrt{~x^2+y^2}<1\}$. I am aware that the aforementioned transformations have the specific form $f(z)=Ke^{cz}$, $\forall K\in\mathbb{C}$ or $f(z)=z^2$, where such mappings exist by the Riemann Mapping Theorem.

Answer 1

Let $\mathbb H$ denote the upper half plane and let $\kappa: \mathbb H \to \mathbb D$ be the Cayley transform $$ z \mapsto {z - i \over z + i}$$

Let $f_2: \mathbb H \to \mathbb H , z \mapsto \sqrt{z}$. Note that $f_2$ maps the second quadrant conformally onto $\mathbb H$.

Hence $\kappa \circ f_2$ maps $G_2$ conformally onto $\mathbb D$.

Now you use similar maps $f_1, f_3, f_4$ for the other $G_i$.