Find the Möbius transformation that maps the region $D_1 = \{ z \in \mathbb{C} \mid |z| < 1 \land \operatorname{Im} z > 0 \}$ to $D_2 = \{ z \in \mathbb{C} \mid \operatorname{Re} z > 0 \land \operatorname{Im} z > 0 \}$.
Attempt: To find the Möbius transformation that maps the upper half of the unit disk $ D_1 = \{ z \in \mathbb{C} \mid |z| < 1 \wedge \operatorname{Im} z > 0 \} $ to the first quadrant $ D_2 = \{ z \in \mathbb{C} \mid \operatorname{Re} z > 0 \wedge \operatorname{Im} z > 0 \} $, I use the transformation $ f(z) = i \frac{1 – z}{1 + z} $. This transformation maps the key point $ z = i $ to $ w = 1 $ and $ z = 0 $ to $ w = i $, effectively mapping $ D_1 $ to $ D_2 $. Is this conclusion ok?
Best Answer
The idea is to show that for the upper half of the unit disc as input ($D_1$) to $f(z)$ you have only values in the 1st quadrant as an output ($D_2$).
For $z$ being in the upper half of the unit disc we have
$$|z| < 1 $$
and
$$\operatorname{Im} z > 0$$
$$z = x + iy$$
Meaning we define this restrictions to the input:
$$x^2 + y^2 < 1$$
The previous inequality can be written as: $$1-x^2 - y^2 >0 $$
And we also have the restriction over $y$:
$$y > 0$$
So $f(z)$ can be expressed in terms of $x$ and $y$:
$$ f(z) = i \frac{1 - z}{1 + z} $$
$$f(z) = i \frac{(1 - x) - iy}{(1 + x) + iy}$$
$$f(z) = i \frac{(1 - x) - iy}{(1 + x) + iy}\frac{(1 + x) - iy}{(1 + x) -iy}$$
$$f(z) = i \frac{((1 - x) - iy)((1 + x) -iy)}{(1 + x)^2 + y^2}$$
$$f(z) = i \frac{(1 - x^2)-iy(1 - x)-iy(1+x)-y^2}{(1 + x)^2 + y^2}$$
$$f(z) = i \frac{1 - x^2-iy + iyx-iy -iyx-y^2}{(1 + x)^2 + y^2}$$
$$f(z) = i \frac{1 - x^2 -y^2 -i2y }{(1 + x)^2 + y^2}$$
$$f(z) = \frac{ 2y+i(1 - x^2 -y^2) }{(1 + x)^2 + y^2}$$
Arriving to the final expression:
$$f(z) = \frac{ 2y }{(1 + x)^2 + y^2}+i\frac{ 1 - x^2 -y^2 }{(1 + x)^2 + y^2}$$
From here is easy to see that given $y>0$ and $1 - x^2 -y^2>0$ we are going to have both the real and imaginary parts of $f(z)$ in the first quadrant.