Proposition

Find a Mobius transformation that maps the unit disk to $\{ x + iy \in \mathbb{C} : x + y > 0 \}$.

Solution

First, $f(z) = 1 / (z - 1)$ maps the unit circle to the line $\Re z = -1/2$ because

Thus every point $f(z)$ is equidistant from $0$ and $-1$ for every $\abs{z} = 1$. Using the argument above, $\abs{w - 0}\abs{z} = \abs{w - (-1)}$. Thus if $\abs{z} < 1$, then $w$ is closer to 0 than -1, and vice versa. Therefore, the unit disk is mapped to $\{ x + iy \in \mathbb{C} : x \leq -1/2 \}$.

Let $g(z) = z + 1/2, h(z) = -z, k(z) = e^{\pi i / 4}z = \frac{(1 + i)z}{\sqrt{2}}$.

Then $k(h(g(f(z)))) = \frac{(1 + i)z + (1 + i)}{(-2\sqrt{2})z + 2\sqrt{2}}$ maps the unit disk into $\{ x + iy \in \mathbb{C} : x + y > 0 \}$.