Consider the unit-disk $\mathbb{D} = \{ z : |z|\leq 1 \}$.
I need to find a Möbius Transformation $w=Tz$ that maps $\mathbb{D}$ to the upper-half plane $\mathbb{H} = \{ w : Im(w) \geq 0\}$.
I have searched and found that the linear fractional transformation $f(z) = \frac{i (1+ z)}{1-z}$ maps $\mathbb{D}$ to $\mathbb{H}$.
But I am not sure about the logic behind how one can come to this result. If any hints could be given, it would be much appreciated. Thanks.
On
You may establish the explicit map as follows. Let $w= \frac{i (1+ z)}{1-z}$. Then, $z= \frac{w-i }{w+i}$. Given the unit disk for $z$, we have $|z|=r\le 1$, or
$$\frac{w-i }{w+i} \frac{\bar w+i }{\bar w-i }=r^2$$
Rearrange
$$|w^2| -i \frac{1+r^2}{1-r^2}\bar w + i \frac{1+r^2}{1-r^2}w +1=0 $$ or,
$$| w- i \frac{1+r^2}{1-r^2}|^2 = (\frac{2r}{1-r^2})^2 $$
which represents a circle with center $ c=\frac{i (1+r^2)}{1-r^2}$ and radius $R= \frac{2r}{1-r^2}$. Thus, a circle of radius $r$ for $z$ maps to a circle in the upper plane with center $c$ along the vertical axis and radius $R$. The map covers the entire upper plane as $r$ varies from $0$ to $1$. (See the plot below.)
Geometrically $\{z\in \Bbb C | \ \ |z+1|=|z-1|\}$ is locus of all $z$ which is equidistant fron $1$ and $-1$, which is nothing but imaginary axis in complex plane.
Whereas $A=\{z\in \Bbb C | \ \ |z+1| \geq |z-1|\}$ is right half plane. $\tag{1}$
Hence the map $f:A \to \Bbb D$ defined as $$z \mapsto \frac{z-1}{z+1}$$ and from $(1)$ we can infer that $f$ is well-defined and is bijective map.
$g:A \to \Bbb H$ defined as $z\mapsto iz$ rotates the right half plane to upper half plane.
Hence the desired map $g\circ f^{-1}: \Bbb D \to \Bbb H$.