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.


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 $g\circ f^{-1}: \Bbb D \to \Bbb H$.

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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

$$\frac{w-i }{w+i} \frac{\bar w+i }{\bar w-i }=r^2$$


$$|w^2| -i \frac{1+r^2}{1-r^2}\bar w + i \frac{1+r^2}{1-r^2}w +1=0 $$ 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.)

enter image description here

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