# Möbius Transformation from a unit disk to the upper half plane

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

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.) 