# [SOLVED]example about uhp iso to unit disc

#### dwsmith

##### Well-known member
I am trying to understand this example:
Let H be the upper half plane. The map
$$f:z\mapsto\frac{z - i}{z + i}$$
is an isomorphism of H with the unit disc.

proof:
Let $w=f(z)$ and $z=x+yi$. Then
$$f(z) = \frac{x + (y-1)i}{x+(y+1)i}.$$
Since $z\in H$, $y>0$, it follows that $(y-1)^2<(y+1)^2$ whence
$$x^2+(y-1)^2=|z-i|^2<x^2+(y+1)^2=|z+i|^2$$
and therefore
$$|z-i|<|z+i|,$$
(I understand the above)
so $f$ maps the upper half plane into the unit disc (I don't understand why we can make this statement now? How does the above allow for this?). Since
$$w=\frac{z-i}{z+i},$$
we can solve for z in terms of w, because $wz+wi = z-i$, so that
$$z=-i\frac{w+1}{w-1}.$$
Write $w=u+iv$. By computing directly the real part of $(w+1)/(w-1)$, and so the imaginary part of
$$-i\frac{w+1}{w-1}$$
you will find that this imaginary part is > 0 if $|w| < 1$ (why is this?).
So I computed the imaginary part and obtained
$$-i\frac{(u+1)(u-1)+v^2}{(u-1)^2+v^2}$$
Hence the map
$$h:w\mapsto -i\frac{w+1}{w-1}$$
sends the unit disc into the upper half plane. Since by construction $f$ and $h$ are inverse to each other, it follows that they are inverse isomorphisms of the upper half plane and the disc.