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[SOLVED] example about uhp iso to unit disc

dwsmith

Well-known member
Feb 1, 2012
1,673
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.