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- #1

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.