Show the function $g(z)=\frac{1+z}{1-z}$ maps $\mathbb{D}$ onto $\{ z \in \mathbb{C} :\operatorname{Im}(z) >0\}$ Show the function $g(z)=\frac{1+z}{1-z}$ maps $\mathbb{D}$ onto $\{ z \in \mathbb{C} :\operatorname{Im}(z) >0\}$
First, we know that $g$ has a pole at $z=1$, which is on the boundary of the unit circle. So the circle is being mapped to a line.
Now we need to figure out which line.
Note that we have $g(1)=dne, g(-1)=0, g(i)=i, g(-i)=-i, g(0)=1$.
This is the part I'm confused about. Since we want to show that  $g$ maps $\mathbb{D}$ onto $\{ z \in \mathbb{C} :\operatorname{Im}(z) >0\}$, shouldn't $g(0)$  be somewhere in $\{ z \in \mathbb{C} :\operatorname{Im}(z) >0\}$? And shouldn't we have any $z$ with $|z|=1$ being mapped to the real-axis?
Aside: I tried to find a function that maps $\mathbb{D}$ onto $\{ z \in \mathbb{C} :\operatorname{Im}(z) >0\}$ as an exercise on my own.
I tried by using the cross-ratio method, and I got that $f(z)= \frac{(z-1)(i+1)}{(z+1)(i-1)}$ is such a function. Could someone please verify this?
Thanks!
 A: The transformation you have,  as pointed out in the comments,  is close,  in that it maps the unit disk to the right half plane.   If you multiply by $i=e^{\pi/2 i}$, it will rotate it onto the upper half plane.
So you get: $$\dfrac{iz+i}{-z+1}$$.

The inverse you're looking for is quite famous,  the "Cayley transformation":
$$\dfrac{z-i}{z+i}$$
The way to do these is actually quite simple (enough that even I can do it) :  check three points to see what happens to the boundary,  and then a test point to see which side goes to which.
So, note that $\infty\to1, 1\to -i,--1\to i$.  That already establishes that the $x$-axis goes to the circle $\lvert z\rvert =1$ (because,  as we know generalized circles go to generalized circles).
Now the test point:   $i$ will work nicely.   It's in the upper half plane.  And, $i\to0$.  That's enough to conclude,  by continuity, that the entire upper half plane must go to the interior of the disk.
To get this inverse, you can interchange $w$ and $z$ in $w=f(z)$, getting $z=f(w)$, and then solve for $w$.
A: $w=\frac{1+z}{1-z}\implies z=\frac{w-1}{w+1}$
I hope $\mathbb D$ is open unit disc i.e. $\{z\in\mathbb C: |z|<1\}$
$\implies \rvert\frac{w-1}{w+1}\lvert<1\\\implies|w-1|<|w+1|\\\implies|(u+iv)-1|<|(u+iv)+1|\\\implies \sqrt{(u-1)^2+v^2}<\sqrt{(u+1)^2+v^2}\\\implies u>0 (on\ simplification)$
which is right half of $w-$ plane.
Further, to prove it is onto, observe that for every $w$ in right half $w-$plane, you have
$u>0\\
\implies \left (\frac{w+\overline w}{2}\right )>0\\
\implies \frac{1+z}{1-z}+\frac{1+\overline z}{1-\overline z}>0\\
\implies 1-z\overline z>0\\
\implies z\overline z<1\\
\implies |z|^2<1\\
\implies |z|<1$
i.e. the pre-image always lie inside $\mathbb D$.
