Show the function $g(z)=\frac{1+z}{1-z}$ maps $\mathbb{D}$ onto $\{ z \in \mathbb{C} :\operatorname{Im}(z) >0\}$

Question

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!

Answer 1

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: $$\dfrac{iz+i}{-z+1}$$.

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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 $w$.

Answer 2

$w=\frac{1+z}{1-z}\implies z=\frac{w-1}{w+1}$

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 $\mathbb D$.