2 conformal maps doing same shape changing seem not equal

Question

$f(z)=\frac{1+z}{1-z}$ would map the upper half disk to the first quadrant. While $g(z)=z^2$ would map the first quadrant to the upper half plane. On the other hand, $h(z)=-\frac{1}{2}(z+\frac{1}{z})$ would also do the same for mapping the upper half disk to the upper half plane. Yet, composition of $f$ followed by $g$ is not the same as $h$. Is it normal?

Answer 1

This is entirely expected. They differ only by an automorphism of the upper half space. Specifically, the map $$ k\colon w\mapsto\frac{w-1}{w+1} $$ satisfies $kh=gf$

How did we come up with this $k$?

Look at the boundary (or rather points near it and take limit). $gf$ has the effect mapping $0\mapsto 1$, $-1\mapsto 0$ and $1\mapsto\infty$, and $h$ has the effect mapping $0\mapsto\infty$, $-1\mapsto 1$ and $1\mapsto -1$ So you want a Mobius map sending $\infty,1,-1$ to $1,0,\infty$ and that is the map $k$ above.

Verifying $kh=gf$

Indeed, you can check $$ \begin{align*} (k\circ h)(z) &=\frac{h(z)-1}{h(z)+1}\\ &=\frac{-2zh(z)+2z}{-2zh(z)-2z}\\ &=\frac{z^2+2z+1}{z^2-2z+1}\\ &=(g\circ f)(z) \end{align*} $$