# Why does $\frac{e^z-1}{e^z+1}$ map the infinite strip $\{z:|\text{Im}(z)|<\pi/2\}$ onto the open unit disk

I'm reading Conway's complex analysis book and on page 53 the author found the analytic function $f:G\to \mathbb C$, where $G=\{z:\text{Re}(z)>0\}$ and $f(G)=D=\{z:|z|<1\}$ defined by $$f(z)=\frac{z-1}{z+1}$$

Afterwards he states the following:

Could someone help me to find where are in the book the results the author is talking about?

It is not as specific as it stated. But it is on page 46, when the author is talking about the function $e^z$.
Notice that this function $e^z$ maps the infinite strip $\{z: |\Im z|<\frac{\pi }{2}\}$ onto the right half plane.
Consider $e^{x+yi}$. In the strip $|y|<\frac{\pi}{2}$. Then $e^{x+yi}=e^x \cos y+i e^x \sin y$ where the real part is $>0$, whereas the imaginary part can assume any value, hence the right half plane.
Now the map $g(z)=\frac{e^z-1}{e^z+1}$ is the composite function of $e^z$ and the $f(z)$. So the maps are also composited.
I think he is referring to point 2.16 at the bottom of page 38, starting with the remarks that $|\exp(z)|=\exp(\hbox{Re } z)$ and $\arg\exp(z)=\hbox{Im } z.$