Take the tour ×  Say you have the half-plane
$\{z\in\mathbb{C}:\Re(z)>0\}$. Is there a rigorous explanation
why the transformation $w=\dfrac{z-1}{z+1}$ maps the half plane
onto $|w|<1$?  share|improve this question add comment  2
Answers  You can also check it explicitly:  $$ \left|
\frac{z-1}{z+1} \right|^2 =
\frac{z-1}{z+1}\cdot\frac{\overline{z}-1}{\overline{z}+1} =
\frac{|z|^2-2 \Re(z) +1}{|z|^2+2 \Re(z)+1} < 1. $$  The last
inequality follows simply because $\Re(z) > 0$ and so the
numerator is smaller than the denominator.  The other way
around: the inverse is given by  $$ z \mapsto \frac{1+z}{1-z} $$
and we can check the real part for $|z| < 1$:  $$
\Re\left(\frac{1+z}{1-z}\right) = \frac{1}{2} \left(
\frac{1+z}{1-z} + \frac{1+\overline{z}}{1-\overline{z}}\right) =
\frac{1-|z|^2}{|1-z|^2} > 0. $$  share|improve this answer add
comment  There is, but I'm not sure what you do or don't know.
But we'll see what we can do.  So you know that it's a
fractional linear transformation. It's continuous, bijective