So I'm trying to find a solution of this function for $z = e^{i\theta}$ and for $x \in \Re$ and $\theta \in \Re$
$$f(z) = \frac {(e^x + 1)z + (e^x - 1)} {(e^x - 1)z + (e^x + 1)}$$
First of all, does this function has a name?
I have the feeling that the result should be on the unit circle but I'm not able to find a satisfying solution by myself.
I've found that when $z \in \Re$ then $f(z) \in \Re$ Also, $f(1) = 1$ and $f(-1) = -1$ and if $x=0$ then $f(z)=z$
I tried to replace $e^{i\theta}$ by $cos(\theta) + i*sin(\theta)$ with no success. Here's what I got :
$f(e^{i\theta}) = \frac {e^{2x} + 1 + cos(\theta)*(e^x + 1)} {(e^x - 1)^2 + cos(\theta)*(e^x - 1)} + i*\frac{2e^x*sin(\theta)} {(e^x - 1)^2 + cos(\theta)*(e^x - 1)}$
Is there a way to simplify the writing of the function when $z=e^{i\theta}$ ? Is there a way to know if the solution lie on the unit circle?
It is a fact proven in complex analysis that the Moebius transforms of the form $$S(z):=e^{i\alpha}{z-c\over 1-\bar c z},\qquad \alpha\in{\mathbb R},\quad |c|<1,$$ map the unit disc conformally onto itself, whereby the (complex) point $c$ is mapped to the origin. By continuity such an $S$ will map the unit circle $\partial D$ bijectively onto itself.
In the case at hand we have $\alpha =0$ and $c={1-e^t\over 1+e^t}=-\tanh{t\over2}$: $$f(z):={(e^t+1)z+(e^t-1)\over(e^t-1)z+(e^t+1)}={z-c\over 1-\bar c z}\ .$$ It follows that points $z:=e^{i\theta}$ are mapped to $\partial D$ by the given $f$.