Show that all complex numbers $z \not= -1$ can be written as $z=\frac{1+it}{1-it}$ Show that for all complex number $z \not= -1$, with $|z| = 1$, can be written as $z=\frac{1+it}{1-it}$, with $t \in \mathbb{R}$.
I think this structure seems to be a stereographic projection from the unit circle in the complex plane. This observation is clear when we regard this equation : $\frac{1+it}{1-it}= \frac{1-t ^2}{1+t^2} + i \frac{2t}{1+t^2}$.
Instead of this solution, could we use the fact that $\frac{1+i \tan \alpha}{1-i \tan \alpha} = (1-\tan \alpha) + i 2 \tan \alpha$ for $\alpha \in (\frac{-\pi}{2}, \frac{\pi}{2})$?
BTW, this website has the same initial problem, but not the same final question
 A: If $z = \frac{1+it}{1-it}$ then $z -itz = 1 +it$. Simplify to obtain $$t= \frac{z-1}{i+iz}.$$ Now I suggest to show that, if $|z|=1$ then $\frac{z-1}{i+iz}$ is real. 
A: Taking a slightly different approach, since you're interested in an approach using $\tan$:
Division of complex numbers means, when regarding complex numbers as vectors, dividing lengths and subtracting angles: $\dfrac{r_0e^{\theta_0i}}{r_1e^{\theta_1i}} = \frac{r_0}{r_1}e^{(\theta_0-\theta_1)i}$.
You know that $|1+it|=|1-it|$, so $|\frac{1+it}{1-it}| = 1$ for all $t$.
You also know that any angle $\theta$ in the range $(-\frac{1}{2}\pi, \frac{1}{2}\pi)$ can be represented by $1+i\tan\theta$, with the negated angle $-\theta$ representable by $1+i\tan-\theta=1-i\tan\theta$. It should be easy to prove from this that $\frac{1+it}{1-it}$ can represent any angle in the range $(-\pi,\pi)$.
Combine these two and you have what you're after.
A: For all bijection $f$ of $\mathbb C$ you can do $$z=f(t)$$ In particular $f(t)=\frac{1+it}{1-it}$ is a bijection whose inverse is $f^{-1}(z)=\frac{-iz+i}{z+1}$ and it is not the only homography you can use to the same purpose, of course.
