TITLE: If $|z|=1$, $z\neq-1$, show that $z$ may be expressed in the form $ z=\frac{1+it}{1-it}$ where $t\in \mathbb{R}$.
QUESTION [6 upvotes]: If $|z|=1$, $z\neq-1$, show that $z$ may be expressed in the form
$$ z=\frac{1+it}{1-it},$$
where $t\in \mathbb{R}$.
I don't know, how to begin. I started with the given conditions. Given that $|z|=1 \text{ and } z\neq-1\implies z=e^{it}, t\in[0,2\pi]/\{\pi\}.$

REPLY [3 votes]: Taking the modulus of both sides and applying the fact that $|z|=1$ gives
$$\frac{\left|1+\mathrm i t\right|}{\left|1-\mathrm i t\right|} = 1$$
This implies that $\left|1+\mathrm i t\right|=\left|1-\mathrm i t\right|$, i.e.
$\left|(\mathrm i)(t-\mathrm i)\right| = \left|(-\mathrm i)(t+\mathrm i)\right|$, i.e.
$$\left|t- \mathrm i\right|=\left| t+ \mathrm i \right|$$
This is the perpendicular bisector of $t=\mathrm i$ and $t=-\mathrm i$, i.e. the real axis.
This tells us that the Mobius transformation $\mathbb f : \widehat{\mathbb C}_t \to \widehat{\mathbb C}_z$, given by
$$z=\mathbb f(t) = \frac{1+\mathrm it}{1-\mathrm it}$$
takes the real axis $t \in \mathbb R$ onto the circle $|z|=1$, and vice-versa.
(Non-degenerate Mobius transformations are automorphisms of the Riemann Sphere $\widehat{\mathbb C}$) 
Your question mentions $z \neq -1$. That is not a problem on the Riemann Sphere: $\mathrm f(\infty)=-1$.